NEET UG 2019 — Physics Waves & Oscillations

The distance covered by the mass falling from height ' $h$h ' during its time of fall ' $t$t ' is given by $$\begin{array}{r} \mathrm{s}=\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^2 \\ \text { As, } \mathrm{u}=0 \Rightarrow \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \Rightarrow \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \end{array}$$s=h=ut+21​gt2 As, u=0⇒h=21​gt2⇒t= g2 h​​​ The time period of simple pendulum is $$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$$T=2πgl​​ where, lis the length of the pendulum. From Eq. (i) and (ii), since ' $h$h ' and ' $l$l ' are constant so, we can conclude that, $$\mathrm{t} \propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}} \text { and } \mathrm{T} \propto \frac{\mathrm{l}}{\sqrt{\mathrm{g}}} \quad \therefore \frac{\mathrm{t}}{\mathrm{T}}=\mathrm{l}$$t∝g​l​ and T∝g​l​∴Tt​=l Thus, the ratio of time of fall and time period of pendulum is independent of value of gravity $(\mathrm{g})$(g) or any other parameter like mass and radius of the planet. Thus, the relation between $\mathrm{t}^{\prime}$t′ and $\mathrm{T}^{\prime}$T′ on another planet irrespective of its mass or radius will remains same as it was on earth i.e. $$\mathrm{t}^{\prime}=2 \mathrm{~T}^{\prime}$$t′=2 T′