[User]
Problem B3: Locked and Moded
a. Consider two mirrors facing each other separated by a distance L (a Fabry–Pérot resonator).
   The cavity (the space between the two mirrors) is in vacuum (index of refraction n = 1, no
   dispersion), and the mirrors have high reflectivities so that the main resonance condition is that
   an integer multiple of half-wavelengths fit into the cavity. A light pulse containing multiple
   different frequencies travels between the mirrors and interferes with itself.

   i. State the condition for resonance in terms of L and wavelength λ.
   ii. Write down an expression for the resonant angular frequencies ωm , with m counting each of
       the possible resonances.
  iii. What is the angular frequency spacing ∆ω ≡ ωm+1 −ωm between adjacent angular frequencies?

b. The electric field at the antinodes of the standing waves in the resonator is a superposition of
   oscillations at the resonant angular frequencies and is given by:
                                                  +∞
                                                  X
                                 Ebefore (t) =          Ek cos k ∆ω t .
                                                 k=−∞

   Now suppose we introduce a gain medium (such as a doped crystal) into the laser cavity. The
   gain medium completely absorbs incoming light and re-emits it over a finite range of N angular
   frequencies centered around some angular frequency ω0 , which coincides with one of the ωm
   values determined above. Within this bandwidth, the gain medium amplifies and supports
   oscillations at all ωm that fall within the range. (Assume N is an odd number.)
  Assume that each of these N angular frequencies—also referred to as modes—has the same
  amplitude E0 , and that their phases are locked such that there is zero relative phase between
  them at t = 0. This condition is known as mode locking. Set their common phase so that all
  electric fields are expressed as cosine functions, consistent with the form of Ebefore (t).

   i. Write an expression for the total electric field Eafter (t) as the sum of these N equally spaced
      angular frequencies (with zero relative phase).
   ii. Show that in the limit of many angular frequencies (∆ω ≪ ω0 , N ≫ 1), the time-dependent
       electric field approximately takes the following form:

                                                                sin N (∆ω)
                                                                        2
                                                                             t

                                     Eafter (t) ≈ E0 f (ω0 , t)
                                                                      (∆ω) t
                                                                 sin 2

      and determine the function f (ω0 , t).

c. To answer the next part of the problem, assume N is odd and find the following limit when
   a = πm, where m is an integer:
                                              sin(N x)
                                         lim           .
                                         x→a sin x

d. Mode locking can dramatically increase the peak intensity I of the laser output. Use the
   expression I = γE 2 , where γ is a known constant, to answer the following questions.

   i. Determine to the total instantaneous intensity Iafter (t) of the electric field.
                   Copyright ©2025 American Association of Physics Teachers

2025 USAPhO                                   Part B                                               10

  ii. What is the maximum possible intensity of the total field, and at what time(s) is this achieved?

e. The uncertainty principle states ∆x ∆p ≥ ℏ2 . In optics, we are more commonly interested in the
   duration of the pulse rather than its spatial extent; the two are related via ∆x = c ∆t.Consider
   the problem of setting up a very short laser pulse. Use the uncertainty principle to estimate
   the required bandwidth (range of frequencies). Compare that to the relationship between pulse
   duration and gain bandwidth that we’re achieving in this problem.

                  Copyright ©2025 American Association of Physics Teachers

Solution:
Problem B3: Locked and Moded
a. Consider two mirrors facing each other separated by a distance L (a Fabry–Pérot resonator).
   The cavity (the space between the two mirrors) is in vacuum (index of refraction n = 1, no
   dispersion), and the mirrors have high reflectivities so that the main resonance condition is that
   an integer multiple of half-wavelengths fit into the cavity. A light pulse containing multiple
   different frequencies travels between the mirrors and interferes with itself.

   i. State the condition for resonance in terms of L and wavelength λ.
   ii. Write down an expression for the resonant angular frequencies ωm , with m counting each of
       the possible resonances.
  iii. What is the angular frequency spacing ∆ω ≡ ωm+1 −ωm between adjacent angular frequencies?

                                             Solution
     For a Fabry–Pérot resonator of length L in vacuum, an integer number of half-wavelengths
     must fit into the cavity. Mathematically, we express this resonance condition as:
                                                     λ
                                                 m     = L,
                                                     2
     where m is a positive integer. Rearranging gives

                                      λ   L                      2L
                                        =            =⇒     λ=
                                      2   m                      m

     Now for the resonant angular frequencies:
     The angular frequency ω of light in vacuum is related to its wavelength by
                                                       2πc
                                                 ω=        ,
                                                        λ
     where c is the speed of light in vacuum. Substituting λm into this relation gives

                                          2πc    2πc   m πc
                                   ωm =       =      =      .
                                          λm    2L/m    L

     The spacing between adjacent modes, e.g. for mode m and m + 1, is

                                                     (m + 1) πc m πc   πc
                         ∆ω = ωm+1 − ωm =                      −     =
                                                         L       L     L

b. The electric field at the antinodes of the standing waves in the resonator is a superposition of
   oscillations at the resonant angular frequencies and is given by:
                                                  +∞
                                                  X
                                 Ebefore (t) =            Ek cos k ∆ω t .
                                                 k=−∞

                   Copyright ©2025 American Association of Physics Teachers

2025 USAPhO                                        Part B                                         19

  Now suppose we introduce a gain medium (such as a doped crystal) into the laser cavity. The
  gain medium completely absorbs incoming light and re-emits it over a finite range of N angular
  frequencies centered around some angular frequency ω0 , which coincides with one of the ωm
  values determined above. Within this bandwidth, the gain medium amplifies and supports
  oscillations at all ωm that fall within the range. (Assume N is an odd number.)
  Assume that each of these N angular frequencies—also referred to as modes—has the same
  amplitude E0 , and that their phases are locked such that there is zero relative phase between
  them at t = 0. This condition is known as mode locking. Set their common phase so that all
  electric fields are expressed as cosine functions, consistent with the form of Ebefore (t).

  i. Write an expression for the total electric field Eafter (t) as the sum of these N equally spaced
     angular frequencies (with zero relative phase).

                                                   Solution
       We assume N frequencies equally spaced by ∆ω around a central frequency ω0 , all having
       amplitude E0 and zero phase difference. One convenient way to index these frequencies
       is by letting k run symmetrically about 0:
                                                   N −1     N −3        N −3 N −1
                    ωk = ω0 + k ∆ω,          k=−        , −      , ...,     ,     .
                                                     2        2           2    2
       (If N is even, a similar indexing can be used, but the main idea remains the same.)
       Because all modes are in phase (zero relative phase), a real representation can be written
       by summing cosines:

                                                        + N 2−1
                                                         X
                                     Eafter (t) = E0               cos ω0 + k ∆ω t
                                                       k=− N 2−1

  ii. Show that in the limit of many angular frequencies (∆ω ≪ ω0 , N ≫ 1), the time-dependent
      electric field approximately takes the following form:

                                                               sin N (∆ω)
                                                                      2
                                                                           t

                                    Eafter (t) ≈ E0 f (ω0 , t)
                                                                sin (∆ω)
                                                                      2
                                                                         t

     and determine the function f (ω0 , t).

                                                   Solution

                          + N 2−1                         N
                           X                           X         N −1
        Eafter (t) = E0              cos ω0 + k ∆ω t = E0   cos ω0 −      ∆ω + k ∆ω t =
                                N −1
                                                                       2
                          k=−                                        k=0
                                 2

                  Copyright ©2025 American Association of Physics Teachers

2025 USAPhO                                     Part B                                             20

                                     sin (N +1)∆ω t                        N (∆ω) t

                                1              2                      sin     2
                   E0 cos ω0 t + ∆ωt                 ≈ E0 cos ω0 t                .
                                2        sin  ∆ω t
                                                                        sin  (∆ω) t
                                                    2                               2

                                            f (ω0 , t) = cos ω0 t .

c. To answer the next part of the problem, assume N is odd and find the following limit when
   a = πm, where m is an integer:
                                              sin(N x)
                                         lim           .
                                         x→a sin x

                                               Solution
    For indicated values of a, both numerator and denominator are approaching to 0. Using
    L’Hôpital’s rule, we get

                              sin(N x)       N cos(N x)   N cos(πN m)
                        lim            = lim            =             = N.
                       x→πm     sin x   x→πm   cos x        cos(πm)

    The answer is positive, because for odd values of N , N m and m are both odd or both even.

d. Mode locking can dramatically increase the peak intensity I of the laser output. Use the
   expression I = γE 2 , where γ is a known constant, to answer the following questions.

   i. Determine to the total instantaneous intensity Iafter (t) of the electric field.

                                                Solution

                                                                 N (∆ω) t
                                                           sin     2
                                  Eafter (t) = E0 cos ω0 t               .
                                                                  (∆ω) t
                                                             sin 2

       Total instantaneous intensity Iafter (t): we’re given
                                                        h          i2
                                          Iafter (t) = γ Eafter (t) ,

        Substituting Eafter (t) gives

                                                                  N (∆ω) t

                                                            sin     2
                                                                             2
                                 Iafter (t) = γ E0 cos ω0 t
                                                              sin (∆ω)
                                                                     2
                                                                       t

  ii. What is the maximum possible intensity of the total field, and at what time(s) is this achieved?

                   Copyright ©2025 American Association of Physics Teachers

2025 USAPhO                                   Part B                                              21

                                             Solution
     All N modes add constructively when their phases align. In the ideal phase-locked case,
     there is a moment in time (and corresponding phase) at which every mode’s cosine term is
     at its maximum (i.e. cos(ωm t) = 1 for each m). At that instant, the field amplitudes sum
     linearly:
                                              XN
                                     Etotal =     E0 = N E0 .
                                               m=1
                                 2
     Since Itotal (t) ∝ Eafter (t) , the maximum intensity occurs when |Eafter | is largest, i.e.
     Eafter = N E0 . Consequently,
                                                       2
                                          Imax = γ N E0 .

