Question

The bob of a swinging seconds pendulum (one whose time period is $2s$) has a small speed $v_0$ at its lowest point. Its height from this lowest point, $2.25s$ after passing through it is

• A. $\frac{v_0^2}{2g}$
• B. $\frac{v_0^2}{g}$
• C. $\frac{v_0^2}{4g}$
• D. $\frac{9v_0^2}{4g}$

Answer 1

The correct option is C $\frac{v_0^2}{4g}$

The bob will reach its lowest point after $2s$.
As it is traveling further for $\frac{1}{4}$ sec, i.e., $t=2.25s$
Hence $v=A\omega cos\omega t=v_{0}cos\frac{\pi }{4}=\frac{v_0}{\sqrt{2}}$. ($\because$ $A\omega =v_0=v_{max}$).
Applying law of conservation of energy
$\Rightarrow \frac{1}{2}m{v}_0^2=\frac{1}{2}m\frac{v_0^2}{2}+mgh\Rightarrow h=\frac{v_0^2}{4g}$