A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2…

A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2 seconds its velocity is $\pi \mathrm{m} / \mathrm{s}$. Amplitude of the oscillation is $\left[\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3} / 2\right]$
  1. 6 m
  2. 12 m
  3. $12 \sqrt{3} \mathrm{~m}$
  4. $6 \sqrt{3} \mathrm{~m}$

Solution

Displacement of the particle, $x=A \sin \omega t$ Velocity of the particle, $\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{A} \omega \cos \omega \mathrm{t}...(i)$
Given that, $\mathrm{v}=\pi \mathrm{m} / \mathrm{s}, \mathrm{~T}=12 \mathrm{~s},$ $\therefore \quad \omega=\frac{2 \pi}{\mathrm{~T}}=\frac{\pi}{6} \mathrm{rad} / \mathrm{s}$ Substituting in equation (i), we get, $\begin{aligned} & \pi & =\mathrm{A} \times \frac{\pi}{6} \times \cos \left(\frac{\pi}{6} \times 2\right) \\ \therefore & 1 & =\frac{A}{6} \cos \left(\frac{\pi}{3}\right)=\frac{A}{6} \times \frac{1}{2} \\ \therefore \quad & A & =12 \mathrm{~m} \end{aligned}$

Asked in: MHT CET 2024 (11 May Shift 1)

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