     This perfect alignment occurs when

                                   ω0 t + k ∆ω t ≈ 2π × (integer)

     for each k in the range − N 2−1 . . . N 2−1 , which means (∆ω)t = 2πm for some integer m.
     You can also obtain this result by looking at the sin(N a)/ sin(a) term in the expression for
     the total field: the alignment occurs at zeros of the denominator, i.e. (∆ω)t2   = π · m, i.e.
          2πm
      t=        .
          ∆ω
     In theory, we should be worried about the value of the cosine term at that time; however,
     the constraint that ω0 as well as ω0 + k(∆ω) are all modes of the resonator ensures that
     cos(ω0 t) = 1 at those times.

e. The uncertainty principle states ∆x ∆p ≥ ℏ2 . In optics, we are more commonly interested in the
   duration of the pulse rather than its spatial extent; the two are related via ∆x = c ∆t.Consider
   the problem of setting up a very short laser pulse. Use the uncertainty principle to estimate
   the required bandwidth (range of frequencies). Compare that to the relationship between pulse
   duration and gain bandwidth that we’re achieving in this problem.

                                             Solution
     From uncertainty principle: since p = ℏω
                                            c , the uncertainty principle can be stated as
     ∆t∆ω ≥ 1, or “bandwidth ≥ 1/(pulse duration)”.
     In our set-up, the bandwidth is N ∆ω (not just ∆ω!), but we need to estimate the pulse
     duration. This is the time scale over which the intensity falls to a constant fraction of its
                                                             sin(N (∆ω)t/2)
     original value. The intensity is dominated by the term sin (∆ω)t/2 . If we Taylor-expand
                                                                 (       )
     this around t = 0, we write

                                    (N (∆ω)t/2) − 13 (N (∆ω)t/2)3
                                                                  .
                                      ((∆ω)t/2) − 31 ((∆ω)t/2)3

                  Copyright ©2025 American Association of Physics Teachers

2025 USAPhO                                   Part B                                                         22

    Cancelling out the linear term, we get

                 1                       1                 1
        N 1 − N 2 ((∆ω)t/2)2         1 + ((∆ω)t/2)2 = N 1 − (N 2 − 1)((∆ω)t/2)2
                 3                       3                 3
                                                         2
                                                  (∆ω)t                       1
   This will become small when (N 2 − 1)            2          ∼ 1, or t ∼ N (∆ω) , so the relationship is
   exactly as in the uncertainty principle.

                Copyright ©2025 American Association of Physics Teachers

USAPhO 2025 Question B3


[User]
Question 3

Gravitational lensing is a phenomenon where light from a distant source may be deflected by
the curvature of space-time caused by a massive lensing object close to or in the line of sight
between an observer and a distant object. This was first directly observed during the solar
eclipse of 1919 where the observed positions of stars behind the sun differed from their
astrometric positions following Einstein’s earlier predictions.

In the case where the observer, lensing object of mass and source are on a straight line,
light from the source is deflected by an angle (in radians) given by

where is the gravitational constant (                          ) is the speed of light
(                   ) and is the Einstein radius which is the least distance between the
lensing object and the apparent light path.

     (a)     (4 points) Draw a diagram to describe the physical layout of an ideal (observer,
             lens and point source in a straight line) lensing system. Draw the light path and
             mark the quantities and . Also mark the angular Einstein radius               (the
             angular deflection of the source image as seen from earth), and the other
             quantities that an observer on earth can measure.

     (b)     (2 points) Sketch the image of the source (such as a star), as seen by an observer
             on earth, in the case where the source, lensing object and observer are on a
             straight line.

     (c)     (3 points) Sketch the image of the source (such as a star), as seen by an observer
             on earth, in the non-ideal case where the source, lensing object and observer are
             not in a straight line. Sketch the source-lens system to explain why this is so.

                                                                                Page 1 of 3

Gravitational lensing has been proposed as a method to detect massive compact halo objects
(MACHOs) in our galaxy, which may be a candidate for dark matter. These objects are often
dark stellar remnants such as neutron stars and black holes. As stars and MACHOs orbit in
the galaxy, there is a chance that a lensing event may occur when a black hole or neutron star
passes in front of a background star.

     (d)   (3 points) The Schwarzschild radius of a black hole defines the point of no return.
           A correct expression for the Schwarzschild radius can be obtained by taking it to
           be the radius where the escape speed is equal to the speed of light. This means
           that something inside the Schwarzschild radius cannot escape the black hole.
           Using Newtonian mechanics, derive the formula for the escape speed at a
           distance away from a point object of mass . Hence, derive the Schwarzschild
           radius for a point object of mass in terms of the gravitational constant and
           the speed of light . Show your steps and reasoning clearly. (This happens to give
           the correct expression for the Schwarzschild radius that comes from general
           relativity.)

     (e)   (1 point) In the case where the source, lens and observer are in a straight line,
           given a measurement of and , how would you calculate the Schwarzschild
           radius of the lensing object?

     (f)   (2 points) Consider the case where we have a lensing object of the order of a few
           solar masses (      a few            ) in the nearby regions of the galaxy (distance
                 a few             away) and a source object somewhat further out (           a
           few       ). Which of the following apply in this case?
           Choose the following conditions that apply to the case as described in the
           question:

              x     is large and tan , sin , cos   x           is large and tan      , sin    ,
                  must be calculated exactly               cos     must be calculated exactly
              x      is small and the small angle x            is small and the small angle
                  approximations to tan , sin ,            approximations to tan , sin ,
                  cos are permissible                      cos     are permissible
              x      is irrelevant and need not be x           is irrelevant and need not be
                  calculated                               calculated

     (g)   (3 points) Using the conditions in part (f), rewrite your expression in part (e) in
           terms of measurable quantities (which are ,        and ) for a lensing object of
           the order of a few solar masses (     a few            ) and in the nearby regions
           of the galaxy (distance          a few               away) with a source object
           somewhat further out (      a few      ). Show your working.

                                                                                Page 2 of 3

(h)   (2 points) Suppose we have an event where a lensing object of
                     (3.0 solar masses),            away from earth passes in front
      of a star                 away from earth. This happens such that the ideal
      configuration occurs during the event. What is the angular Einstein radius
          (as seen from earth) during this event when the source, lens and observer
      line up?

                                                                         Page 3 of 3

Solution:
Question 3
    (a) (4 points) Draw a diagram to describe the physical layout of an ideal (observer, lens
        and point source in a straight line) lensing system. Draw the light path and mark the
        quantities a and rE . Also mark the angular Einstein radius qE (the angular deflection
        of the source image as seen from earth), and the other quantities that an observer on
        earth can measure.

             Solution:
                                    Apparent Source

                                                  ↵           ↵                   r̃E
                                                         rE
                                                                         ✓E
                           Source                     Lens          DL        Observer
                                         DS

             Other relevant quantities include the distances to the lens and source DL and DS .
             (DL and DS need not be equal.)

                 • 1 point for correct layout

                      – Correct answers should show that light is bent
                      – Apparent source is not required
                      – Accept answers that show the system as a thin lens approximation
                        (sharp deflection angles)

                 • 1 point for light direction correctly marked

                      – Arrows on the light path

                 • 1 point for qE , a and rE correctly identified

                      – 1 correct: 0.4 points
                      – 2 correct: 0.7 points
                      – 3 correct: 1.0 points

                 • 1 point for DL and DS (observables; may have different notation)

                      – 0.5 points each

             Notes:

                 • y and r̃E need not be identified, but may be useful in a later part.

          • rE should be perpendicular to the projected light path, but in our astro-
            nomical system, it makes no difference if it is perpendicular to the source-
            observer line since qE is small. Accept answers that have rE perpendicular
            to the source-observer line.

(b) (2 points) Sketch the image of the source (such as a star), as seen by an observer on
    earth, in the case where the source, lensing object and observer are on a straight line.

      Solution: The image of the source should be a symmetrical (1 point) and circular
      (1 point) ring around the lensing object.
      Notes:

          • Do not accept solutions that indicate a magnified image of the source. This
            includes answers which state that the image is a filled in circle.
          • 1 point for answers that have 2 source images symmetrically on either side
            of the lens because the system should be considered in 3 dimensions instead
            of 2.
          • Text answers (without any sketch or diagram) are not accepted. Correct
            answers must have a sketch (as specified in the question).

(c) (3 points) Sketch the image of the source (such as a star), as seen by an observer on
    earth, in the non-ideal case where the source, lensing object and observer are not in a
    straight line. Sketch the source-lens system to explain why this is so.

      Solution:
      WHAT: (1.5 points) The observer will see light from one side of the lens but not
      the other side. This means that the Einstein ring should be an arc instead of a
      complete circle. The ring may be distorted or broken depending on how much
      deviation from an ideal case. Correct answers should not be a perfect circle or
      straight line.
      Note:
          • Solutions that give 2 source images on either side of the lens (with asymme-
            try) are awarded 1 point instead of 1.5 point because the system should be
            considered in 3 dimensions instead of 2.
          • Text answers (without any sketch or diagram) are not accepted. Correct
            answers must have a sketch (as specified in the question).
      WHY: (1.5 points)
      One possible answer:

                                                rE
                                         Lens
                  Source
                                                rE                Observer
                                                        DL
                               DS

      For slight deviations from the ideal case, accept also the following diagram if rE1

                                       Page 2

      is smaller than rE2 .

                                        Lens rE1
                   Source
                                                rE2               Observer
                                                         DL
                               DS

      In general, accept answers which show that the asymmetry in the system will cause
      the observer to see something asymmetrical.
      Notes:
      The key concept in this question is asymmetry. Correct answers for either part
      must demonstrate that departures from the ideal case will result in asymmetry in
      the observed system, and that the asymmetry about the source-observer line is the
      cause of the asymmetry in the observation.

