Question 273: Oscillations

Question

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 $\left[\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3} / 2\right]$

Accepted Answer

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

Answer

6 m

Answer

12 m

Answer

$12 \sqrt{3} \mathrm{~m}$

Answer

$6 \sqrt{3} \mathrm{~m}$

Oscillations

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]

Choose the correct answer: