Question

A simple harmonic motion is represented by $y=5(\sin 3\pi t+\sqrt{3}\cos 3\pi t)cm$ The amplitude and time period of the motion are :

• A. $10cm,\frac{2}{3}s$
• B. $5cm,\frac{2}{3}s$
• C. $10cm,\frac{3}{2}s$
• D. $5cm,\frac{3}{2}s$

Answer 1

The correct option is A $10cm,\frac{2}{3}s$

$y=5(\sin 3\pi t+\sqrt{3}\cos 3\pi t)$

$y=5\times 2(\frac{1}{2}\sin 3\pi t+\frac{\sqrt{3}}{2}\cos 3\pi t)$

$y=10(\cos \frac{\pi }{3}\sin 3\pi t-\sin \frac{\pi }{3}\cos 3\pi t)$

$y=10\sin (3\pi t+\frac{\pi }{3})$

Comparing it with $y=A\sin (\omega t+\varphi )$

$\Rightarrow A=10$ i.e., amplitude is $10$cm

$\Rightarrow T=\frac{2\pi }{\omega }=\frac{2\pi }{3\pi }=\frac{2}{3}$ sec. i.e, time period is $\frac{2}{3}$ sec.