Question

Two particles A and B execute simple harmonic motion with periods of T and $\frac{5T}{4}$ respectively .They start simultaneously from mean position. The phase difference between them when A completes one oscillation will be –

• A. 0
• B. $\frac{\pi}{2}$
• C. $\frac{\pi}{4}$
• D. $\frac{2\pi}{5}$

Answer 1

Answer: (D)

$\frac{2\pi}{5}$

Answer 2

Df = (w₁ – w₂)t = $\left( \frac{2\pi}{T}–\frac{2\pi}{5T/4} \right)$ T = $\frac{2\pi}{5}$