Physics - SHM Question with Solution | TestHub

The time period ($T$) of a simple pendulum length$l$, is given by

$T=2\pi \sqrt{\frac{l}{g}}=\frac{1}{\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\left(n\right)}$

where$\mathrm{g}$is acceleration due to gravity.

$\therefore$$\frac{n_1}{n_2}=\sqrt{\frac{l_2}{l_1}}$

$\Rightarrow$$\frac{l_1}{l_2}={\left(\frac{n_2}{n_1}\right)}^2$

Given,$\frac{n_2}{n_1}=\frac{3}{2}$

$\therefore$$\frac{l_1}{l_2}={\left(\frac{3}{2}\right)}^2=\frac{9}{4}$