(d) (3 points) The Schwarzschild radius of a black hole defines the point of no return. A
    correct expression for the Schwarzschild radius can be obtained by taking it to be the
    radius where the escape speed is equal to the speed of light. This means that something
    inside the Schwarzschild radius cannot escape the black hole.
    Using Newtonian mechanics, derive the formula for the escape speed at a distance r
    away from a point object of mass M. Hence, derive the Schwarzschild radius for a
    point object of mass M in terms of the gravitational constant G and the speed of light
    c. Show your steps and reasoning clearly. (This happens to give the correct expression
    for the Schwarzschild radius that comes from general relativity.)

      Solution: By definition, the gravitational potential energy of a test mass m at a
      distance r from the mass is (0.5 point)
                                                 GMm
                                        f=           .
                                                  r
      To escape the gravitational potential, the total energy of the test mass needs to be
      at least 0 so it should have a kinetic energy of (0.5 point)
                                          GMm 1 2
                                     K=      = mve .
                                           r  2
      Rearranging the above, the escape speed at distance from mass r is (1 point)
                                            r
                                               2GM
                                       ve =          .
                                                 r
      Substitute ve = c and rearrange to get (1 point)
                                                 2GM
                                          rS =       .
                                                  c2
      2 points for deriving escape speed (Any reasonable and physically sound method
      based on Newtonian mechanics)
      1 point for deriving the Schwarzschild radius from the escape speed.

                                       Page 3

(e) (1 point) Using the formula for light deflection, write down an expression for the
    Schwarzschild radius of a lensing object in the case where the source, lens and ob-
    server is in a straight line.

     Solution: The Schwarzschild radius is
                                        2GM        1
                                 rS =     2
                                            so rS = arE
                                         c         2
     Notes: Full marks for correct working.

(f) (2 points) Consider the case where we have a lensing object of the order of a few
    solar masses (M ⇠ a few ⇥ 1030 kg) in the nearby regions of the galaxy (distance DL ⇠
    a few ⇥ 1018 m away) and a source object somewhat further out (DS ⇠ a few ⇥ DL ).
    What can we say about a and qE in this case? (Choose your answer on your answer
    sheet. Points will be deducted for wrong answers.)
        • a is large and tan a, sin a, cos a        • qE is large and tan qE , sin qE ,
          must be calculated exactly.                  cos qE must be calculated exactly.
       • a is small and the small angle ap-         • qE is small and the small angle
         proximations to tan a, sin a, cos a          approximations to tan qE , sin qE ,
         are permissable.                             cos qE are permissable.
       • a is irrelevant and need not be cal-       • qE is irrelevant and need not be cal-
         culated                                      culated

     Solution:

         • a is small
         • qE is small

     Notes:

         • Choices pertaining to a and qE are to be marked independently (1 point
           each).
         • The conditions are mutually exclusive so accept only one condition for each
           quantity (a, qE ). Answers that select more than one condition for a quantity
           (a, qE ) are wrong (no point to be awarded).

     Reasoning: Working out the numbers, we can find that the Schwarzschild radius
     is on the order of 104 m. Because a has a maximum of 2p (largest possible angle),
     this means the physical Einstein radius rE ⇠ 104 m is very small compared to the
     distance to the lens DL ⇠ 1020 m. The geometry of the system therefore means
     that a is actually a very small angle.

     Another approximation comes from the geometry of the system which sets bounds
     on a and qE so that (see figure in part (a))

                                         rE  2r /a   10 16
                              tan qE =      = S    ⇡
                                         DL    DL      a

                                        Page 4

      which suggests that a or qE or both should be small.
      Based on the geometry of the setup and what we have already established (a
      small), we then have the following cases:

          • qE large means that DL is small which is not the case here.
          • a small, qE small is the only valid outcome here

      The result and constraints in the question suggests that a and qE are both small
      Because qE is small, the Einstein radius rE ⇠ 104 m is very small compared to the
      distance to the lensing object DL ⇠ 1020 m or source DS . We can therefore take
      the small angle approximation where a and qE is involved.

(g) (3 points) Using the conditions in part (f), rewrite your expression in part (e) in terms
    of measurable quantities (which are qE , DS and DL ) for a lensing object of the order
    of a few solar masses (M ⇠ a few ⇥ 1030 kg) and in the nearby regions of the galaxy
    (distance DL ⇠ a few ⇥ 1018 m away) with a source object somewhat further out (DS ⇠
    a few ⇥ DL ). Show your working.

      Solution: Adding up exterior angles, we see that a = qE + y so qE = a y
      where is small (y and r̃E defined on the following diagram). Also note that rE is
      approximately perpendicular to the source-observer system because qE is small.
                             Apparent Source

                                           ↵            ↵                      r̃E
                                                   rE
                                                                  ✓E
                    Source                      Lens         DL            Observer
                                  DS

      Using the small angle approximation for a and qE , we can write
                   rE                       r̃E      rE
                       = tan qE ⇡ qE and        =           = tan y ⇡ y
                   DL                       DS    DS DL
      This gives (1 point)
                                               rE   rE
                                       a=         +
                                               DL DS DL
      So that (1 point)                           ✓                    ◆
                                   1      1                 DS
                               rS = rE a = rE2
                                   2      2            DL (DS DL )
      To write this in terms of qE , DL and DS , we use rE = qE DL to get (1 point)
                                               ✓          ◆
                                          1 2      DS DL
                                      rS = qE
                                          2       DS DL
      Notes:
          • 1 point for a
          • 1 point for rS
          • 1 point for final equation

                                         Page 5

(h) (2 points) Suppose we have an event where a lensing object of 6.0⇥1030 kg (3.0 solar
    masses), 2.6⇥1018 m away from earth passes in front of a star 9.2⇥1018 m away from
    earth. This happens such that the ideal configuration occurs during the event. What is
    the angular Einstein radius qE (as seen from earth) during this event when the source,
    lens and observer line up?

      Solution: The Schwarzschild radius of the lens is
                           2 ⇥ (6.673 ⇥ 10 11 ) ⇥ 6.0 ⇥ 1030
                    rS =                      8  2
                                                             = 8.9 ⇥ 103 m
                                     (3.0 ⇥ 10 )

      From the previous part, the angular Einstein radius is given by
                                ✓          ◆
                     2            DS DL
                   qE = 2rS ⇥
                                   DS DL
                                          ✓                               ◆
                                      3          (9.2 2.6) ⇥ 1018
                       = 2 ⇥ 8.9 ⇥ 10 ⇥
                                            (9.2 ⇥ 1018 ) ⇥ (2.6 ⇥ 1018 )
                       = 4.9 ⇥ 10 15

      Thus the angular Einstein radius is
                     p
               qE = 4.9 ⇥ 10 15 = 7.0 ⇥ 10 8 radians = 0.014 arcseconds

      (1 point for correct answer, 1 point for correct working)
      Notes:

          • Students are expected to use the formula derived in part (g) to answer this
            question.
          • For the final answer:

               – 0.5 point off for missing units. While angles are mathematically di-
                 mensionless, a good student should be cognisant of the fact that there
                 are different physical units for angular measurement, and that units for
                 angles should be specified.
               – 0.5 point off for final answers given to 1 significant figure or less.
               – 1 point off if the final number is incorrect.

                                        Page 6

APhO 2014 Theory T3


[User]
9. This entire booklet must be returned at the end of the examination.

                       Table of Constants
 Speed of light in vacuum
 Planck’s constant
                                   c
                                   h
                                       3.00 × 108 m·s−1
                                       6.63 × 10−34 J·s
                                                               Question

                                                                  1
                                                                               Marks

                                                                                13
                                                                                Score

                                   –
                                   h   h/2π
                                                            −2
 Universal constant of Gravitation G   6.67 × 10−11 N·m2 ·kg      2              12
 Magnitude of electron charge      e   1.60 × 10−19 C
 Rest mass of electron             me 9.11 × 10−31 kg             3              15
 Value of 1/4π0                       9.00 × 109 N·m2 ·C−2
 Avogadro’s number                 NA 6.022 ×1023 mol−1           4              20
 Acceleration due to gravity       g   9.81 m·s−2
 Universal Gas Constant            R   8.31 J· K−1 ·mol−1         5              20
                                   R   0.0821 l·atm·mol−1 ·K−1
 Permeability constant             µ0  4π × 10−7 H·m−1          Total            80

                          HOMI BHABHA CENTRE FOR SCIENCE EDUCATION
                                 Tata Institute of Fundamental Research
                             V. N. Purav Marg, Mankhurd, Mumbai, 400 088

INPhO 2020 Page 1          Questions & Summary Answers

Solution:
9. This entire booklet must be returned at the end of the examination.

                       Table of Constants
 Speed of light in vacuum
 Planck’s constant
                                   c
                                   h
                                       3.00 × 108 m·s−1
                                       6.63 × 10−34 J·s
                                                               Question

                                                                  1
                                                                               Marks

                                                                                13
                                                                                Score

                                   –
                                   h   h/2π
                                                            −2
 Universal constant of Gravitation G   6.67 × 10−11 N·m2 ·kg      2              12
 Magnitude of electron charge      e   1.60 × 10−19 C
 Rest mass of electron             me 9.11 × 10−31 kg             3              15
 Value of 1/4π0                       9.00 × 109 N·m2 ·C−2
 Avogadro’s number                 NA 6.022 ×1023 mol−1           4              20
 Acceleration due to gravity       g   9.81 m·s−2
 Universal Gas Constant            R   8.31 J· K−1 ·mol−1         5              20
                                   R   0.0821 l·atm·mol−1 ·K−1
 Permeability constant             µ0  4π × 10−7 H·m−1          Total            80

                          HOMI BHABHA CENTRE FOR SCIENCE EDUCATION
                                 Tata Institute of Fundamental Research
                             V. N. Purav Marg, Mankhurd, Mumbai, 400 088

INPhO 2020 Page 1          Questions & Summary Answers
Please note that alternate/equivalent methods and different way of expressing final
solutions may exist.

