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 $5t/4)x_2=A_2\sin \left({\omega }_{2}t\right)$
${\omega }_2=2\pi \times 4/5t=8\pi /5t$
after $5t/4x_1=A_1\sin (5\pi /2)$
and $x_2=A_2\sin (2\pi )$
so, the phase difference is $\pi /2=90^{\circ }$