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 $ta{n}^{-1}\left(\frac{a}{g}\right)$ for any value of $a\le g.$
• D. angular amplitude of oscillation will be $ta{n}^{-1}\left(\frac{a}{g}\right)$ only when $a<<g.$

Answer 1

Answer: (B), (C)

The correct options are
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 $ta{n}^{-1}\left(\frac{a}{g}\right)$ for any value of $a\le g.$

Time period and angular amplitude is given as per the theory. Hence, B & C are correct.