Question

A pendulum is made of a massless string of length \( L \) and a small bob of negligible size and mass \( m \). It is released making an angle \( \theta_0 \) (<< 1 rad) from the vertical. When passing through the vertical, the string slips a bit from the pivot so that its length increases by a small amount \( \delta \) (\( \delta \ll L \)) in negligible time. If it swings up to angle \( \theta_1 \) on the other side before starting to swing back, then to a good approximation, which of the following expressions is correct? 

• A. \( \theta_1 = \theta_0 \)
• B. \( \theta_1 = \theta_0 \left( 1 - \frac{\delta}{L} \right) \)
• C. \( \theta_1 = \theta_0 \left( 1 - \frac{\delta}{L} \right)^2 \)
• D. \( \theta_1 = \theta_0 \left( 1 - \frac{3 \delta}{2L} \right) \)

Hint:

Small changes in the length of a pendulum affect the amplitude of its swing in a non-linear fashion, with a correction factor that is proportional to \( \frac{\delta}{L} \).

Answer 1

The Correct Option is D

Step 1: Understanding the change in length.
When the length of the pendulum increases by a small amount \( \delta \), the restoring force changes, and the angle \( \theta_1 \) reached after the swing will depend on this change. To a good approximation, the new angle is related to the initial angle by \( \theta_1 = \theta_0 \left( 1 - \frac{3 \delta}{2L} \right) \), accounting for the impact of both the length increase and the reduced amplitude of the swing.
Step 2: Conclusion.
Therefore, the correct answer is option (D).