Question

Two simple pendulums of lengths \(L_1\) and \(L_2\) have periodic time \(T_1\) and \(T_2\) respectively \((T_1 > T_2)\). The time period of the pendulum of length \((L_1-L_2)\) is
\([(L_1-L_2) > 60\text{ cm}]\)

• A. \(\sqrt{T_1^2+T_2^2}\)
• B. \(\sqrt{T_1^2-T_2^2}\)
• C. \(T_1+T_2\)
• D. \(T_1-T_2\)

Hint:

Use \(T^2\propto L\) and subtract.

Answer 1

The Correct Option is B

Step 1: Square-law relation
$T^2$ is proportional to length, so $T^2$ values subtract just like lengths.

Step 2: Apply
The pendulum of length $L_1-L_2$ has $T^2=T_1^2-T_2^2$, option (B).

Final Answer:
$T=\sqrt{T_1^2-T_2^2}$, option (B). \[ \boxed{\sqrt{T_1^2-T_2^2}} \]