Question

A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2 seconds its velocity is $\pi$m/s. Amplitude of the oscillation is $\left[sin 30^\circ = cos 60^\circ = 0.5, sin 60^\circ = cos 30^\circ = \sqrt{3}/2\right]$

• A. 6 m
• B. 12 m
• C. $12\sqrt{3}$ m
• D. $6\sqrt{3}$ m

Answer 1

Answer: (B)

12 m

Answer 2

Given the SHM position as

$$x(t)=A\sin (\omega t)$$

with time period $T=12$ s, we find

$$\omega = \frac{2\pi}{T}=\frac{2\pi}{12}=\frac{\pi}{6}\,\text{rad/s}.$$

The velocity is

$$v(t)=A\omega\cos (\omega t).$$

At $t=2$ s, the velocity is given as $\pi$ m/s:

$$v(2)=A\left(\frac{\pi}{6}\right)\cos\left(\frac{\pi}{6}\times2\right)=A\left(\frac{\pi}{6}\right)\cos\left(\frac{\pi}{3}\right).$$

Since $\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$,

$$v(2)=A\left(\frac{\pi}{6}\right)\times\frac{1}{2}=\frac{A\pi}{12}.$$

Equate to the given velocity:

$$\frac{A\pi}{12}=\pi \quad \Longrightarrow \quad A=12\,\text{m}.$$