Question

A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is_______ cm. [Assume that the mass of the pendulum remains same]}

• A. 7.5
• B. 120
• C. 0.75
• D. 15

Hint:

To double the number of oscillations in the same time (double the frequency), you must reduce the length of the pendulum to one-fourth of its original value.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The time period $T$ of a simple pendulum is the time taken for one oscillation. It depends on the length of the string $L$ and the acceleration due to gravity $g$. If the number of oscillations $n$ in a fixed time $t$ changes, the frequency and time period change accordingly.
Step 2: Key Formula or Approach:
1. Time period $T = 2\pi\sqrt{\frac{L}{g}}$.
2. Also, $T = \frac{\text{Total Time}}{\text{Number of Oscillations}} = \frac{t}{n}$.
3. Therefore, $\frac{t}{n} \propto \sqrt{L} \implies n \propto \frac{1}{\sqrt{L}}$ for a constant time $t$.
Step 3: Detailed Explanation:
Let $n_1 = 20$, $L_1 = 30$ cm.
Let $n_2 = 40$, $L_2 = ?$.
Since $n \propto \frac{1}{\sqrt{L}}$, we can write: \[ \frac{n_1}{n_2} = \sqrt{\frac{L_2}{L_1}} \] \[ \frac{20}{40} = \sqrt{\frac{L_2}{30}} \] \[ \frac{1}{2} = \sqrt{\frac{L_2}{30}} \] Squaring both sides: \[ \frac{1}{4} = \frac{L_2}{30} \] \[ L_2 = \frac{30}{4} = 7.5 \text{ cm} \]
Step 4: Final Answer:
The required length of the string is 7.5 cm.