Question

A simple pendulum of length \(l_1\) has periodic time \(2.4\) s. Another simple pendulum of length \(l_2 < l_1\), has periodic time \(1.8\) s. The periodic time of simple pendulum of length \((l_1-l_2)\) is nearly

• A. \(1.2\) s
• B. \(1.4\) s
• C. \(1.6\) s
• D. \(1.8\) s

Hint:

T squared is proportional to length, so lengths subtract in T squared.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a simple pendulum \(T = 2\pi\sqrt{\frac lg}\), so \(l = \frac{gT^2}{4\pi^2}\), meaning \(l \propto T^2\).

Step 2: Subtract lengths:
\(l_1 - l_2 \propto T_1^2 - T_2^2\). So the period \(T\) of the pendulum with length \(l_1 - l_2\) satisfies
\[ T^2 = T_1^2 - T_2^2 = (2.4)^2 - (1.8)^2 = 5.76 - 3.24 = 2.52 \]

Step 3: Take the root:
\(T = \sqrt{2.52} \approx 1.59\) s, which is nearly \(1.6\) s.

Step 4: Why the other options are wrong.
1.2 s is \(2.4 - 1.8\) taken directly as a difference in periods, which is wrong because T is not linear in length. 1.4 s and 1.8 s do not equal \(\sqrt{2.52}\).

Final Answer:
The period is nearly \(1.6\) s, option (C). \[ \boxed{1.6\text{ s}} \]