Question

In the figure shown, a lift goes downward with a constant retardation. An observer in the lift observes a simple pendulum with time period $2s$. The length of the string which is used to suspend the ball from the ceiling is $2.0m$. If the retardation of the lift is $\alpha m/s^2$ then find the value of $\frac{\alpha }{2}$.
[use: ${\pi }^2=10$ and $g=10m/s^2$ ]

Answer 1

Answer: 5

The lift goes down with retardation means acceleration is upward, let it be $\alpha$.
$T=2\pi \sqrt{\frac{h}{g_{\text{eff }}}}=2\pi \sqrt{\frac{h}{g+\alpha }}$
$\Rightarrow 2=2\pi \sqrt{\frac{2}{10+\alpha }}$
$\Rightarrow \alpha =10$