Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :

Physics Problem: Phase Synchronization of Two Pendulums

Two pendulums of length \( 121 \, \text{cm} \) and \( 100 \, \text{cm} \) start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

• (a) 11
• (b) 9
• (c) 10
• (d) 8

Detailed Solution

The time period \( T \) of a simple pendulum is given by:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

For the two pendulums:

\[ T_1 = 2\pi \sqrt{\frac{121}{g}} \quad \text{and} \quad T_2 = 2\pi \sqrt{\frac{100}{g}} \]

The ratio of their time periods is:

\[ \frac{T_1}{T_2} = \sqrt{\frac{121}{100}} = \frac{11}{10} \]

This means the longer pendulum completes \( 10 \) oscillations while the shorter pendulum completes \( 11 \) oscillations in the same time.

For the two pendulums to be in phase again at the mean position, the shorter pendulum must complete an integer number of oscillations that is a multiple of the ratio \( \frac{11}{10} \). The smallest such integer is \( 11 \).

Thus, the minimum number of vibrations of the shorter pendulum required is \( 11 \).

Answer: (a) 11