Question

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:

• A. 11
• B. 9
• C. 10
• D. 8

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For two pendulums to be back in phase, the time elapsed must be an integral multiple of both their time periods.
Since the shorter pendulum moves faster, it will complete exactly one more vibration than the longer one to reach the same phase.
Key Formula or Approach:
1. Time period \(T = 2\pi\sqrt{\frac{L}{g}}\). So, \(T \propto \sqrt{L}\).
2. Let the longer pendulum make \(n\) vibrations and the shorter one make \((n+1)\) vibrations.
\[ n T_{\text{long}} = (n+1) T_{\text{short}} \]
Step 2: Detailed Explanation:
1. Calculate ratio of time periods:
\[ \frac{T_{\text{long}}}{T_{\text{short}}} = \sqrt{\frac{L_{\text{long}}}{L_{\text{short}}}} = \sqrt{\frac{121}{100}} = \frac{11}{10} \]
2. Equate total time for phase synchrony:
\[ n T_{\text{long}} = (n+1) T_{\text{short}} \implies \frac{T_{\text{long}}}{T_{\text{short}}} = \frac{n+1}{n} \]
\[ \frac{11}{10} = \frac{n+1}{n} \]
3. Solve for \(n\):
\[ 11n = 10(n+1) \]
\[ 11n = 10n + 10 \]
\[ n = 10 \]
This means the longer pendulum completes 10 vibrations.
4. Find vibrations of the shorter pendulum:
Vibrations of shorter pendulum = \(n + 1 = 10 + 1 = 11\).
Step 3: Final Answer:
The minimum number of vibrations of the shorter pendulum is 11.