Question

The amplitude of the vibrating particle due to super position of two simple harmonic motions of $y_{1}=\sin \left(\omega t+\frac{\pi}{3}\right)$ and $y_{2}=\sin (\omega t)$ will be :

• A. $2$
• B. $ \sqrt{3} $
• C. $ \sqrt{2} $
• D. $1$

Answer 1

Answer: (B)

For a wave travelling with amplitude $a$ and angular velocity $\omega$ at time $t$, the displacement $y$ is given by
$y=a \sin \omega t$
For phase difference $\phi$, the general equation is
$y=a \sin (\omega t+\phi)$
Given equations are
$y_{1}=\sin \left(\omega t+\frac{\pi}{3}\right)$
and $y_{2}=\sin \omega t$
Comparing with standard equation, we have
$a_{1}=1, a_{2}=1, \phi=\frac{\pi}{3}=60^{\circ}$
Therefore, resultant amplitude is
$a=\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi}$
$a=\sqrt{1+1+2 \times 0.5}=\sqrt{3}$