Question

A bob of simple pendulum is suspended by a metallic wire. If the length of simple pendulum is increased four times of its initial value. Find the time period with respect to its previous value.

• A. $\text{not be different}$
• B. $\text{be halved}$
• C. $\text{be}\sqrt{2}\text{times}$
• D. $\text{becomes twice}$

Answer 1

The correct option is D $\text{becomes twice}$

Given,
Initial length $=l$
Final length $\left(l^{\prime }\right)=4l$
Initial time period $\left(T\right)=2\pi \sqrt{\frac{l}{g}}$
Final time period $\left(T^{\prime }\right)=2\pi \sqrt{\frac{l^{\prime }}{g}}$
$\Rightarrow T^{\prime }=2\pi \sqrt{\frac{4l}{g}}$
$\Rightarrow T^{\prime }=2\left(2\pi \sqrt{\frac{l}{g}}\right)$
$\therefore T^{\prime }=2T$
Hence final time becomes twice of initial value