0

This scenario is taken from the 2018 NZ Scholarship physics exam:

enter image description here

(Just in case the image doesn't load, it involves a wagon on a curved track (concave up). The wagon moves down from one side, to the other, then back again (with simple harmonic motion of period 60s). The question later goes onto say that the track is the arc of a circle.

The answers do not take this method, but I am wondering if it is possible to equate the motion of this wagon on this track, to the motion of some mass attached to a pendulum - that gives a SHM period of 60s, where the length of the pendulum is the radius of the circle this arc track is from. Is it possible to do so and why/why not?

The paper goes onto ask these 3 questions:

enter image description here enter image description here enter image description here

Solutions (without assuming this can be modelled with pendulum):

enter image description here enter image description here

We see the circle has a radius of 901m. However, if we assume it can be modelled by a pendulum, the radius / length of pendulum, will be 849m (found with $T = 2pi*sqrt(l/g)$. This suggests we cannot model this as a pendulum, why is this?

Qmechanic
  • 201,751
Jaynindu
  • 23
  • Welcome to PSE. Your question is unclear because you didn't post the problem as in the 2018 NZ Scholarship physics exam wherein we read : (a) (i) State the conditions that must apply for the motion to be simple harmonic motion. (ii) Show that the maximum speed attained is $:15.7m/s$. (b) When the wagon is halfway between $:\rm B:$ and $:\rm C:$, calculate its approximate height above $:\rm B$.(c) (i)$\cdots\cdots$(ii)$\cdots\cdots$ – Frobenius Nov 28 '21 at 23:46
  • .... (d) The track $:\rm A:$ to $:\rm C:$ is an arc of a circle. By first calculating the radius of the circle,discuss whether the original assumption that this motion is simple harmonic motion is valid. – Frobenius Nov 28 '21 at 23:46
  • @Frobenius I don't think OP is asking for the solution to the problem exactly as stated, but rather whether the setup described therein is equivalent to a mass on a string. – noah Nov 28 '21 at 23:49
  • Try to solve and answer the problem as it had been given and after that wonder ... if it is possible to equate the motion of this wagon on this track, to the motion of some mass attached to a pendulum... – Frobenius Nov 28 '21 at 23:49
  • @noah : we must answer the (a) and (d) parts of the exercise first. – Frobenius Nov 28 '21 at 23:52
  • @Frobenius I don't really understand why. – noah Nov 28 '21 at 23:56
  • @Frobenius I attempted the approach you have suggested: completing a, and d, with both approaches and seeing if they give the same answer. They do not (with my calculations), and I was wondering if I had made some sort of error in my assumptions - will edit original post. – Jaynindu Nov 29 '21 at 00:12
  • OK, well done, Jaynindu. – Frobenius Nov 29 '21 at 00:14
  • You might care to read the answers to the recent (day old?) question "Why does the equation for time period of a simple pendulum become less accurate at angular displacements greater than 20 degrees? " [Sorry that I don't know how to do links.] – Philip Wood Nov 29 '21 at 00:35
  • @PhilipWood surely that is due to the path of the pendulum not being linear, making sense why the curved track may not give exact simple harmonic motion. However, surely this model exactly replicates the motion of a pendulum (with all the errors with SHM it brings)? – Jaynindu Nov 29 '21 at 00:39
  • Why does the equation for time period of a simple pendulum become less accurate at angular displacements greater than 20 degrees?. – Frobenius Nov 29 '21 at 00:59
  • @PhilipWood : [any title you want](the link). For example [Why does the equation for time period of a simple pendulum become less accurate at angular displacements greater than 20 degrees?](https://physics.stackexchange.com/questions/679446/why-does-the-equation-for-time-period-of-a-simple-pendulum-become-less-accurate). – Frobenius Nov 29 '21 at 01:03
  • Note that a difficult condition for the motion to be simple harmonic is the track to be part of a cycloid, see here : What shape must a bowl be to have a ball rolling in the bowl execute perfect simple harmonic motion?. Of course the examination doesn't ask to have such an answer. – Frobenius Nov 29 '21 at 01:25
  • ...and if cycloid the SHM could be simulated by Huygens pendulum (Horologium Oscillatorium). – Frobenius Nov 29 '21 at 06:57
  • @Frobenius so simply being the arc of a circle (not a cycloid) would not mean it would be able to be modelled by SHM? – Jaynindu Nov 29 '21 at 08:11
  • @Jaynindu Look at Noah's answer; for the circle you need small angle approximation. For cycloid you do not. – BioPhysicist Nov 29 '21 at 10:54
  • Hello! It is preferable to type out screenshots or images of text; for formulae, one can use MathJax. Thanks! – jng224 Dec 02 '21 at 18:58

2 Answers2

1

If the track is a segment of a circle, the given setup is exactly equivalent to a regular pendulum on a string (sometimes called a mathematical pendulum when we ignore the mass of the string). The motion of the mass follows the exact same path, and the involved forces are equivalent. The only difference is that for a mass on a string, the force is provided by tension in the string pulling radially inward, while in this example it is provided by the rail tracks pushing, but also radially towards the center of the circle (since there is no friction).

Whether or not this can be treated as a harmonic oscillator depends on how large the amplitude is and how accurate the result should be. When solving a mathematical pendulum, we usually resort to approximating

$$ \sin{x} \approx x$$

for small values of $x$. Then the motion becomes a simple harmonic oscillation. If the amplitude is too large, this approximation doesn't hold very well.

noah
  • 10,324
0

There is no reason this motion cannot be modeled as a simple pendulum (given meaningful data). In part (a)(ii) the 2π/T is the angular frequency of oscillation which is not the same as the maximum angular velocity. And, the given width tells you nothing about the radius. The given solution is nonsense.

R.W. Bird
  • 12,139
  • I believe they are using the formula: v = -Aωsin(ωt). At the point where v is a maximum, sin(ωt) = 1, hence: v = Aω. Then they use a similar approach as in part b to work out the "y coordinates" of the ends of the track, then substituting into the equation for a circle, find its radius. I am unable to see where the given solution (or my modelling by a pendulum) fails. – Jaynindu Nov 29 '21 at 08:11
  • OK, If the angle stays small, you can say that the horizontal component of motion is close to being simple harmonic with $v_{max}$ = Aω, where (A) is the amplitude of horizontal motion. Then conservation of energy does give h = 12.576 m. I did not follow the logic for the half way point, but if you write 150 = R sin(θ) and h = R(1 – cos(θ)) and solve you do get R = 900.85m and θ = 9.585 degrees. These figures do agree with the motion of a simple pendulum (within approximations). – R.W. Bird Nov 29 '21 at 17:34