Two identical bodies, each of mass m, are connected by a spring having spring constant k and they are placed on a frictionless floor. The spring is compressed a little and then released. What will be the frequency of oscillation of the system?

Explanation

If the maximum compression of the spring from its position of equlilbrium is A, then restoring force = -kA. In this condition, the whole energy of the spring is its potential energy = `1/2kA^(2)`.
Again, during oscillation, when the two bodies just cross the position of equilibrium, the potential energy becomes zero and the total energy is then equal to the kinetic energy of the two bodies. At this stage, velocity of each body = maximum velocity = `omegaA/2`, where `omega` = angular frequency and `A/2` = amplitude of vibration of each body.
`therefore" "` Kinetic energy of the two bodies
`=1/2momega^(2)(A/2)^(2)+1/2momega^(2)(A/2)^(2)=1/4momega^(2)A^(2)`.
According to the principle of conservation of energy,
`1/2kA^(2)=1/4momega^(2)A^(2)or,omega=sqrt((2k)/m)`.
`therefore` Frequency of oscillation of the system = `omega/(2pi)=1/(2pi)sqrt((2k)/m)`.