Question

A spring pendulum performs SHM with a frequency v. If it is taken in a lift slowly accelerating upward, the frequency will

• A. increase
• B. decrease
• C. remains same
• D. becomes zero

Answer 1

The correct option is C remains same

$\text{Given-}\text{A lift slowly accelerating upwards.}$

$\text{formula used}\to [\nu =\frac{1}{2\pi }\cdot \sqrt{\frac{g_{eff}}{l}}]$

$\text{let the acieleration of lift upworrds be}\left(a_0\right)$

$\Rightarrow g_{eff}=g+a_0$

$\begin{array}{c}\text{Since the lift is accelerating very slowly-}\\ \Rightarrow g_{\text{eff}}\approx g\end{array}$

$\Rightarrow \nu =\frac{1}{2\pi }\sqrt{\frac{g}{\ell }}$

$\text{Hence there is no significant change in the frequency (}\nu \text{)}$