Question

A simple pendulum has a length, $l$. The inertial and gravitational masses of the bob are $m_1$ and $m_2$ respectively. Then, the time period, $T$, is given by

• A. $T=2\pi \sqrt{\frac{m_{2}l}{m_{1}g}}$
• B. $T=2\pi \sqrt{\frac{m_{1}l}{m_{2}g}}$
• C. $T=2\pi \sqrt{\left(m_1{m}_2\frac{l}{g}\right)}$
• D. $T=2\pi \sqrt{\left(\frac{l}{m_1}{m}_{2}g\right)}$

Answer 1

The correct option is B $T=2\pi \sqrt{\frac{m_{1}l}{m_{2}g}}$

Here, $T\cos \theta =m_{2}g$
Restoring force, $F=-T\sin \theta$
or $m_{1}a=-\left(\frac{m_{2}g}{\cos \theta }\right)\sin \theta =-m_{2}g\tan \theta$
or $a=-\frac{m_{2}g}{m_1}.\frac{x}{l}$ (for small oscillation $\tan \theta \sim x/l$)
so, $\omega =\sqrt{\frac{m_{2}g}{m_{1}l}}$
Time period $T=2\pi \sqrt{\frac{m_{1}l}{m_{2}g}}$