Question

A pendulum bob is hanging from the roof of an elevator with the help of a light string. When the elevator moves up with a uniform acceleration ${}^{\prime }{a}^{\prime }$, the tension in the string is $T_1$. When the elevator moves down with the same acceleration, the tension in the string is $T_2$. If the elevator was stationary, the tension in the string would be

• A. $\frac{T_1+T_2}{2}$
• B. $\sqrt{T_1{T}_2}$
• C. $\frac{T_1{T}_2}{T_1+T_2}$
• D. $\frac{2T_1{T}_2}{T_1+T_2}$

Answer 1

The correct option is A $\frac{T_1+T_2}{2}$

For case-1, $ma=T_1-mg$
$T_1=ma+mg.....\left(1\right)$
For case-2, $ma=mg-T_2$
$T_2=mg-ma......$(2)
when elevator is stationary, then only gravitational force of bob is the tension of the string, i.e $T=mg$
$\left(1\right)+\left(2\right)$, we get

$mg=\frac{T_1+T_2}{2}$
Therefore, $T=\frac{T_1+T_2}{2}$