The displacement of a particle along the $x$ axis is given by $\mathrm{x}=\mathrm{a} \sin ^2 \omega…

The displacement of a particle along the $x$ axis is given by $\mathrm{x}=\mathrm{a} \sin ^2 \omega \mathrm{t}$. The motion of the particle corresponds to
  1. simple harmonic motion of frequency $\omega / \pi$
  2. simple harmonic motion of frequency $3 \omega / 2 \pi$
  3. non simple harmonic motion
  4. simple harmonic motion of frequency $\omega / 2 \pi$

Solution

For a particle executing SHM acceleration (a) $\propto-\omega^2$ displacement $(x)$ Given $\mathrm{x}=\mathrm{a} \sin ^2 \omega \mathrm{t}$ Differentiating the above equation w.r.t, we get $\frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{a} \omega(\sin \omega \mathrm{t})(\cos \omega \mathrm{t})$ Again differentiating, we get $\begin{aligned} & \frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}=\mathrm{a}=2 \mathrm{a} \omega^2\left[\cos ^2 \omega \mathrm{t}-\sin ^2 \omega \mathrm{t}\right] \\ & =2 \mathrm{a} \omega^2 \cos 2 \omega \mathrm{t} \end{aligned}$ The given equation does not satisfy the condition for SHM [Eq. (i)] . Therefore, motion is not simple harmonic. *

Asked in: NEET 2010 (Screening)

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