A block of mass m, attacted to a string of spring constant k, oscillates on a smooth horizontal table. The other end of the spring is fixed to a wall. The block has a speed v when the spring is at its natural length. Before coming to an instantaneous rest. If the block moves a distance x from the mean position, then

Options

A. `x=sqrt((mv)/(K))`
B. `x=vsqrt((m)/(K))`
C. `x=(1)/(v)sqrt((m)/(K))`
D. `x=sqrt((m)/(K))`

Answer

B

Explanation

In this case, the spring is oscillating horizontally on a horizontal table. Hence the weight of the block is balanced by the normal reaction R. The free end of its natural length is the position of equilibrium of the block. It is treated as a SHM particle.
When it is at its mean position, its energy `=(1)/(2)mv^(2)` and at a dis"tan"ce x from the mean position, it stops or it is at the extreme position. Its total energy
`=P.E. =(1)/(2)Kx^(2)=P.E.` of the spring.
`therefore` By the principle of conservation of energy.
`(1)/(2)mv^(2)=(1)/(2)Kx^(2) therefore x^(2)=(mv^(2))/(K) therefore x= v sqrt((m)/(K))`