Question

A spring mass system is hanging from the ceiling of an elevactor which is at rest. The elevator starts accelerating upwards with acceleration a. Its motion is observed with respect to the elevator

• A. Amplitude of oscillation is $\frac{ma}{2k}$
• B. Amplitude of oscillation is $\frac{m(g+a)}{k}$
• C. Time period of oscillation is $2\pi \sqrt{\frac{m}{k}}$
• D. Time period of oscillation is $2\pi \sqrt{\frac{2m}{k}}$

Answer 1

The correct option is C Time period of oscillation is $2\pi \sqrt{\frac{m}{k}}$

When elevator is at rest, mg = $k{x}_0-\left(1\right)$ Apply pseudo of ma. The mean position shifts downwards such that the distance between initial and new equilibrium positions is amplitude of oscillation A.
$k(x_0+A)=mg+ma\therefore A=\frac{ma}{k}--\left(2\right)$
Net restoring force when the block is x units below the mean position = - k $(x_0+A+x)$ + mg + ma = - kx
$\therefore -kx=ma=-m{\omega }^{2}x\therefore \omega =\sqrt{\frac{k}{m}}\therefore T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}$.