Physics - SHM Question with Solution | TestHub

The distance covered by the mass falling from height ' $h$ ' during its time of fall ' $t$ ' is given by $$\begin{array}{r}\mathrm{s}=\mathrm{h}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2\\ \text{ As, }\mathrm{u}=0\Rightarrow \mathrm{h}=\frac{1}{2}{\mathrm{gt}}^2\Rightarrow \mathrm{t}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}\end{array}$$ The time period of simple pendulum is $$\mathrm{T}=2\pi \sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$$ where, lis the length of the pendulum. From Eq. (i) and (ii), since ' $h$ ' and ' $l$ ' are constant so, we can conclude that, $$\mathrm{t}\propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}}\text{ and }\mathrm{T}\propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}}\therefore \frac{\mathrm{t}}{\mathrm{T}}=\mathrm{l}$$ Thus, the ratio of time of fall and time period of pendulum is independent of value of gravity $(\mathrm{g})$ or any other parameter like mass and radius of the planet. Thus, the relation between ${\mathrm{t}}^{\prime }$ and ${\mathrm{T}}^{\prime }$ on another planet irrespective of its mass or radius will remains same as it was on earth i.e. $${\mathrm{t}}^{\prime }=2{\mathrm{T}}^{\prime }$$