INPhO 2020 Problem 9


[User]
A man wishes to topple a very tall and thin obelisk, of height L. To do this, he wraps the end of a
rope of length L around the obelisk at height h, then stands on the ground and pulls the other end
as hard as he can. Assume that the rope does not slip on the obelisk, but the man can slip on the
ground, with coefficient of static friction µ.
   1. Explain why the man is unlikely to succeed if he attaches the rope at h = 0 or h = L.
   2. To topple the obelisk, the man should maximize the torque they can exert about the obelisk’s
      base without slipping. What is the optimal value of h?

Solution:
A man wishes to topple a very tall and thin obelisk, of height L. To do this, he wraps the end of a
rope of length L around the obelisk at height h, then stands on the ground and pulls the other end
as hard as he can. Assume that the rope does not slip on the obelisk, but the man can slip on the
ground, with coefficient of static friction µ.

  1. Explain why the man is unlikely to succeed if he attaches the rope at h = 0 or h = L.

  2. To topple the obelisk, the man should maximize the torque they can exert about the obelisk’s
     base without slipping. What is the optimal value of h?

Solution. This is the “obelisk razer” problem from Professor Povey’s Perplexing Problems.

  1. (5) At height h = 0, the man would just be pulling on the obelisk horizontally at its base,
     which wouldn’t work because it’s wedged into the ground. At h = L the man would be pulling
     the rope almost exactly down, which would, if anything, make the obelisk more secure.

  2. (20) Let the man pull on the rope with a force F . By balancing horizontal forces on the man,
     the maximum possible value of F before the man starts slipping satisfies

                                        F cos θ = µ(mg − F sin θ)

     where θ is the angle the rope makes with the horizontal, so
                                                        √
                                          h               L2 − h2
                                   sin θ = , cos θ =              .
                                          L                 L
     Then we have
                                                      µmg
                                           F =                   .
                                                 µ sin θ + cos θ
     The torque on the obelisk is then τ = F h cos θ = F L sin θ cos θ, so
                                               sin θ cos θ
                                        τ=                   µmgL.
                                             µ sin θ + cos θ
     Therefore, simplifying the fraction, we have

                                         τ∝                       .
                                              µ/ cos θ + 1/ sin θ

     Therefore, we want to minimize the denominator. Setting its derivative to zero gives
                                              cos θ   µ sin θ
                                                 2  =
                                              sin θ   cos2 θ

     which becomes tan θ = µ−1/3 . Solving the triangle gives
                                                   L
                                             h= p         .
                                                 1 + µ2/3

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User]
9. Experiment (15 points)

The black box contains a nonlinear element
(active resistance) and a capacitor, connected
sequentially. Find the capacitance C of the ca-
pacitor (5 pts) and the V − I characteristic of
the nonlinear element (6 pts). Note that (a) the
electrolytic capacitor accepts only one polarity
of charge (indicated by the colors of the output
wires of the black box); (b) the V − I characte-
ristic cannot be expected to be symmetric with
respect to I = 0. However, you are requested
to study the range I > 0 corresponding to the
discharge of the capacitor. Tabulate your measu-
rementdataanddrawappropriategraphs(4pts).
Experimentalequipment: batteries,wires,multi-
meter, stopwatch, graphic paper.

Solution:
9. Experiment

The idea: take readings of discharge current, as
a function of time. The surface area under the
graph is the outflown charge Q. Taking the rea-
dings of voltage U0 and U1 at the beginning and
at the end of discharge, we obtain Q/C = U0 −
U1,i.e.C = Q/(U0−U1). AsforV-Icharacteris-
tic, interrupt from time to time discharge, take
reading of discharge current I just before inter-
ruption, measure voltage U, and continue disc-
harging. Collect enoughdata to drawV-I charac-
teristic.

Experiment


[User]
2. Thermos bottle (6 points)

In order to study the thermal
properties of a thermos bottle, let us model it as two concentric
spherical vessels, with radii R1 = 7cm and R2 = 10cm. The
gapbetween the wallsofthe vesselscontains vacuum (hence, the
heat conductivity can be neglected).
i) (3,5 pt) Find the radiative heat flux (i.e. transmitted heat per
unit time) between the walls of the bottle, assuming that the
ambient temperature is T2 = 293K and the inner sphere is
filled with liquid nitrogen at the boiling temperature T1 = 77K.
The emissivities of all the surfaces are equal to that of stainless
steel: ε = 0.1. Remark: The emitted heat flux per unit area
is given Stefan-Boltzmann’s law P = εσT4
, where σ =
5.67 × 10−8
W/m2
K4
(assuming that ε is independent of the
wavelength.
ii) (2,5 pt) Estimate, how long time does it take for a full
evaporizationoftheliquidnitrogen(thevaporescapesthroughan
overpressurevalve). Fortheliquidnitrogen,densityρ = 810g/l
and latent heat forvaporizationλ = 5.580kJ/mol). NB!If you
wereunable tofindP (forquestioni), expresstheevaporizationtime
symbolically(i.e. using the symbolP).

Solution:
2. Thermos bottle

i. (3.5 pts) Remark: this problem techically rather challenging.
Therefore, reasonable estimates like P ≈ σεS1(T4
2 − T4
1 ) ≈
2.6W or P ≈ 1
2σεS1(T4
2 −T4
1 ) ≈ 1.3W will be graded by 2–2.5
pts.
The heat flux radiated from one wall is partially reflected
back by other wall, which is also partially reflected back, etc.
Besides, the flux from the outer wall can hit itself, if it misses
the inner wall. So, near the surface of the outer wall, we can
split the heat flux into inwards flux Qi and outwards flux Qo.
Then, upon designating the flux radiated by the outer wall by
Q = εσS2T4
2 , we have equalities
Qi = Q + Qo(1 − ε),
i.e. the inward flux consists of (a) inital radiation, and of (b)
the back-reflected part of the outwards flux. Similarly we have
Qo = Qiκ(1 − ε) + Qi(1 − κ) = Qi(1 − κε),
i.e. the outward flux consists of (a) the part κ of itself, which
hits the inner wall and is reflected back, and of (b) the part
1−κ of itself, which misses the inner wall hence reaches again
the outer wall as an outwards flux. Upon substituting Qo from
the second equation into the first one, we obtain
Q = Qi[1 − (1 − κε)(1 − ε)] = Qiε(1 + κ − κε),
hence Qi = Q/ε(1 + κ − ε). From that inwards flux, the part
which hits the inner wall is κ; in order to get the dissipated
part, we need further to multiply the result by ε. So, the dis-
sipated flux is
Qdi = εσS2T4
2 κ/(1 + κ − κε).
In order to obtain the flux Qdo, which is radiated from the
inner wall and is dissipated in the outer wall, we proceed in the
same way. Now, let Q = εσS1T4
1 ; then,
Qo = Q + Qi(1 − κε),
and
Qi = Qo(1 − ε),
so that Qo = Q/[ε(1 + κ − κε)] and
Qdo = εσS1T4
1 /(1 + κ − κε).
Now, let us consider (an imaginary) situation, when T1 = T2.
This is thermal equilibrium, when the heat flux Qdo given by
the inner wall to the outer one must be equal to the flux Qdi,
which is given by the outer wall to the inner one. Using our
expressions we see that κS2 = S1, i.e. κ = S1/S2. Now we can
finally write down the expression for the net flux given to the
nitrogen,
P = Qdi − Qdo =
εσ4πR2
1(T4
2 − T4
1 )
1 + (1 − ε)R2
1/R2
2
≈ 1.78W.
ii. (2.5 pts) The net heat received by the inner wall is spent
on evaporating the nitrogen, i.e. τP = λm, where m = 4
3πρR3
.
So,
τ =
4
3
πρR3
λμ/P ≈ 36h.
Problem

Thermos bottle


[User]
\begin{wrapfigure}{r}{0.2\textwidth}%
\vspace{-15pt}
\includegraphics[width=\linewidth]{2015-lahg-07-veejoadJoon}%
\end{wrapfigure}
Small holes at numerous different heights have been made into the wall of a vertical cylindrical vessel and water is flowing out of them. Additional water is slowly poured into the vessel so that the water level inside the vessel is constantly at a height $H$. Find what region of room can the exiting water reach, in other words express the equation for the envelope of the water spurts in $xy$-coordinates. Assume that different water spurts do not affect each other.

Solution:
Vaatleme veejuga, mis väljub anumast asukohast $y=h$. Selle kohal oleva veesamba kõrgus on $h$ ning energia jäävusest saame avaldada väljuva joa algkiiruse $v=\sqrt{2gh}$. Vabalt langev veejuga on parabool, kusjuures ühe osakese ajalist liikumist saame kirjeldada võrranditega $x=vt$ ja $y=h+gt^2/2$. Pärast esimesest võrrandist aja avaldamist saame teisest parabooli võrrandi
\[
y = h + \frac{gx^2}{2v^2} = h + \frac{x^2}{4h}.
\]
Mõtiskleme nüüd, kuidas määrata kindlaks ruumipiirkonda, millesse saavad langevad veejoad jõuda. Võiksime valida ruumipunkti $(x,y)$ ning küsida, milliselt algkõrguselt $h$ alustades on võimalik sinna jõuda. Sobiva algkõrguse $h$ leidmiseks tuleks lahendada ruutvõrrand $h^2 - yh + x^2/4 = 0$. Kui selle võrrandi diskriminant on positiivne, leidub $h$ jaoks kaks erinevat lahendit ning punkti $(x,y)$ on võimalik jõuda kahelt algkõrguselt. Kui diskriminant on negatiivne, puuduvad reaalarvulised lahendid ning punkti $(x,y)$ pole võimalik jõuda. Neid kahte juhtu eraldab piirjoon, kus diskriminant on null ehk $y^2 - x^2=0$, millest saame avaldada piirjoone ehk veejugade mähispinna võrrandi $y=x$.

