Physics - SHM Question with Solution | TestHub

The time period of a simple pendulum is given by $T=2\pi \sqrt{\frac{l}{g_{eff}}}$.

When the lift is at rest, the effective acceleration due to gravity is $g$.

$T_1=2\pi \sqrt{\frac{l}{g}}\Rightarrow \frac{1}{T_1^2}=\frac{g}{4{\pi }^{2}l}$

When the lift moves up with acceleration $a$, the effective acceleration due to gravity is $g+a$.

$T_2=2\pi \sqrt{\frac{l}{g+a}}\Rightarrow \frac{1}{T_2^2}=\frac{g+a}{4{\pi }^{2}l}$

When the lift moves down with acceleration $a$, the effective acceleration due to gravity is $g-a$.

$T_3=2\pi \sqrt{\frac{l}{g-a}}\Rightarrow \frac{1}{T_3^2}=\frac{g-a}{4{\pi }^{2}l}$

Adding the equations for $\frac{1}{T_2^2}$ and $\frac{1}{T_3^2}$:

$\frac{1}{T_2^2}+\frac{1}{T_3^2}=\frac{g+a}{4{\pi }^{2}l}+\frac{g-a}{4{\pi }^{2}l}=\frac{2g}{4{\pi }^{2}l}$

Substituting $\frac{g}{4{\pi }^{2}l}=\frac{1}{T_1^2}$:

$\frac{1}{T_2^2}+\frac{1}{T_3^2}=\frac{2}{T_1^2}$

$\frac{T_2^2+T_3^2}{T_2^2{T}_3^2}=\frac{2}{T_1^2}$

$T_1^2=\frac{2T_2^2{T}_3^2}{T_2^2+T_3^2}$

$T_1=\frac{\sqrt{2}{T}_2{T}_3}{\sqrt{T_2^2+T_3^2}}$

Answer: $T_1=\frac{\sqrt{2}{T}_2{T}_3}{\sqrt{T_2^2+T_3^2}}$