Question

Two particles execute SHM with amplitude A and 2A and angular frequency $\omega$ and $2\omega$ respectively. At t=0, they start with some initial phase difference. At $t=\frac{2\pi }{3\omega }$, they are in same phase. Their initial phase difference is

• A. $\frac{\pi }{3}$
• B. $\frac{2\pi }{3}$
• C. $\frac{4\pi }{3}$
• D. $\pi$

Answer 1

The correct option is A $\frac{\pi }{3}$

$A_1=A$

$A_2=2A$

${\omega }_1=\omega$

${\omega }_2=2\omega$

At $t=0,$

${\varphi }_1=0$

${\varphi }_2=\varphi$

At $t=\frac{2\pi }{3w}$

${\theta }_1=\theta$

$x=a\sin (\omega t+\varphi )$

$x_1=A\sin \left(\omega t\right)$

$x_2=2A\sin \left(2\omega t\right)$

At $t=\frac{2\pi }{3\omega }\Rightarrow \sin \omega t=\sin (2\omega +\varphi )$

$sin\frac{2\pi }{3}=\sin (\frac{\pi }{3}+\varphi )$

$\frac{2\pi }{3}=(\frac{\pi }{3}+\varphi )$

$\therefore$ $\frac{2\pi }{3}-\frac{\pi }{3}=\varphi$ $\Rightarrow \varphi =\frac{\pi }{3}$