Question

A simple pendulum is executing simple harmonic motion with a time period T; If the length of the pendulum. Is increased by 21%, the Increase in the time period of the pendulum of Increased length is

Answer 1

We know that,

The time period is

$T_1=2\pi \sqrt{\frac{l}{g}}$

Now, length is increased by 21%

So, new length is

$l=l+l\frac{21}{100}=\frac{121l}{100}$

Now put the value of l in equation (I)

$T_2=2\pi \sqrt{\frac{\frac{121l}{100}}{g}}$

$T_2=2\pi \sqrt{\frac{121l}{100g}}$

$T_2=\frac{11}{10}\times 2\pi \sqrt{\frac{l}{g}}$

$T_2=\frac{11}{10}{T}_1$

Now, change in time period is

$\Delta T=T_2-T_1$

$\Delta T=\frac{11}{10}{T}_1-T_1$

$\Delta T=\frac{1}{10}{T}_1$

Now,

Percentage change

$=\frac{\Delta T}{T}\times 100$

$=\frac{\frac{T_1}{10}}{T_1}\times 100$

$=10$ %

Hence, this is the required solution