A body of mass `m=0.50 kg `is suspended from a rubber cord with elasticity coefficient `k=50N//M`. Find the maximum distance over which the body can be pulled down for the body's oscillations to remain harmonic. What is the energy of oscillation in this case ?

Explanation

There is an important difference between a rubber cord or steel coire and a spring. A spring can be ppulled or compressed in both cases, obey's Hooke's law. But a rubber cord becomes loose when one tries to compress it and does not then obey Hooke's law. Thus if we suspend a body by a rubber cord it stretches by a distance `m//k` in reaching the equilibrium configureation. If we further strech it by a distance `Deltah` it will execute harmonic oscillations when released if `Deltahle mg//k` because only in this case will the cord remain taut and obey Hooke's law.
Thus `Deltah_(max)=mg//k`
The energy of oscillation in this case is
`(1)/(2)k (Deltah_(max))^(2)=(1)/(2)(m^(2)g^(2))/(k)`