Question

The period of oscillation of a simple pendulum is $T$ in a stationary lift. If the lift moves upwards with acceleration of $8g$, the period will:

• A. remain the same
• B. decreases by $\frac{T}{2}$
• C. increase by $\frac{T}{3}$
• D. none of the above

Answer 1

Answer: (C)

When lift moves upwards net acceleration of lift increases.
When lift moves upwards that net force on it is in the upward direction.
Therefore, from Newton's second law, the net force
$F-mg=ma$
$\Rightarrow F=m(g+a)$
Given, $a=8g$
$\therefore g^{\prime }=g+8g=9g$
Also time period $(T)=2\pi \sqrt{\frac{l}{g}}$
$\Rightarrow T\propto \frac{1}{\sqrt{g}}$
Since $T^{2}g=$ constant
$\therefore T_1^{2}g=T_2^2\times 9g$
$\Rightarrow T_2=\frac{T_1}{3}=\frac{T}{3}$