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

Answer 1

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

Formula used:

$\sin (A+B)=\sin A\cos B+\cos A\sin B$
$Y=A\sin (\omega t+\varphi ),T=\frac{2\pi }{\omega }$

Formula used:

$\sin (A+B)=\sin A\cos B+\cos A\sin B$
$Y=A\sin (\omega t+\varphi ),T=\frac{2\pi }{\omega }$
Given amplitude,
$\Rightarrow y=5(\sin 3\pi t+\sqrt{3}\cos 3\pi t)$
Multiply and divided by $2$ above equation
$\Rightarrow y=5\times 2(\frac{1}{2}\sin 3\pi t+\frac{\sqrt{3}}{2}\cos 3\pi t)$
$y=10(\cos \frac{\pi }{3}\sin 3\pi t+\frac{\pi }{3}\cos 3\pi t)$
$\Rightarrow y=10\sin (3\pi t+\frac{\pi }{3})$
Compare this with the wave equation,
$y=A\sin (\omega t+\varphi )$
We get the values,
Amplitude $A=10$
$\omega =3\pi$
The time period,
$T=\frac{2\pi }{\omega }=\frac{2\pi }{3\pi }=\frac{2}{3}sec$

Final Answer: $\left(d\right)$