Question

A simple pendulum is kept suspended vertically in a stationary bus. The bus starts moving with an acceleration a towards left. As observed inside the bus: (Neglect frictional forces on pendulum and assume size of the ball to be very small.):

• A. Time period of oscillation of the pendulum will be $2\pi \sqrt{\frac{l}{\sqrt{a^2+g^2}}}$ for any value of a
• B. Time period of oscillation of the pendulum will be $2\pi \sqrt{\frac{l}{\sqrt{a^2+g^2}}}$ only when $a<<g$
• C. Angular amplitude of oscillation will be ${\tan }^{-1}\left(\frac{a}{g}\right)$ for any value of $a$
• D. Angular amplitude of oscillation will be ${\tan }^{-1}\left(\frac{a}{g}\right)$ only when $a<<g$

Answer 1

Answer: (B), (C)

Bob will oscillate about equilibrium position
with amplitude $\theta ={\tan }^{-1}\left(\frac{a}{g}\right)$ for any value of $a$.
If $a<<g$, motion will be SHM, and then

time period will be $2\pi \sqrt{\frac{l}{\sqrt{a^2+g^2}}}$.