For a pendulum's motion to be simple harmonic motion (S.H.M.) is it necessary for a pendulum to have small amplitude or S.H.M. can be produced at large amplitudes as well? If it is really necessary for an S.H.M. to have small amplitudes then why is it? because even at large amplitudes there is restoring force pulling the pendulum toward mean position and its acceleration is directly proportional to the displacement.
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3 Answers
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As you stated, in order to have simple harmonic motion, you need to have an acceleration that is proportional to the displacement. For a pendulum, if you work out the details, you will find that
$\frac{d^2\theta}{dt^2} \propto -\sin(\theta)$
where $\theta$ is the angle the pendulum makes with the vertical. For small angles, $\sin(\theta)\sim\theta$, which would then lead to simple harmonic motion. For large angles, this approximation no longer holds, and the motion is not considered to be simple harmonic motion.
answered Oct 14, 2015 at 19:42
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$\begingroup$ It means that the increment in the angle would not produce S.H.M. it would be just motion but not S.H.M. and in that case when it will no longer be S.H.M. will its acceleration be directly proportional to the displacement? $\endgroup$ Commented Oct 14, 2015 at 19:49
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1$\begingroup$ Just to clarify, are you asking when the approximation is no longer valid? That would really depend on what kind of accuracy you are trying to achieve. The motion would be oscillatory, but not simple harmonic motion, because simple harmonic motion implies sinusoidal behavior, where the second derivative of the displacement is proportional to the displacement as you pointed out. $\endgroup$ Commented Oct 14, 2015 at 19:54
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$\begingroup$ The point of the simple harmonic oscillator is to have a precise single harmonic frequency. If you were somehow able to design the pendulum to achieve that goal you could have a very accurate clock. But the non linearity in your basic pendulum design results in harmonic distortion. In terms of the spectrum, it no longer appears as a line. The power is spread over a range of frequency from the fundamental - and it's dependent on the amplitude of the swing. $\endgroup$ Commented Oct 15, 2015 at 2:59
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In case of pendulum motion, when the angle of displacement is large(as shown in fig.), the direction of restoring force$(mg. sin \theta)$ is not exactly in the direction of equilibrium position. But the condition of S.H.M. is the restoring force must directed to the equilibrium position in all instant. So in case of large angular displacement, this condition is violated. Hence the motion of the pendulum no more remains simple harmonic in that case.
For this cause the angular displacement of S.H.M. must be kept smaller than $4$ degree.
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2$\begingroup$ At $2^\circ$ it's not perfectly SHM. Why do you choose $4^\circ$? Even at that small angle, the solution has deviated from pure SHM. But it isn't even a 0.5% effect until $>16^\circ$. It's a small bit over 0.1% off from SHM at $8^\circ$. Your $4^\circ$ choice is arbitrary. $\endgroup$– Bill NCommented Oct 14, 2015 at 20:21
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$\begingroup$ I've seen in many books that generally the angle kept smaller than $ 4∘$. Actually the aim is to keep the pendulum motion in approximately straight path. In that case restoring force always directed to the equilibrium position. Now this curvature of the path not only depends on angular displacement only but also depends on the radius of curvature or length of the pendulum thread. Higher thread length and smaller angular displacement keeps the pendulum motion along approximately straight path. $\endgroup$ Commented Oct 15, 2015 at 4:25
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$\begingroup$ The correction factor applied to $\omega_{SHM}$ does not depend on pendulum length. It depends on maximum angle. Introducing the idea of curvature is a diversion at best. Yes, for shorter lengths a given horizontal displacement will create a larger correction factor, but that's because the angle is larger. The correction arises because $\sin\theta \ne\theta$, but $\sin\theta=\theta-\theta^3/2!+\theta^5/5!-\theta^7/7! + \ldots$. $\endgroup$– Bill NCommented Oct 15, 2015 at 16:55
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$\begingroup$ Sir with respect, I never taking about the correction factor. I've said the curvature of the pendulum path(i.e. in which path the pendulum ball moving) is depends on the length of the thread of pendulum. I'm not going to mathematical derivation. I'm trying to give ans of this question physically. $\endgroup$ Commented Oct 15, 2015 at 17:17
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It's just because at large angular displacements, it does not approximate the SHM of, say, a block on a spring with no friction. The restoring force is not in the direction of the displacement; therefore it does not act like SHM.
