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Why if a mass tied to an ideal string is given a small displacement, its motion is SHM. However, for a large displacement it is not SHM but oscillatory?

Yukti
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  • Hi. Could you elaborate your experimental setup? Are you referring to a mass suspended using a string like a pendulum? Thanks! – Vishnu Jan 26 '20 at 09:07
  • This might interest you: https://physics.stackexchange.com/questions/57623/what-is-the-period-of-a-physical-pendulum-without-using-small-angle-approximatio – Semoi Jan 26 '20 at 09:47

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If you are talking about the motion of a pendulum, its equation is given by $\frac{d^2 \theta}{d t^2} = - \frac{g}{L} \sin \theta$. However, the simple harmonic motion is described by $\frac{d^2 \theta}{d t^2} = - \frac{g}{L}\theta$ - note the absence of the sine here. For small displacements we can approximate $\sin \theta \simeq \theta$, using a Taylor series expansion, which means that for small angular displacements the motion is indeed simple harmonic to a reasonable degree of accuracy. As $\theta$ becomes larger, the non-linearity of the sine function become more pronounced, and the motion departs from the simple harmonic form. This has a solution in terms of elliptic integrals.

Clara Diaz Sanchez
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  • But how it is oscillatory? – Yukti Jan 26 '20 at 10:14
  • It is still oscillatory - the torque acts to move the pendulum towards its point of equilibrium. But the motion is not simple harmonic, so the period depends on the amplitude of the swing. Another way of saying it is that the pendulum is moving (oscillating) around a potential minimum, but the potential is not parabolic, as it would be for SHM. – Clara Diaz Sanchez Jan 26 '20 at 10:46