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

$y=5(\sin3\pi t+\sqrt{3}\cos 3\pi t)$

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

$y=10\left(\cos \dfrac{\pi}{3}\sin 3\pi t-\sin \dfrac{\pi}{3}\cos 3\pi t\right)$

$y=10\sin\left(3\pi t+\dfrac{\pi}{3}\right)$

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

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

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