Two bodies of mass m_(1)andm_(2) are suspended from a weightless spring. The force constant of the spring is k. When the bodies are in equilibrium position, the body of mass m_(1) is taken away from the system such that the equilibrium condition of the system is not disturbed at that very moment. Determine the angular frequency and the amplitude of motion for the body of mass m_(2).

Two bodies of mass m_(1)andm_(2) are suspended from a weightless sprin

1.	Two bodies of mass m_(1)andm_(2) are suspended from a weightless spring. The force constant of the spring is k. When the bodies are in equilibrium position, the body of mass m_(1) is taken away from the system such that the equilibrium condition of the system is not disturbed at that very moment. Determine the angular frequency and the amplitude of motion for the body of mass m_(2).
Answer» Solution :If increase in length of the spring DUE to the two masses `m_(1)andm_(2)` is l, then `k=((m_(1)+m_(2))g)/lor,l=((m_(1)+m_(2))g)/k` SIMILARLY for the mass `m_(2)`, increase in length, `l_(2)=(m_(2)g)/k`. This `l_(2)` is the increase in length for the final equilibrium position. So with the mass `m_(1)`, displacement from the equilibrium position = amplitude of motion = `l-l_(2)=((m_(1)+m_(2))g)/k-(m_(2)g)/k=(m_(1)g)/k` Since only the mass `m_(2)` vibrates, the angular FREQUENCY `omega` is GIVEN by, `omega^(2)=k/m_(2)or,omega=sqrt(k/m_(2))`.