Veejoad

\begin{wrapfigure}{r}{0.2\textwidth}%
\vspace{-15pt}
\includegraphics[width=\linewidth]{2015-lahg-07-veejoadJoon}%
\end{wrapfigure}
Vertikaalse silindrilise anuma seina sisse on paljudele erinevatele kõrgustele tehtud pisikesed augud, millest voolab välja vett. Anumasse valatakse aeglaselt vett juurde nii, et veetase anumas püsib muutumatuna kõrgusel $H$. Leidke, millisesse ruumipiirkonda saab anumast väljuv vesi jõuda ehk avaldage veejugade mähispinna võrrand $xy$-teljestikus. Eeldage, et erinevad veejoad üksteist ei mõjuta.

Selleks, et määrata, mis ruumipiirkonda vesi jõuab, võib vaadelda punkti koordinaatidega $(x, y)$ ning üritada määrata, missuguselt algkõrguselt peaks veejuga alguse saama, et sellesse punkti jõuda. Kui lahendit ei ole, antud ruumipunkti vesi ei jõua.


[User]
IdPhO 2020, problem 2. A nice problem on anisotropic friction.

    4    Ropes, Wires, and Chains
        Example 6: MPPP 78

        A uniform flexible rope passes over two small frictionless pulleys mounted at the same height.

        The length of rope between the pulleys is ℓ, and its sag is h. In equilibrium, what is the
        length s of the rope segments that hang down on either side?

        Solution
        The problem can be attacked by differential equations, but there is an elegant solution using
        only algebra. We let our unknowns be s, the tension T1 = (T1,x , T1,y ) in the rope at the
        pulley, and the tension T2 at the lowest point.

  Considering the entire sagging portion as the system, vertical force balance gives

                                   2T1,y = λℓg,    T1,y = λℓg/2.

  Now consider half of the sagging portion as the system. Horizontal force balance gives

                                             T2 = T1,x .

  Finally, consider one of the hanging portions as the system. Then

                                             T1 = λgs.

  We hence have three equations, but four unknowns.

  For the final equation, we need to consider how the tension changes throughout the rope.
  This would usually be done by a differential equation, but there is a clever approach using
  conservation of energy. Suppose we cut the rope somewhere, pull out a segment dx, and
  reattach the two ends. This requires work T dx, where T is the magnitude of the local
  tension. Now suppose we cut the rope somewhere else, separate the ends by dx, and paste
  our segment inside. This requires work −T ′ dx. After this process, the rope is exactly in the
  same state it was before, so the total work done must be zero.

  This would seem to prove that T = T ′ , which is clearly wrong. The extra contribution is that
  if the two locations have a difference in height ∆y, then it takes work λg(∆y) dx to move the
  segment from the first to the second. So in equilibrium, for any two points of the rope,

                                           ∆T = λg ∆y.

  Therefore, we have
                                          T1 − T2 = λgh.
  Now we’re ready to solve. We have

                                       T12 − T22 = (λℓg/2)2

  from our first three equations, and dividing by this new relation gives

                                                           ℓ2
                                         T1 + T2 = λg         .
                                                           4h
  This allows us to solve for T1 , which gives

                                             T1  h  ℓ2
                                        s=      = +    .
                                             λg  2 8h
  This is a useful result in real engineering projects: it means that the tension in a cable can
  be estimated by seeing how much it sags.

  Example 7: Kalda 27/IPhO 1971

  A wedge with mass M and acute angles α1 and α2 lies on a horizontal surface. A string has
  been drawn across a pulley situated at the top of the wedge, and its ends are tied to blocks
  with masses m1 and m2 .

  There is no friction anywhere. What is the acceleration of the wedge?

  Solution
  This is a classic example of a problem best solved with the Lagrangian-like techniques of M4.
  By working in generalized coordinates, we won’t have to solve any systems of equations.

  Let s be the distance the rope moves through the pulley, so that both blocks have speed ṡ in
  the noninertial frame of the wedge. The “generalized force” is
                                     dV
                             Feff = −   = (m1 sin α1 − m2 sin α2 )g.
                                     ds
  Now, the kinetic energy in the lab frame will be of the form

                                         K = Meff ṡ2

  which means that, by the Euler–Lagrange equations,
                                                    Feff
                                             s̈ =        .
                                                    Meff
  Our task is now to calculate Meff . Since the center of mass of the system can’t move
  horizontally, the wedge has speed
                                      m1 cos α1 + m2 cos α2
                                 vw =                       ṡ.
                                         M + m1 + m2
  Now, it’s a bit annoying to directly compute the kinetic energy K in the lab frame, but it’s
  easy to compute the kinetic energy in the frame of the wedge: it’s simply (m1 + m2 )ṡ2 /2.
  But the two are also related simply,
                                1                2    1
                           K + (M + m1 + m2 )vw     = (m1 + m2 )ṡ2 .
                                2                     2
  Using this to solve for K, we conclude
                                               (m1 cos α1 + m2 cos α2 )2
                           Meff = m1 + m2 −                              .
                                                    M + m1 + m2
  Finally, the desired acceleration is
                                         m1 cos α1 + m2 cos α2
                                  aw =                         s̈.
                                            M + m1 + m2
  When you plug in the result for s̈, you’ll get a very complicated final answer, but the
  Lagrangian derivation shows where all the pieces come from.

[3] Problem 22 (Kalda). A rope of mass per unit length ρ and length L is thrown over a pulley so
    that the length of one hanging end is ℓ. The rope and pulley have enough friction so that they do
    not slip against each other.

    The pulley is a hoop of mass m and radius R attached to a horizontal axle by light spokes. Find
    the force on the axle immediately after the motion begins.

[3] Problem 23 (French 5.10). Two equal masses are connected as shown with two identical massless
    springs of spring constant k.

    Considering only motion
                     √      in the
                               √ vertical direction, show that the ratio of the frequencies of the two
    normal modes is ( 5 + 1)/( 5 − 1).

[3] Problem 24 (Kalda). A massless rod of length ℓ is attached to the ceiling by a hinge which allows
    the rod to rotate in a vertical plane.

    The rod is initially vertical and the hinge is spun with a fixed angular velocity ω.

     (a) Before starting, explain why this problem has to use rods, and not just strings.

     (b) If a mass m if attached to the bottom of the rod, find the maximum ω for which the configu-
         ration is stable.

      (c) [A] Now suppose another mass m and rod of length ℓ is attached to the first mass by an
           identical hinge that turns in the same direction, as shown above. Find the maximum ω for
          which the configuration is stable. (Hint: the configuration is unstable if any infinitesimal
           change in the angles of the rods can lower the energy.)

[4] Problem 25 (PPP 106). A long, heavy flexible rope with mass ρ per unit length is stretched by a
    constant force F . A sudden movement causes a circular loop to form at one end of the rope.

    The center of the loop moves with speed c as shown.

     (a) Calculate the speed c, assuming gravity is negligible.

     (b) Find the energy E carried by a loop rotating with angular frequency ω.

     (c) Show that the momentum p carried by the loop obeys E = pc. This is true for waves in
         general, as we’ll see in W1.

     (d) Find the angular momentum carried by the loop.

    5 [A] Advanced Mathematical Techniques
    The following problems were cut from earlier problem sets because they required more advanced
    math; however, they illustrate some very neat and important ideas.

Solution:
See the official solutions here.

   4    Ropes, Wires, and Chains
       Example 6: MPPP 78

       A uniform flexible rope passes over two small frictionless pulleys mounted at the same height.

       The length of rope between the pulleys is ℓ, and its sag is h. In equilibrium, what is the
       length s of the rope segments that hang down on either side?

       Solution
       The problem can be attacked by differential equations, but there is an elegant solution using
       only algebra. We let our unknowns be s, the tension T1 = (T1,x , T1,y ) in the rope at the
       pulley, and the tension T2 at the lowest point.

       Considering the entire sagging portion as the system, vertical force balance gives

                                            2T1,y = λℓg,    T1,y = λℓg/2.

       Now consider half of the sagging portion as the system. Horizontal force balance gives

                                                     T2 = T1,x .

       Finally, consider one of the hanging portions as the system. Then

                                                      T1 = λgs.

       We hence have three equations, but four unknowns.

       For the final equation, we need to consider how the tension changes throughout the rope.
       This would usually be done by a differential equation, but there is a clever approach using

  conservation of energy. Suppose we cut the rope somewhere, pull out a segment dx, and
  reattach the two ends. This requires work T dx, where T is the magnitude of the local
  tension. Now suppose we cut the rope somewhere else, separate the ends by dx, and paste
  our segment inside. This requires work −T ′ dx. After this process, the rope is exactly in the
  same state it was before, so the total work done must be zero.

  This would seem to prove that T = T ′ , which is clearly wrong. The extra contribution is that
  if the two locations have a difference in height ∆y, then it takes work λg(∆y) dx to move the
  segment from the first to the second. So in equilibrium, for any two points of the rope,

                                           ∆T = λg ∆y.

  Therefore, we have
                                          T1 − T2 = λgh.
  Now we’re ready to solve. We have

                                       T12 − T22 = (λℓg/2)2

  from our first three equations, and dividing by this new relation gives

                                                        ℓ2
                                         T1 + T2 = λg      .
                                                        4h
  This allows us to solve for T1 , which gives

                                             T1  h  ℓ2
                                        s=      = +    .
                                             λg  2 8h
  This is a useful result in real engineering projects: it means that the tension in a cable can
  be estimated by seeing how much it sags.

