Question

A mass m is hung on an ideal massless spring another equal mass is connected to the other end of the spring the whole system is at rest at t=0 m is released and the system falls freely under gravity assume that natural length of the spring is `L_(0)` its initial stretched length is L and the acceleration due to gravity is g what is distance between masses as function of time

A. `L_(0)+(L-L_(0))cos sqrt((2k)/m)t`
B. `L_(0)cos sqrt((2k)/m)t`
C. `L_(0)sinsqrt((2k)/m)t`
D. `L_(0)+(L-L_(0))sin sqrt((2k)/m)t`

Answer 1

Correct Answer - A
In CM frame both the masses execute SHM with
`omega=sqrt(k/mu)=sqrt ((2k)/m)`
initially particles are at extreme
`distance = L_(0)+(L-L_(0))cos sqrt((2k)/m)t`