Thank you for reporting, we will resolve it shortly
Q.
The length of a seconds pendulum on the surface of earth is $1 \,m$. Its length on the surface of the moon is
Oscillations
Solution:
Time period of seconds pendulum is $2 \,s$
$\therefore 2=2\pi \sqrt{\frac{L}{g}}$
On earth, $L=1 \,m$, $g=g_{e}$
$\therefore 2=2\pi \sqrt{\frac{1}{g_{e}}}$ $\ldots\left(i\right)$
On moon, $L=L', g=\frac{g_{e}}{6}$
$\therefore 2=2\pi \sqrt{\frac{L'}{g_{e} /6}}=2\pi \sqrt{\frac{6L'}{ge}} \ldots\left(ii\right)$
From(i) and (ii), we get
$6L'=1$ or $L' =\frac{1}{6}\, m$