Question

Two pendulums have time periods $t$ and $ 5t/4 $ . They starts SHM at the same time from the mean position. What will be the phase difference between them after the bigger pendulum completed one oscillation?

• A. $45^{\circ}$
• B. $90^{\circ}$
• C. $60^{\circ}$
• D. $30^{\circ}$

Answer 1

Answer: (B)

let the S.H.M. equation of the ist pendulum (of time period $t) x_{1}=A_{1} \sin \left(\omega_{1} t\right)$
$\omega_{1}=2 \pi / t$
S.H.M. equation of $2^{nd}$ pendulum (of time period $5 t / 4) x _{2}= A _{2} \sin \left(\omega_{2} t \right)$
$\omega_{2}=2 \pi \times 4 / 5 t =8 \pi / 5 t$
after $5 t / 4 x _{1}= A _{1} \sin (5 \pi / 2)$
and $x _{2}= A _{2} \sin (2 \pi)$
so, the phase difference is $\pi / 2=90^{\circ}$