Physics - SHM Question with Solution | TestHub

When the mass is displaced slightly out of the plane containing $P$, $Q$, and $R$, it will oscillate about the horizontal axis passing through $P$ and $Q$.

The system behaves as a simple pendulum whose effective length $l_{eff}$ is the perpendicular distance from the mass at $R$ to the axis $PQ$.

Let $O$ be the midpoint of $PQ$. Since $PR=QR=L$, the line $RO$ is perpendicular to $PQ$.

The distance $PO=\frac{PQ}{2}=\frac{2d}{2}=d$.

In the right-angled triangle $POR$, using Pythagoras theorem:

$RO=\sqrt{P{R}^2-P{O}^2}$

$RO=\sqrt{L^2-d^2}=(L^2-d^2{)}^{1/2}$

This distance $RO$ is the effective length of the pendulum, so $l_{eff}=(L^2-d^2{)}^{1/2}$.

The time period of a simple pendulum is given by:

$T=2\pi \sqrt{\frac{l_{eff}}{g}}$

Substituting the value of $l_{eff}$, we get:

$T=2\pi \sqrt{\frac{(L^2-d^2{)}^{1/2}}{g}}$

Answer: $2\pi \sqrt{\frac{{\left(L^2-d^2\right)}^{1/2}}{g}}$