Question

Simple pendulum is executing simple harmonic motion with time period $T$. If the length of the pendulum is increased by $21\%$, then the increase in the time period of the pendulum of the increased length is :

• A. $22\%$
• B. $13\%$
• C. $50\%$
• D. $10\%$

Answer 1

The correct option is D $10\%$

Time period, $T=2\pi \sqrt{\frac{l}{g}}\Rightarrow T^2=4{\pi }^2\frac{l}{g}$

Taking log on both sides,

$2log$ $T=log4{\pi }^2+logl-logg$

Differentiating on both sides,

$\frac{2}{T}dT=0+\frac{1}{l}dl-0$
$\frac{dT}{T}=\frac{1}{2}.\frac{dl}{l}\Rightarrow \frac{dT}{T}\times 100=\frac{1}{2}.\frac{dl}{l}\times 100$

Percent increase in time period
$=\frac{1}{2}\%$ increase in length

$=\frac{1}{2}\times 21=10.5\%$
Percent increase in time period $\cong 10\%$