Answer the following questions:
$(a)$ The time period of a particle in $SHM$ depends on the force constant $k$ and mass $m$ of the particle: $T=2 \pi \sqrt{\frac{m}{k}}$. $A$ simple pendulum executes $SHM$ approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
$(b)$ The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation,a more involved analysis shows that $T$ is greater than $2 \pi \sqrt{\frac{l}{g}}$. Think of a qualitative argument to appreciate this result.
$(c)$ $A$ man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
$(d)$ What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

(N/A) For a simple pendulum,the restoring force is $F = -mg \sin \theta$. For small angles,$\sin \theta \approx \theta$,so $F \approx -mg \theta = -mg (x/l) = -(mg/l)x$. Comparing this with $F = -kx$,we get $k = mg/l$. Substituting this into $T = 2 \pi \sqrt{m/k}$,we get $T = 2 \pi \sqrt{m / (mg/l)} = 2 \pi \sqrt{l/g}$. Thus,$T$ is independent of mass $m$.
$(b)$ For larger angles,$\sin \theta < \theta$. The restoring force $F = -mg \sin \theta$ is smaller than the linear approximation $F = -mg \theta$. $A$ smaller restoring force leads to a slower motion and a longer time period $T$.
$(c)$ Yes. $A$ wristwatch works on the principle of spring-driven mechanical oscillations or quartz crystal vibrations,which are independent of the acceleration due to gravity $g$. Thus,it gives the correct time during free fall.
$(d)$ In a freely falling cabin,the effective acceleration due to gravity $g_{eff} = g - a = g - g = 0$. The time period $T = 2 \pi \sqrt{l/g_{eff}} = \infty$. Therefore,the frequency $f = 1/T = 0$.