Question

A vertical spring mass system has the same time-period as simple pendulum undergoing small oscillations.Now both of then are put in a elevator going downwards with an acceleration 5m/s^(2) .The ratio of the-period of the spring mass system to the time-period of the pendilum is (Assume g=10m/s^(2) )

Answer 1

In a elevator going downwards:

$$a'=g-a=10-5=5\frac{m}{s^2}$$

$$T=2\pi\sqrt{\frac{l}{g}},T'=2\pi\sqrt{\frac{l}{a'}}$$

$$\frac{T}{T'}=\sqrt{\frac{a'}{g}}=\sqrt{\frac{5}{10}}=\frac{1}{\sqrt{2}}$$