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This question already has an answer here:

Why simple harmonic motion time period equation is independent from displacement from mean position?

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marked as duplicate by JMac, ZeroTheHero, Kyle Kanos, stafusa, Emilio Pisanty Sep 19 '17 at 10:31

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The expression for time period does not contain displacement in it and that is why it does not depend on displacement. But, that however is not what you are looking for I think. So, think about the the speed of the bob. It is variable. The further you displace it, more the potential energy it acquires and hence collects more speed when coming back towards mean position.

$$mgh = \frac{mv^2}{2}$$

Here, $h$ is the vertical displacement form mean position.

So, as the displacement gets bigger, so does the velocity. Thus the time period remains constant.

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  • $\begingroup$ Sorry but what is $h$ here, and why would the gravity $g$ enter into a spring-and-mass problem? $\endgroup$ – ZeroTheHero Sep 19 '17 at 9:34
  • $\begingroup$ @ZeroTheHero Why does this have to be a spring and mass problem? This is an example of a pendulum motion. $\endgroup$ – JMac Sep 19 '17 at 12:25
  • $\begingroup$ @JMac I never read this as a pendulum problem, for which the period is only approximately independent of the amplitude. SHM usually refers to spring-and-mass problem, so if the OP had pendulum in mind the question is unclear, or the answer requires clarifications and qualifications. $\endgroup$ – ZeroTheHero Sep 19 '17 at 13:57
  • $\begingroup$ @ZeroTheHero If you treat it like an ideal linear pendulum (as it has been in this answer, and is necessary for SHM), then it is exactly independent of amplitude. OP didn't specify what system they were referring to, so any type of SHM is fair game. This answer could be more general; but it does address the question. It seems like you interpreted it as spring potential; but nothing in the question makes that definitive. $\endgroup$ – JMac Sep 19 '17 at 14:04
  • $\begingroup$ @JMac Agreed that in the linear approximation it is SHM. Your accepted answer to the alternate question was completely clear on this point (as was the alternate question.) $\endgroup$ – ZeroTheHero Sep 19 '17 at 15:43

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