Question

A simple harmonic motion is represented by:

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

The amplitude and time period of the motion are:

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

Answer 1

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

$\text{Given:}y=5\left[\sin \right(3\pi t)+\sqrt{3}\cos (3\pi t\left)\right]$
$y=5\sin \left(3\pi t\right)+5\sqrt{3}\sin (3\pi t+\frac{\pi }{2})$
The above expression can be considered as the superposition of two SHM with same frequency and with a phase difference of $\frac{\pi }{2}$,

Therefore, the resultant amplitude is given by,

$A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$

$\text{Here},A_1=5\text{cm},A_2=5\sqrt{3}\text{cm},\varphi =\frac{\pi }{2}$

$\Rightarrow A=\sqrt{25+75+0}=10\text{cm}$
$\text{Time period,}T=\frac{2\pi }{\omega }=\frac{2\pi }{3\pi }=\frac{2}{3}\text{s}$

Hence, option $\left(A\right)$ is correct.

Alternate solution: $y=5\left[\sin \right(3\pi t)+\sqrt{3}\cos (3\pi t\left)\right]$ $y=10\left[\frac{1}{2}\sin \right(3\pi t)+\frac{\sqrt{3}}{2}\cos (3\pi t\left)\right]$ $y=10\left[\sin \right(3\pi t\left)\cos \right(\pi /3)+\cos (3\pi t\left)\sin \right(\pi /3\left)\right]$ $y=10\sin (3\pi t+\pi /3)$ $\therefore A=10\text{cm}\&T=\frac{2\pi }{\omega }=\frac{2}{3}\text{s}$