Question

Two particles A and B execute simple harmonic motions of periods T and 5T / 4. They start from mean position. The phase difference between them when the particle A complete one oscillation will be:

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{2\pi }{5}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is C $\frac{2\pi }{5}$

Equations of motion of the practicals are

$X_1=A_{1}sin\frac{2\pi }{T_1}t$

and $X_2=A_{2}sin\frac{25}{T_2}t$

$\therefore$ Phase difference $\Delta \varphi =(\frac{2\pi }{T_1}-\frac{2\pi }{T_2})t$

$=(\frac{2\pi }{T}-\frac{2\pi }{5T/4})t$

at t=T

$\Delta \varphi =(2\pi -\frac{4\times 2\pi }{5})\frac{T}{T}=\frac{2\pi }{5}$