Question

Time periods of pendulums \( A \) and \( B \) are \( T \) and \( \frac{5T}{2} \). If they start executing S.H.M. at the same time from the mean position, the phase difference between them after the bigger pendulum has completed one oscillation is

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{2} \)
• C. \( \frac{\pi}{8} \)
• D. \( \frac{\pi}{16} \)
• E. \( \pi \)

Hint:

Phase difference = difference in angular frequencies × time.

Answer 1

Answer: (E)

Concept: \[ \phi = \omega t \]

Step 1: Time taken.
Larger period pendulum: \[ t = \frac{5T}{2} \]

Step 2: Angular frequencies.
\[ \omega_A = \frac{2\pi}{T}, \quad \omega_B = \frac{2\pi}{5T/2} = \frac{4\pi}{5T} \]

Step 3: Phase difference.
\[ \Delta \phi = (\omega_A - \omega_B)t = \left(\frac{2\pi}{T} - \frac{4\pi}{5T}\right)\cdot \frac{5T}{2} \] \[ = \pi \]