Question

Two pendulums have 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 ons oscillation?

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

Answer 1

Answer: (B)

${90}^o$

let the S.H.M. equation of the 1st 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 2nd 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/4 $x_1$=$A_1\sin \left(5\pi /2\right)$

and $x_2$ =$A_2\sin \left(2\pi \right)$

so, the phase difference is $\pi /2$=${90}^0$