Question

The length of a simple pendulum executing simple harmonic motion is increased by $10\%$. The percentage increase in the time period of the pendulum due to increased length will be:

• A. $10\%$
• B. $5\%$
• C. $15\%$
• D. $20\%$

Answer 1

The correct option is B $5\%$

Let the length of simple pendulum be ${}^{\prime }{l}^{\prime }$.
Let the time period of oscillation of simple pendulum be ${}^{\prime }{T}^{\prime }$.
$T=2\pi \sqrt{\frac{l}{g}}$

$\because T\propto l^{\frac{1}{2}}$
$\Rightarrow \frac{\Delta T}{T}=\frac{1}{2}\left(\frac{\Delta l}{l}\right)...\left(i\right)$
Since fractional change in length of pendulum is,
$\frac{\Delta l}{l}=\frac{10}{100}=0.1$
substituting in Eq.$\left(i\right)$
$\Rightarrow \frac{\Delta T}{T}=0.05$
Hence $\%$ change in time period of pendulum is.
$\frac{\Delta T}{T}\times 100=5\%$
The percentage increase in the time period of the pendulum will be $5\%$