Why simple harmonic motion time period equation is independent from displacement from mean position?
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1$\begingroup$ Would you — intuitively — think that the period should be larger or smaller, if the displacement were larger? $\endgroup$– pelaCommented Sep 19, 2017 at 7:41
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5$\begingroup$ Possible duplicate of Intuition - why does the period not depend on the amplitude in a pendulum? $\endgroup$– JMacCommented Sep 19, 2017 at 9:25
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$\begingroup$ I have posted an answer which I think answers your question here. physics.stackexchange.com/a/358061/104696 $\endgroup$– FarcherCommented Sep 19, 2017 at 12:16
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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? $\endgroup$ Commented Sep 19, 2017 at 9:34
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$\begingroup$ @ZeroTheHero Why does this have to be a spring and mass problem? This is an example of a pendulum motion. $\endgroup$– JMacCommented Sep 19, 2017 at 12:25
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$\begingroup$ @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. $\endgroup$ Commented Sep 19, 2017 at 13:57
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$\begingroup$ @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. $\endgroup$– JMacCommented Sep 19, 2017 at 14:04
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$\begingroup$ @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.) $\endgroup$ Commented Sep 19, 2017 at 15:43