Question

If the inertial mass $m_i$ of the bob of a simple pendulum of length ${;}^{\prime }{l}^{\prime }$ is not equal to the gravitational mass $m_g$ then its time period is:

• A. $T=2\pi \sqrt{\frac{m_{i}l}{m_g.g}}$
• B. $T=2\pi \sqrt{\frac{m_g.l}{m_i.g}}$
• C. $T=2\pi \sqrt{\frac{l}{g}}$
• D. $T=2\pi \sqrt{\frac{(m_i+m_g)}{(m_i-m_g)}.\frac{l}{g}}$

Answer 1

The correct option is A $T=2\pi \sqrt{\frac{m_{i}l}{m_g.g}}$

Inertial mass is defined by the Newton’s Law of motion and its given as,

$F=ma$whereas, the Gravitational mass is defined as the force of Gravitation which is given as $F=\frac{GmM}{r^2}$

Restoring force is directly proportional to displacement

$F=-mg\theta$

$\Rightarrow \theta =\frac{s}{l}$

There is no physical difference found between the two masses gravitational and inertial hence the time period if the pendulum will remain same in the two cases but since its mentioned that the two masses are different the time period will be

Time period is $T=2\pi \sqrt{\frac{m_i}{k}}$

$k=\frac{m_{g}g}{l}$

By substituting in the above quation we get

$T=2\pi \sqrt{\frac{m_{i}l}{m_{g}g}}$