Physics - SHM Question with Solution | TestHub

Time period of a simple pendulum is $\mathrm{T}=2\pi \sqrt{\left(\frac{l}{\mathrm{a}}\right)}$ In stationary lift, the value of acceleration is $\mathrm{a}=\mathrm{g}$ $$\mathrm{T}=2\pi \sqrt{\left(\frac{l}{\mathrm{g}}\right)}$$ When the lift is accelerating in an upward direction, there is a pseudo force acting in a downward direction. $$\therefore \mathrm{ma}=\mathrm{mg}+\frac{\mathrm{mg}}{3}$$ $\therefore \mathrm{a}=\mathrm{g}+\frac{\mathrm{g}}{3}$ $\therefore a=\frac{4g}{3}$ $\therefore$ Period for a pendulum in accelerating lift is $$\begin{array}{rl}{\mathrm{T}}^{\prime }& =2\pi \sqrt{\frac{l}{\mathrm{a}}}=2\pi \sqrt{\frac{3l}{4\mathrm{g}}}\\ \therefore {\mathrm{T}}^{\prime }& =\frac{\sqrt{3}}{2}\left(2\pi \sqrt{\frac{l}{\mathrm{g}}}\right)\\ \therefore {\mathrm{T}}^{\prime }& =\frac{\sqrt{3}}{2}\mathrm{T}\end{array}$$