Question

If $T$ is the time period of a simple pendulum, then at a time $\frac{T}{6}$ the pendulum passes its mean position

• A. Kinetic and potential energies of the pendulum are equal
• B. The displacement of the pendulum is half of its amplitude
• C. Acceleration of the pendulum is half of the maximum acceleration
• D. The velocity of the pendulum is half of its maximum velocity

Hint:

In SHM, the velocity is maximum at the mean position and zero at the extremes, varying as a cosine function of the phase.

Answer 1

The Correct Option is D

The displacement of a simple pendulum is given by \( x = A \sin(\omega t + \phi) \). At the mean position (\( x = 0 \)), the phase \( \omega t + \phi = 0 \) or \( \pi \). The time period \( T = \frac{2\pi}{\omega} \), so \( \omega = \frac{2\pi}{T} \). At \( t = \frac{T}{6} \), the phase is \( \omega t = \frac{2\pi}{T} \times \frac{T}{6} = \frac{\pi}{3} \) (assuming \( \phi = 0 \) at the mean position for simplicity).

- Option 1: At \( t = \frac{T}{4} \), the phase is \( \frac{\pi}{2} \), where kinetic and potential energies are equal (\( x = A \), \( v = 0 \)). At \( \frac{T}{6} \), this is not true.
- Option 2: \( x = A \sin\left(\frac{\pi}{3}\right) = A \frac{\sqrt{3}}{2} \approx 0.866A \), not \( \frac{A}{2} \).
- Option 3: Acceleration \( a = -\omega^2 x = -\omega^2 A \sin\left(\frac{\pi}{3}\right) \). Maximum acceleration is \( \omega^2 A \), so \( a = \omega^2 A \frac{\sqrt{3}}{2} \approx 0.866 \) of the maximum, not half.
- Option 4: Velocity \( v = A\omega \cos\left(\frac{\pi}{3}\right) = A\omega \frac{1}{2} \). Maximum velocity is \( A\omega \), so \( v = \frac{1}{2} \) of the maximum velocity.

Thus, the correct answer appears to be option 4, but since it's not marked, please confirm.