Q. A simple pendulum has a length $l.$ The inertial and gravitational masses of the bob are $m_{i\ }and\ m_{g}$ respectively. Then the time period T is given by
A) $T = 2\pi\sqrt{\frac{m_{g}l}{m_{i}g}}$
B) $T = 2\pi\sqrt{\frac{m_{i}l}{m_{g}g}}$
C) $T = 2\pi\sqrt{\frac{m_{i} \times m_{g} \times l}{g}}$
D) $T = 2\pi\sqrt{\frac{l}{m_{i} \times m_{g} \times g}}$
Explanation
(b) Torque acting on the bob
$= I\alpha = - (mg)l\sin\theta\ $
Or $\left( m_{i}l^{2} \right)\alpha = - \left( m_{g}g \right)l\theta$
Or $\alpha = - \left( \frac{m_{g}g}{m_{i}l} \right)\theta = - \omega^{2}\theta$;
Where $\omega^{2} = \frac{m_{g}g}{m_{i}l}$
$\therefore\ T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m_{i}l}{m_{g}g}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.