What is the Relationship Between Amplitude and Speed in SHM?

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Discussion Overview

The discussion revolves around the relationship between amplitude and speed in simple harmonic motion (SHM), specifically examining how halving the amplitude of a pendulum affects its speed at a given position, x = ± A/4. The focus is on the mathematical reasoning behind the changes in speed due to alterations in amplitude.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents a problem regarding a pendulum oscillating in SHM and questions the speed at a specific position after halving the amplitude.
  • Another participant notes that the speed at a given phase point is affected by the halving of the amplitude, suggesting that the speed will also be halved.
  • A further contribution emphasizes the need to compare velocities at different amplitudes and positions, indicating that the phase relationship changes with the new amplitude.
  • One participant suggests performing calculations using the SHM equations to derive the velocities for the specified position, implying that the relationship is not intuitive and requires mathematical analysis.

Areas of Agreement / Disagreement

Participants express a shared understanding that calculations are necessary to grasp the relationship between amplitude and speed, but there is no consensus on an intuitive explanation for the changes observed.

Contextual Notes

The discussion highlights the dependence on mathematical derivations and the phase relationship in SHM, indicating that intuitive understanding may be limited without calculations.

ual8658
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So in class we were given a problem.
A pendulum is oscillating in SHM with amplitude A. After a while you slow it down so that its amplitude is halved. What happens to its speed at x = ± A/4 ?

The answer is 1/sqrt(5) and I don't get why.
 
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There are two things that have changed. One is that, because the amplitude has halved from ##A## to ##A/2##, the speed for a given phase point has also halved.

The other is that, because the amplitude has changed, the point ##x=A/4## is at a different phase from what it was when the amplitude was ##A##.

You need to compare the velocity when position is at half the amplitude (the new situation) with the velocity when position is one quarter of the amplitude (the old situation).
 
ual8658 said:
So in class we were given a problem.
A pendulum is oscillating in SHM with amplitude A. After a while you slow it down so that its amplitude is halved. What happens to its speed at x = ± A/4 ?

The answer is 1/sqrt(5) and I don't get why.
You need to do the calculation; it isn't an intuitive thing. If you write out the equation for SHM it will show you how the position varies with time. If you differentiate it wrt time, you will get the variation of velocity with time. So you can put your two values of Amplitude into that velocity equation and see what the velocities are for x= A/4. AS the differential of sin is cos, life is pretty easy for you.
 
Thank you both. I thought there'd be a way to see it intuitively but the calculations show the reason why.
 

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