A block of mass `m` is suspended by a rubber cord of natural length `l=(mg)/(k)`, where k is force constant of the cord. The block is lifted upwards so that the cord becomes just tight and then block is released suddently. Which of the following will not be true?

Options

A. Block performs periodic motion with amplitude greater than l.
B. Block performs SHM with amplitude equal to l.
C. Blocks will never return to the position from where it was released.
D. Angular frequency `omega` is equal to `1(rad)/(s)`

Answer

A, C, D

Explanation

When the block is released suddenly, it starts to move down. During its downwards motion the rubber coed elongates. Hece, a tension is developed in it but the blocks continues to accelerate downwards till tension becomes equal to weight `mg` of the block. After this moment, the block continues to move down due to its velocity and rubber cord further elongates. Therefore tension becomes greater than the weight, hence, the block now retards and comes to an instantaneous rest. At lowest position of the block, strain energy in the cord equals loss of potential energy of the block. Suppose the block comes to an instantaneous rest when elongation of the rubber cord is equl to y. Then
`(1)/(2)ky^2=mgyimpliesy=(2mg)/(k)` and 0
Hence block will be instantaneously at rest, at `y=0` and at `y=(2mg)/(k)`. If fact, the block oscillates between these two values. Since the rubber cord is elastic, tension in it is directily proportional to elongation therefore, the block will perform SHM. Its amplitude will be equal to half of the distance between these extreme positions of the block or amplitude is
`(1)/(2)xx(2mg)/(k)=(mg)/(k)=l`
Hence option (b) is correct.
The angular frequency of its SHM will be equal to
`omega=sqrt((k)/(m))`
Since k and m are not given in the question, it cannot be calculated. Hence option (d) is not correct.