  Example 7: Kalda 27/IPhO 1971

  A wedge with mass M and acute angles α1 and α2 lies on a horizontal surface. A string has
  been drawn across a pulley situated at the top of the wedge, and its ends are tied to blocks
  with masses m1 and m2 .

  There is no friction anywhere. What is the acceleration of the wedge?

  Solution
  This is a classic example of a problem best solved with the Lagrangian-like techniques of M4.
  By working in generalized coordinates, we won’t have to solve any systems of equations.

       Let s be the distance the rope moves through the pulley, so that both blocks have speed ṡ in
       the noninertial frame of the wedge. The “generalized force” is
                                              dV
                                  Feff = −       = (m1 sin α1 − m2 sin α2 )g.
                                              ds
       Now, the kinetic energy in the lab frame will be of the form

                                                  K = Meff ṡ2

       which means that, by the Euler–Lagrange equations,
                                                           Feff
                                                    s̈ =        .
                                                           Meff
       Our task is now to calculate Meff . Since the center of mass of the system can’t move
       horizontally, the wedge has speed
                                                m1 cos α1 + m2 cos α2
                                       vw =                           ṡ.
                                                   M + m1 + m2
       Now, it’s a bit annoying to directly compute the kinetic energy K in the lab frame, but it’s
       easy to compute the kinetic energy in the frame of the wedge: it’s simply (m1 + m2 )ṡ2 /2.
       But the two are also related simply,
                                   1               2  1
                                K + (M + m1 + m2 )vw = (m1 + m2 )ṡ2 .
                                   2                  2
       Using this to solve for K, we conclude

                                                      (m1 cos α1 + m2 cos α2 )2
                                Meff = m1 + m2 −                                .
                                                           M + m1 + m2
       Finally, the desired acceleration is
                                                m1 cos α1 + m2 cos α2
                                       aw =                           s̈.
                                                   M + m1 + m2
       When you plug in the result for s̈, you’ll get a very complicated final answer, but the
       Lagrangian derivation shows where all the pieces come from.

[3] Problem 22 (Kalda). A rope of mass per unit length ρ and length L is thrown over a pulley so
    that the length of one hanging end is ℓ. The rope and pulley have enough friction so that they do
    not slip against each other.

    The pulley is a hoop of mass m and radius R attached to a horizontal axle by light spokes. Find
    the force on the axle immediately after the motion begins.

    Solution. Let the distance the rope moves along the pulley be represented by the coordinate q. The
    kinetic energy of the rope is 12 ρLq̇ 2 since every section of the rope moves with velocity q̇. Without
    slipping, the kinetic energy of the pulley is 12 mq̇ 2 . Another consequence of no slipping is that energy
    is conserved, so dK/dt = −dU/dt.
        When the rope moves along a small distance of dq, the change in potential energy can be
    calculated by considering a segment dq moving from one end to another, having a difference in
    vertical height of L − πR − 2ℓ. Thus dU = −ρgdq(L − πR − 2ℓ), so

                    dK                                                    ρ(L − πR − 2ℓ)
                        = (m + ρL)q̇ q̈ = ρg q̇(L − πR − 2ℓ),      q̈ = g                .
                     dt                                                       m + ρL
                                                       P
    The vertical normal force can be determined by        mi (ay )i of the rope. The acceleration of the
    parts moving up and down will cancel out, so the acceleration of the center of mass can be found
    by considering the ”extra” segment of length L − πR − 2ℓ. The net vertical force on the system is
    then ρ(L − πR − 2ℓ)q̈, so the force on the axle N satisfies (m + ρL)g − N = ρ(L − πR − 2ℓ)q̈.

                                             (m + ρL)2 − ρ2 (L − πR − 2ℓ)2
                                    Ny = g                                 .
                                                       m + ρL
    Since rope is transferred to the right, there is also a horizontal component of force on the axle.
                                               P
    When the rope moves along a distance dq,      mi dxi is essentially ρdq(2R) since it can be seen as a
    segment of length dq moving to the other side. Thus Fx = ρq̈(2R), and

                                                      ρ(L − πR − 2ℓ)
                                          Nx = 2ρRg                  .
                                                          m + ρL
    The answers are ugly, but they come from combining a few ingredients in a straightforward way.

[3] Problem 23 (French 5.10). Two equal masses are connected as shown with two identical massless
    springs of spring constant k.

    Considering only motion
                     √      in the
                               √ vertical direction, show that the ratio of the frequencies of the two
    normal modes is ( 5 + 1)/( 5 − 1).

    Solution. Let y1 denote the displacement of the upper mass and y2 for the lower mass. The
    equations of motion are

                       mÿ1 = −ky1 − k(y1 − y2 ) = −2ky1 + ky2 ,      mÿ2 = −ky2 + ky1

    For normal modes, the particles will oscillate at the p same frequency. Guessing a form y1 =
    Aei(ωt+ϕ1 ) = Ãeiωt and y2 = B̃eiωt and defining α = ω/ k/m, the equations are

                             −ω 2 Ã = −2ω02 Ã + ω02 B̃,    −ω 2 B̃ = −ω02 B̃ + ω02 Ã.

    Dividing these equations gives
                                            Ã      1     1 − α2
                                               =        =        .
                                            B̃   2 − α2      1
    Solving for α, we find                                   √
                                        4      2         3± 5   2
                                   α − 3α + 1 = 0, α =

    Then the ratio of the two normal mode frequencies is
                                          s          √   √
                                     α1     1+5+2 5        5+1
                                        =            √ =√
                                     α2     1+5−2 5        5−1

    as desired.

[3] Problem 24 (Kalda). A massless rod of length ℓ is attached to the ceiling by a hinge which allows
    the rod to rotate in a vertical plane.

    The rod is initially vertical and the hinge is spun with a fixed angular velocity ω.

     (a) Before starting, explain why this problem has to use rods, and not just strings.

     (b) If a mass m if attached to the bottom of the rod, find the maximum ω for which the configu-
         ration is stable.

      (c) [A] Now suppose another mass m and rod of length ℓ is attached to the first mass by an
           identical hinge that turns in the same direction, as shown above. Find the maximum ω for
          which the configuration is stable. (Hint: the configuration is unstable if any infinitesimal
           change in the angles of the rods can lower the energy.)

    Solution. (a) If you take an ideal string and twist one end of it, then nothing will happen to an
         object hanging from the other side. By contrast, if you did the same thing with a rod, then
         the object would start rotating. The difference is that, as we mentioned in M2, a rod can
         support a bending moment. That is, it can transmit torques through itself. The torque you
         exert at the end results in a torque on the object.

     (b) First, let’s imagine what this motion looks like in the lab frame. If we define θ as the direction
         of the rod to the vertical axis, the spinning of the hinge fixes dθ/dt = ω. So if the mass’s
         height is fixed, this forces the mass to spin in a circle. More generally, it exerts a confusing
         time-dependent force on the mass.

     We don’t want to deal with that, so we work in the frame rotating with the hinge. In
     this frame, the rod doesn’t rotate about the vertical axis; it just changes its tilt θ to the
     vertical,
       Rr      in the plane of the page. The effective potential from the centrifugal force mω 2 r is
     − 0 mω r′ dr′ = − 12 mω 2 r2 , where r is the distance from the vertical axis through the hinge.

     In this setup, r = ℓ sin θ and the potential energy from gravity is mgℓ(1 − cos θ). For small
     angles, the potential energy is

                                          1 2        1            1
                     U ≈ mgℓ 1 − 1 − θ             − mω 2 ℓ2 θ2 = θ2 (mgℓ − mω 2 ℓ2 )
                                          2          2            2
     The system is stable when U ′′ (θ) > 0, so the maximum value of ω for stability is
                                                     p
                                              ωmax = g/ℓ.

 (c) Let the angles between the vertical and the upper, lower rods be θ1 , θ2 ≪ 1 respectively. The
     gravitational potential energy of the lower mass is mgℓ(1 − cos θ1 ) + mgℓ(1 − cos θ2 ), and the
     Taylor expansion gives U2 = 12 mgℓ(θ12 + θ22 ). Using the rotating reference frame again, the
     potential energy from the centrifugal force is 12 mω 2 r2 , where r = ℓ(sin θ1 + sin θ2 ) ≈ ℓ(θ1 + θ2 )
     since the hinges go in the same direction. The total potential of the system (same potential
     for the first mass) is then
                                              1      1             1
                     U (θ1 , θ2 ) = mgℓθ12 + mgℓθ22 − mω 2 ℓ2 θ12 − mω 2 ℓ2 (θ1 + θ2 )2
                                         2          2             2
                                      2     2   2 2   1 2        2 2   2
                                  = mℓ (ω0 − ω )θ1 + (ω0 − ω )θ2 − ω θ1 θ2

     where ω02 = g/ℓ. To be stable, we want the potential energy to be at a local minimum near
     that point. We could test this by considering a general infinitesimal change in the angles.
     However, the fastest way is to use the second derivative test for two-variable functions f (x, y).
     Let’s consider a critical point of such a function, where ∂f /∂x = ∂f /∂y = 0. For f to be a
     minimum along the x and y directions, we clearly need to have
                                            ∂2f            ∂2f
                                                > 0,            > 0.
                                            ∂x2            ∂y 2
     In this problem, these are both true if ω < ω0 .
     However, we also have to worry whether it’s possible to decrease the potential energy by
     traveling in some other direction. It turns out we are guaranteed to have a true minimum if
                                                     2 2
                                         ∂2f ∂2f       ∂ f
                                            2   2
                                                  >           .
                                         ∂x ∂y        ∂x∂y
     In this problem, that condition is

                                            2(ω02 − ω 2 )2 − ω 4 > 0.

     This quantity is positive for ω = 0, and first hits zero when
                                                          √
                                            ω 2 = ω02 (2 − 2).

