Question

Two simple pendulums of lengths 1 m and 2 m are in phase at the mean position at a certain instant of time. The pendulums will be again in phase at a time of $ \frac{3T}{2} $, where $ T $ is the time period of the shorter pendulum. Then $ x = $?

• A. 4 m
• B. 6 m
• C. 9 m
• D. 12 m

Hint:

When dealing with pendulums, remember that the time period depends on the length of the pendulum, and the longer the pendulum, the longer the time period.

Answer 1

The Correct Option is B

The time period \( T \) of a simple pendulum is given by: \[ T = 2 \pi \sqrt{\frac{L}{g}} \] For the first pendulum with \( L_1 = 1 \, \text{m} \), the time period \( T_1 \) is: \[ T_1 = 2 \pi \sqrt{\frac{1}{10}} = 2 \pi \times 0.316 = 2 \pi \times 0.316 \approx 2 \, \text{s} \] For the second pendulum with \( L_2 = 2 \, \text{m} \), the time period \( T_2 \) is: \[ T_2 = 2 \pi \sqrt{\frac{2}{10}} = 2 \pi \times 0.447 = 2 \pi \times 0.447 \approx 2.8 \, \text{s} \] The two pendulums will be in phase at the time when \( x = 6 \, \text{m} \).