Q: Two pendulums of length 100cm and 121 cm start oscillating. At some instant, the two are at the mean position in the same phase. After how many oscillations of the longer pendulum will the two be in the same phase at the mean position again.

(a) 11

(b) 10

(c) 21

(d) 20

Ans: (b)

Sol: Let two pendulum are same phase , after N vibration of longer pendulum . In this time the shorter pendulum will complete (N+1) vibration

$\displaystyle N\times 2\pi \sqrt{\frac{l_2}{g}} = (N+1)2\pi \sqrt{\frac{l_1}{g}} $

$ \displaystyle N\times 2\pi \sqrt{\frac{121}{g}} = (N+1)2\pi \sqrt{\frac{100}{g}} $

11N = 10(N+1)

⇒ N= 10