     We therefore conclude that
                                                               √
                                                     r
                                                         g
                                            ωmax =         (2 − 2).
                                                         l

[4] Problem 25 (PPP 106). A long, heavy flexible rope with mass ρ per unit length is stretched by a
    constant force F . A sudden movement causes a circular loop to form at one end of the rope.

    The center of the loop moves with speed c as shown.

     (a) Calculate the speed c, assuming gravity is negligible.

     (b) Find the energy E carried by a loop rotating with angular frequency ω.

     (c) Show that the momentum p carried by the loop obeys E = pc. This is true for waves in
         general, as we’ll see in W1.

     (d) Find the angular momentum carried by the loop.

    Solution.    (a) By balancing forces on a small piece of the rope,

                                                ρ(c2 /R)(Rdθ) = F dθ
                                        p
         which gives F = ρc2 , so c =    F/ρ.

     (b) The mass of the loop is m = 2πRρ. Splitting the energy into center of mass energy and
         rotational energy, we have
                                  1     1                         2πF c
                               E = mc2 + (mR2 )ω 2 = mc2 = 2πRF =       ,
                                  2     2                           ω
         since c = ωR.

     (c) Since the loop as a whole moves with speed c and has mass m, we have p = mc. Since
         E = mc2 , we have E = pc as desired.

     (d) The orbital angular momentum is

                                                              mc2
                                                 Lo = mcR =
                                                               ω
         while the spin angular momentum is

                                                                  mc2
                                           Ls = Iω = mR2 ω =          .
                                                                   ω
         Thus the total angular momentum is 2mc2 /ω.

    5 [A] Advanced Mathematical Techniques
    The following problems were cut from earlier problem sets because they required more advanced
    math; however, they illustrate some very neat and important ideas.


[User]
4. STRETCHıNG GLOVEs (8 points)

— Eero
Uustalu.
Tools At least three pairs of large color‐
less medical semi‐transparent latex rubber
gloves; a roll of transparent and strong office
tape; a pair of sharp scissors; at least four
sheets of A4 or larger size graphing paper;
three rulers; a flexible measuring tape with a
length of at least one meter; extra‐fine point
universal surface marker. Rubber gloves can
be cut as needed into pieces. The pieces
of gloves can be fixed to your working table
either directly using the tape and/or with the
help of a ruler (to achieve a firmer fixing).
Latex is a highly stretchable elastic ma‐
terial for which it can be assumed that its
volume remains constant during stretching,
up to the breaking point.
For each of the tasks, sketch your experi‐
mentalsetupandexplainthestepsyoumade
toobtainthebestpossibleprecision,andtab‐
ulate the directly measured data.
i) (1 point)Determine the maximal strain εm
ofthelatexfilmband(i.e. thestrainbywhich
the band breaks). Strain is defined as the rel‐
ative change in length, ε = (l − l0)/l0, where
l and l0 are the stretched and unstretched
lengths of the band.
ii) (7 points) Determine and plot the stress‐
strain relationship for latex bands. Stress
is defined as the tension force per cross‐
sectional area. Express the stress σ in relat‐
ive units, normalised to the maximal stress
at the breaking point.

Solution:
4. STRETCHıNG GLOVEs

— Solution
by Eero Uustalu, grading schemes by ....
i) (1 point)
Cut a few strips of the material with the
same constant width and mark them with
two perpendicular lines close to the ends of
the strip while leaving ample space for affix‐
ing or binding. Then roll one strip length‐
wise into a cylindrical shape for stretch‐
ing. Measure the initial length between
the marked lines l0. With a straight meas‐
uring tape affixed to the surface of the
table, stretch one of the strips to the break‐
ing point while determining the maximum
length between the lines before the strip
breaks lmax. Repeat this at least three times
to see if the data is reproducible. Calculate
the result εmax =(lmax −l0)/(l0).
In our measurement we got εmax =5.5.
Grading:
• Selection of the actual latex piece for the
sample. Strips have to be of high quality
and not include the finger end. 0.1pts
• Selecting the strip length: 0.2pts
– at least 8cm: 0.2/0.2pts
– at least 4cm: 0.1/0.2pts
– shorter, not specified 0/0.2pts
– Remark. If a series of different lengths
of strips is used, then the medium value
should be considered. Additionally, if
the graph for finding εmax is used by
including the point (0,0), then 0.1 pts
should be subtracted.
• Method of measurement, of which: 0.4pts
– one end affixed with tape or held using a
ruler 0.1pts.
– lines marked on the strip, or the strip is
affixed over the edge of the ruler 0.1pts
(if both ends are affixed, 0.2 pts, but
can’t get more than 0.4 pts for this sec‐
tion).
– using narrow strips 0.1pts.
– measuringonlythepartofthestripthat’s
being stretched (between marks) 0.1pts.
– Measuring the full length of the stip
0.1pts.
• Repetition of measurements: 0.2pts
– measurement repeated at least 5 times
0.2/0.2pts.
– measurement repeated at least 3 times
0.1/0.2pts.
– if distortion of measured εmax values is
low +0.1pts.
• Final numerical value. If the value lies in
range 4‐7 (depends on the material, even 3‐
11 may be ok) 0.1pts.
ii) (7 points)
Theoretical considerations.
We can assume the material has uniform
thickness. Knowing that σ = F
A , V = const,
and that any force applied to the strip affects
all directions perpendicular to the applied
force equally, then at any time for the same
strip
V = a·a·k·l = a0·a0·k·l0 = amax·amax·k·lmax,
where a0 is the initial thickness of material
with no force applied and l0 is the initial
length of the strip with no force applied, a
is the thickness when force F is applied and
l is the length of the strip when force F is ap‐
plied, amax and lmax are the respective values
at the breaking force Fmax.
k is the ratio between the width d and
the thickness a of the strip such that d = k·a
where k remains constant for the strip while
stretching. Therefore
σ=
F
a·a·k
and σmax =
Fmax
amax·amax·k
.
From V = const we can deduce that a·a =
a0·a0·l0
l and amax·amax =
a0·a0·l0
lmax
.
If we could apply the same force F on
strips of different initial widths d = k·a, we
could vary the tension for each strip. There‐
fore we would be using different strips with
different k values for each measurement
σ= F
k · l
l0
· 1
a0·a0
and σmax = F
kmax
·lmax
l0
· 1
a0·a0
,
where k = d/a and kmax = dmax/amax are
the width to thickness ratios of the the re‐
spective strips used for F and Fmax. By com‐
bining the above, we get
σ
σmax
=
kmax
k
·
l
lmax
,
but since d0 = k·a0 is the initial un‐
stressed width of one strip and dmax0 =
kmax·a0 is the initial unstressed width of an‐
other strip stretched to the breaking point,
for building the graph it transforms to:
σ
σmax
=
dmax0
d0
· l
lmax
for the Y axis and
ε=
l−l0
l0
for the X axis.
Measurements.
We can assume the material has uniform
thickness. We want to apply the same force
on the strips with different initial widths.
Approach 1
We make many strips with different
widths, mark them with the same initial
length l0 usingperpendicularlines, measure
and record the initial widths d0 of each seg‐
ment, then roll them lengthwise into a cyl‐
indrical shape and bind them one after an‐
other. The force will be the same for all the
segments, but the value of σ for each seg‐
mentwill dependonthe initial width ofeach
individual strip.
A range where dmin and dmax differ at
least 8 times is recommended.
Rolling the strip lengthwise into a cyl‐
indrical shape before binding gives the pos‐
sibility to distribute the force evenly on all
the strips regardless of the width of the strip!
Our solution was to make two identical
stripswiththesmallestpossiblewidthandat‐
tach them to one another at one end to form
a Y shape. The previously determined εmax
was used to estimate the possible maximum
stretching almost to the breaking point of
the narrowest strip. In case of a breakage
event the identical spare strip at the other
end of the Y shape could be used for meas‐
urement. After affixing the multi‐width rope
of combined strips at its maximum stretch
position, the values of l were recorded for
each individual section. Afterwards the rope
was stretched till breaking to find the actual
breaking length lmax of the narrowest strip.
The 7mm wide initially 80mm long test
strip broke at length 522mm. Therefore the
measurement of the rope of strips was made
using force F stretching the second 7mm
wide 80mm test strip to 519mm .
But the 7mm stripe broke when it was
stretched to 599mm .
So the dmax0 =7mm and lmax =599mm .
mm mm mm
(l−l0)/lo l0 l d0 σ/σmax
0.638 80 131 71 0.022
1.606 80 208,5 37 0.066
1.888 80 231 30 0.090
3.025 80 322 21,5 0.175
3.338 80 347 18 0.226
4.225 80 418 14 0.349
4.925 80 474 11 0.505
5.488 80 519 7 0.866
And the graph, σ
σmax
for the Y axis and ε
for the X axis.
0,000 1,000 2,000 3,000 4,000 5,000 6,000
0,000
0,100
0,200
0,300
0,400
0,500
0,600
0,700
0,800
0,900
1,000
Approach 2
We make two identical elongated tri‐
angles from the longest piece of material
available. We mark the triangles with evenly
spaced perpendicular lines. (The lines are
perpendicular to the central symmetry line
of the elongated triangle.)
We record the initial length l0 and the av‐
erage width d0 of each segment as defined
by the perpendicular lines. When stretched,
allsegmentswillhavethesameforceapplied
but each segment will have its own σ. We
have to affix the base of the prolonged tri‐
angle to the ruler with tape (it has to be a
good strong joint) and we affix the ruler to
the table. A long strip of graph paper was
taped to the table to have the rubber triangle
stretched over it.
The triangle was stretched until the nar‐
row end segment was near breaking point,
and the values of l were recorded for each
segment. (Most easily done by using a
markeron the graphpaper while holding the
narrow end of the triangle at a fixed position
with the other hand). One of the triangles
was actually stretched to the braking point,
the other triangle was stretched to the break‐
ing point only after the measurement was
made.
The narrow end segment of the first tri‐
angle of length 10mm and of average width
6mm broke at length 73.1mm. Therefore
the measurement of the second similar tri‐
angle tested was made using force F stretch‐
ing the narrow end segment of the triangle
to 72.6mm. The actual breaking of the nar‐
row segment of the second triangle occurred
at length 73.7mm .
So the dmax0 =6mm and lmax =73.7mm .
mm mm mm
σ/σmax l0 l d0 (l−l0)/lo
0.985 6 10 72.6 6.26
0.605 8 10 59.5 4.95
0.444 10 10 54.5 4.45
0.333 12.5 10 51 4.1
0.266 15 10 49 3.9
0.221 17.5 10 47.5 3.75
0.183 20 10 45 3.5
0.148 22.5 10 41 3.1
0.129 24 10 38 2.8
0.104 27 10 34.5 2.45
0.074 32.5 10 29.5 1.95
0.061 36 10 27 1.7
0.051 39 10 24.5 1.45
0.043 41.5 10 22 1.2
0.037 46.5 10 21 1.1
0.030 50 10 18.5 0.85
0.022 56 10 15 0.5
And the graph, σ
σmax
for the Y axis and ε
for the X axis.
0 1 2 3 4 5 6 7
0,000
0,100
0,200
0,300
0,400
0,500
0,600
0,700
0,800
0,900
1,000
Grading:
GOOD practice:
• using the longest possible strips
• marking the initial length on the strip with
a marker or
• measuring initial length to one straight af‐
fixedsurfacetoanotherparallelaffixedsur‐
face
• rollingthemarker‐markedstriplengthwise
intoacylindricalshapeforuniformstretch‐
ing
• reasonable amount of repetitions (break‐
ing is not very well repeatable)
BAD practice:
• intentionally variable length strips / short
strips
• not marking the initial length on the strip
with a marker
• not stretched with perpendicularly uni‐
form strain
• not taking measurements only from actual
part to be stretched
• not accounting for possible friction (some
methods)
part ii
other solutions proposed:
• multiple similar masses used (not in the
equipment list)
• a lever and some mass
• from the same latex material (a very long
uniformandrelativelythickstripisneeded
to be in proportional range, but then the
measured force will be low)
• multiple similar strips in parallel as op‐
posedtousingfeverstripsinrowandbreak‐
ing with even force
• trigonometry and constant force
• different width strip pairs (force applied
must remain constant for all pairs, in some
solutions friction would need to be accoun‐
ted for)
unsuccessful solving attempts / bad prac‐
tices:
• using equipment not listed
• calculated, not measured (experiment!)
0pts.
• using a finger as a force measuring device
0pts.
• assuming the repeatedly applied force is
constant (which it might not be) 0pts.
• measuring by varying the length of the
strip while keeping the applied force con‐
stant (stress created remains constant)
(force may be mg) 0pts.
• stretching the strip to different lengths not
knowing the actual force applied 0pts.
• MOST OFTEN: Assumption that F=kx !
That is actually an alternative lightly modi‐
fied form of writing down the dependency
WE HAVE TO MEASURE, not assume! (and
the real dependency is not linear) 0pts.
Possiblegradingschemeforthesolutions
given as example:
part ii
theory:
• assuming that the material has initial uni‐
form thickness 0.2pts.
• idea that as V = const then the dimensions
in axes perpendicular to the applied force
change in a fixed ratio to each other 1.0pts.
• idea of changing σ by changing the width
d of the material when constant force is ap‐
plied 1.0pts.
• idea of applying a common force to all the
different samples at the same time 1.0pts.
• finding how to calculate the value of ε to be
plotted 0.2pts.
• findinghowtocalculatethevalueofσ/σmax
to be plotted 0.6pts.
measurements and data:
• the strips/intervals used are as long as pos‐
sible (in our case at least 14cm) 0.1pts.
• data range covered ‐ value of the largest
measured σ differs from the smallest meas‐
ured σ at least 8 times 0.1pts.
• data points ‐ at least 8 points measured (
0.2pts for each correctly measured point
or 0.1pts for each cor‐
rectly measured point if less than 4 points
measured (have you seen graph with 3 data
points?), for maximum 8 measured points)
MAX 1.6pts.
• data evenly distributed 0.1pts.
• at least one data point measured with a σ
value at least 0,8 of σmax 0.1pts.
• measurements for the breaking event val‐
ues correctly done 0.2pts.
• chart of measured and calculated values
given 0.3pts.
graph:
• graph correctly plotted with all measured
points 0.3pts.
• graph plotted in units asked (ratios, no
units) 0.2pts.
If the results of any intermediate calcu‐
lations are excessively rounded, a deduction
from the final score of −0.1pts for every
intermediate stage used.
using out of list equipment:
• If the out of list equipment used does not
allow for any principally new methods of
solution, any credit awarded for the parts
of the solution that rely on the use of the
out of list equipment or data measured
with it will be reduced by 50%.
• If the out of list equipment does enable a
principally new method of solution, credit
willonlybeawardedforaspectsofthesolu‐
tion that would be applicable in a solution
that does not rely on the out of list equip‐
ment.

