A string of a pendulum of length $l$ is displaced through $90^{\circ}$ from its vertical and released. Then,…
A string of a pendulum of length $l$ is displaced through $90^{\circ}$ from its vertical and released. Then, the minimum streng th of the string needed to withstand the tension as, the pendulum passes through its mean position is
$m g$
$3 m g$
$5 \mathrm{mg}$
$6 m g$
Solution
Length of penduIum $=l$
Centre of gravity of pendulum's string $=\frac{l}{2}$
When pendulum is displaced through $90^{\circ}$ and released, then according to law of conservation of energy.
Gravitational potential energy $=$ Kinetic energy
$
\begin{aligned}
\Rightarrow & & m g \cdot \frac{l}{2} & =\frac{1}{2} m v^2 \Rightarrow v^2=g l \\
\Rightarrow & v & =\sqrt{g l} &
\end{aligned}
$
At the mean position, tension in the string is balanced by the weight as weIl as the centrifugal force.
Hence, $T=m g+\frac{m v^2}{l / 2}$
$
\left[\because r=\frac{l}{2}\right]
$
$
\begin{aligned}
& =m g+\frac{2 m}{l} \cdot v^2 \\
& =m g+\frac{2 m}{l} \cdot(\sqrt{g} l)^2 \quad \quad \quad \text { from Eq. } \\
& =m g+\frac{2 m g l}{l}=m g+2 m g=3 m g
\end{aligned}
$
[from Eq. (i)]