Question

Two particles execute SHM with amplitude $A$ and $2A$ and angular frequency $\omega$ and $2\omega$ respectively. At $t=0$ they starts 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

Answer: (B)

$\frac{2\pi }{3}$

Let initial phase of first particle be ${\theta }_1$.

Its phase at time $t$ is ${\theta }_1+\omega t={\theta }_1+2\pi /3$

Let initial phase of first particle be ${\theta }_2$.

Its phase at time $t$ is ${\theta }_2+2\omega t={\theta }_2+4\pi /3$

As two phases are equal at time $t$,

so, $t$ is ${\theta }_1+2\pi /3$$={\theta }_2+4\pi /3$

$\Rightarrow {\theta }_1-{\theta }_2=2\pi /3$, is the phase difference.