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

Step-by-step solution

We know that time period of a spring mass system, T _ l =2 m k where, m= mass of body and k= force constant of the spring Time period of simple pendulum, r ₂=2 l g According to the question, initially time period of a spring mass system, T₁= time period of simple pendulum, T ₂ aligned 2 m k & =2 l g m k & = l 10 aligned (i) [ g=10 ~m / s ^2 ] When both of them are put in an elevator going downwards with an acceleration 5 ~m / s ^2 , then no effect occurs on the time period of spring mass system. i.e., T₁^ =T₁=2 m k