Question

The period of a simple pendulum is found to increase by $50\%$ when the length of the pendulum is increased by $0.6$m. The initial length at a place
where $g=9.8$m/s${}^2$ is:

• A. $0.16m$
• B. $0.32m$
• C. $0.48m$
• D. $0.60m$

Answer 1

Answer: (C)

$0.48m$

If $T_1$ is initial time period and $T_2$ is the final time period, then

$T_1=T$
$T_2=\frac{3}{2}T$

If $L_1$ is initial length and $L_2$ is the final length, then
$L_1=l$
$L_2=(l+0.6)$
$T=2\pi \sqrt{\frac{e}{g}}$
$T\alpha \sqrt{l}$
$\therefore \frac{T_1}{\sqrt{l_1}}=\frac{3}{2\sqrt{l+0.6}}$

$4(l+0.6)=9l$

$5l=2.4$

$l=0.48m$