Question

Two particles $A$ and $B$ execute simple harmonic motion of period $T$ and $5 T / 4$ . They start from mean position. The phase difference between them when the particle $A$ completes one oscillation will be

• A. $\pi / 2$
• B. Zero
• C. $2 \pi / 5$
• D. $\pi / 4$

Answer 1

Answer: (C)

Equation of motion of the particles are
$X_{1} = A_{1} sin \frac{2 \pi }{T_{1}} t$
and $X_{2} = A_{2} sin \frac{2 \pi }{T_{2}} t$
$\therefore $ Phase difference $\Delta \phi = \left(\frac{2 \pi }{T_{1}} - \frac{2 \pi }{T_{2}}\right) t$
$= \left(\frac{2 \pi }{T} - \frac{2 \pi }{5 T / 4}\right) t$
at $t = T$
$\Delta \phi = \left(2 \pi - \frac{4 \times 2 \pi }{5}\right) \frac{T}{T} = \frac{2 \pi }{5}$