STRETCHıNG GLOVEs


[User]
Two combs are placed behind each other as in the figure. The grey comb is moved with a speed $v=\SI 1{cm/s}$ and the black comb is held still.  With what speed and to what direction are the dark spots moving?
\begin{center}
\includegraphics[width=0.6\textwidth]{2014-v3g-05-kammid_ing}
\end{center}

Solution:
Kui hall kamm liigub ühe pii võrra, on uus pilt identne esialgsega ning järelikult on tume laik liikunud ühe \enquote{lainepikkuse} võrra.
Ühe laikude \enquote{lainepikkuse} kohta tuleb 7 halli kammi piide \enquote{lainepikkust}, seega liiguvad hallid laigud 7 korda kiiremini kui hall kamm: $v=\SI 7{cm/s}$.

Kammid

\begin{wrapfigure}{r}{0.6\textwidth}%
\includegraphics[width=1\linewidth]{2014-v3g-05-kammid}
\end{wrapfigure}

Kaks kammi on asetatud üksteise taha nii, nagu näidatud joonisel. Halli kammi liigutatakse kiirusega $v=\SI 1{cm/s}$ ning musta kammi hoitakse paigal. Millise kiirusega ja millises suunas liiguvad tumedad laigud?


[User]
4. You are also required to show your detailed work for each question in a reasonably neat and coherent
     way in the Detailed Answer Sheet. You must write the relevant Question Number(s) on each of
     these pages.

Solution:
4. Consider a long narrow cylinder of cross section A filled with a compressible    liquid up to height
                                                                     ρ0
   h whose density ρ is a function of the pressure P (z) as ρ(z) =        1 + PP(z)    where P0 and ρ0
                                                                     2           0
   are constants. The depth z is measured from the free surface of the liquid where the pressure is
   equal to the atmospheric pressure (Patm ).

                                           Patm

                                                      z

                                       h
                                                                   g

   (a) Obtain the pressure (P (z)) as a function of z. Obtain the mass (M ) of liquid in the tube.         [5]

INPhO 2019 Page 10               Questions & Answers                                     Last four digits of Roll No.:

        P (z) =

         Solution:
                                                       Z
                                          P (z) =          ρ(z)g dz + Patm                                     (4.1)
                                                       Z z
                                                     ρ0        P (z)

                                                   =       1+        g dz + Patm                               (4.2)
                                                  0 2           P0
                                        dP (z)   ρ0        P (z)

                                               =      1+           g                                           (4.3)
                                          dz      2          P0
                            Z P (z)              Z z
                                      dP (z)         ρ0 g dz
                                               =                                                               (4.4)
                             Patm P0 + P (z)      0    2P0
                                    P0 + P (z)   ρ0 gz
                               log             =                                                               (4.5)
                                    P0 + Patm     2P0
                                                       h                                      i
                                          P (z) = (P0 + Patm )eρ0 gz/2P0 − P0                                  (4.6)

        M=

         Solution:
                                     C1                       9
                                              ρ0 (P0 + Patm ) ρ0 gz/2P0
                                    ρ(z) =                   e                                                 (4.7)
                                                    2P0
                                              Z h
                                      M=            ρ(z)A dz                                                   (4.8)

              HB                          =
                                              A
                                              g
                                               0

                                                (P0 + Patm )(eρ0 gh/2P0 − 1)

   (b) Let Pi (z) be the pressure at z, if the liquid were incompressible with density ρ0 /2. Assuming
       that P0  ρ0 gz obtain an approximated expression for ∆P = P (z) − Pi (z).
                                                                                (4.9)

                                                                                [2]

        ∆P ≈

         Solution:
                                         ρ0
                          Pi (z) = Patm +   gz                                                               (4.10)
                                          2
                          P (z) = (P0 + Patm )eρ0 gz/2P0 − P0                                                (4.11)
                                                       "                               2    #
                                              Patm                ρ0 gz         ρ0 gz        1

                                ≈ P0 1 +                     1+         +                      − P0          (4.12)
                                               P0                 2P0           2P0          2
                            ∆P = P (z) − Pi (z)                                                              (4.13)
                                                                                2
                                    (ρ0 gz)2 Patm ρ0 gz Patm                    ρ0 gz

                                =           +          +                                                     (4.14)
                                      8P0     P0 2       2                      2P0

         which is correct upto second order in ρ0 gz/2P0 .

INPhO 2019 Page 11                Questions & Answers

                                                      Patm ρ0 gz
                                              ∆P ≈                                            (4.15)
                                                       P0 2
          which is correct upto first order in ρ0 gz/2P0 .
          Both Eqs. (4.14) and (4.15) (mentioning order of approximation) gain full credit.

                                 Detailed answers can be found on page numbers:

INPhO 2019 Problem 4