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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.For a lift at rest, $g_{eff} = g$, so $T_1 = 2\pi \sqrt{\frac

Vedclass · Accepted Answer

$T_1 = \frac{\sqrt{2}T_2T_3}{\sqrt{T_2^2 + T_3^2}}$

Vedclass · Answer

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.For a lift at rest, $g_{eff} = g$, so $T_1 = 2\pi \sqrt{\frac{l}{g}}$.When the lift moves up with acceleration $a$, $g_{eff} = g + a$, so $T_2 = 2\pi \sqrt{\frac{l}{g+a}}$.When the lift moves down with acceleration $a$, $g_{eff} = g - a$, so $T_3 = 2\pi \sqrt{\frac{l}{g-a}}$.Squaring the equations, we get $T_2^2 = 4\pi^2 \frac{l}{g+a}$ and $T_3^2 = 4\pi^2 \frac{l}{g-a}$.Taking the reciprocals: $\frac{1}{T_2^2} = \frac{g+a}{4\pi^2 l}$ and $\frac{1}{T_3^2} = \frac{g-a}{4\pi^2 l}$.Adding these: $\frac{1}{T_2^2} + \frac{1}{T_3^2} = \frac{g+a+g-a}{4\pi^2 l} = \frac{2g}{4\pi^2 l} = \frac{2}{T_1^2}$.Thus, $\frac{T_3^2 + T_2^2}{T_2^2 T_3^2} = \frac{2}{T_1^2}$, which implies $T_1^2 = \frac{2T_2^2 T_3^2}{T_2^2 + T_3^2}$.Taking the square root, we get $T_1 = \frac{\sqrt{2}T_2T_3}{\sqrt{T_2^2 + T_3^2}}$.

An oscillating pendulum suspended from the roof of a lift which is at rest has time period $T_1$. When the lift moves up with acceleration $a$, its time period is $T_2$. When the lift moves down with acceleration $a$, its time period is $T_3$. The relation between $T_1$, $T_2$, and $T_3$ is:

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