Question

A pendulum suspended from the roof of an elevator at rest has a time period $T_1$, when the elevator moves up with an acceleration a its time period becomes $T_2$, when the elevator moves down with an acceleration a, its time period becomes $T_1$, then.

• A. $T_1=\sqrt{\left(T_2{T}_3\right)}$
• B. $T_1=\sqrt{(T_2^2+T_3^2)}$
• C. $T_1=\frac{T_2{T}_3\sqrt{2}}{\sqrt{T_2^2+T_3^2}}$
• D. None of these

Answer 1

Answer: (C)

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

$T_1=2\pi \sqrt{\frac{l}{g+a}}$ or $\frac{4{\pi }^{2}l}{T_1^2}=g$
$T_2=2\pi \sqrt{\frac{l}{g+a}}$ or $\frac{4{\pi }^{2}l}{T_2^2}=g+a$
$T_3=2\pi \sqrt{\frac{l}{g-a}}$ or $\frac{4{\pi }^{2}l}{T_3^2}=g-a$
$\frac{4{\pi }^{2}l}{T_2^2}+\frac{4{\pi }^{2}l}{T_3^2}=2g$
$\therefore$ where $g=\frac{4{\pi }^{2}l}{T_1^2}$
Solving we get, $T_1=\frac{\sqrt{2}{T}_2{T}_3}{\sqrt{T_2^2+T_3^2}}$.