Question

Two particles $A$ and $B$ execute simple harmonic motion of period $T$ and $5T/4$ . 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 $\Delta \varphi =\left(\frac{2\pi }{T_1}-\frac{2\pi }{T_2}\right)t$
$=\left(\frac{2\pi }{T}-\frac{2\pi }{5T/4}\right)t$
at $t=T$
$\Delta \varphi =\left(2\pi -\frac{4\times 2\pi }{5}\right)\frac{T}{T}=\frac{2\pi }{5}$