A simple pendulum has a length ( $I$, of $1 \mathrm{~m}$. In free vibration the amplitude of its swings falls off by a factor $e$ in 50 swings. The pendulum is set into forced vibration by moving its point of suspension horizontally in SHM with an amplitude of $1 \mathrm{~mm}$.
(a) Show that if the horizontal displacement of the pendulum bob is $x$, and the horizontal displacement of the support is $\xi$, the equation of motion of the bob for small oscillations is
$$
\frac{d^{2} x}{d t^{2}}+\gamma \frac{d x}{d t}+\frac{g}{l} x=\frac{g}{l} \xi
$$
Solve this equation for steady-state motion, if $\xi=\xi_{0} \cos \omega t .$ (Put $\left.\omega_{0}^{2}=g / l .\right)$
(b) At exact resonance, what is the amplitude of the motion of the pendulum bob? (First, use the given information to find $Q .$ )
(c) At what angular frequencies is the amplitude half of its resonant value?