The spring shown in figure is unstretched when a man starts pulling on the cord. The mass of the block is `M`. If the man exerts a constant force `F`, find

(a) the amplitude and the time period of the motion of the block,
(b) the energy stored in the spring when the block passes through the equilibrium position and
(c) the kinetic energy of the block at this position.

Answer

`(F^2/k)`
`1/2F^2/k`

Explanation

`a`. We hasve `f=kx`
`rarr x=F/k`
Acceleration `=F/m`
Time period `T=2pisqrt`(displacement)/Acceleration)
`=2pi (sqrt(F/k)/(F/m))=2sqrt(m/k)`
Amplitude =max displacement
`=F/k`
The energy stored in the springh when the block passes through the equilibrium position
`=1/2kx^2=1/2(F/k)^2`
`=1/2k(F^2/k^2)=(F^2/k)`
b. At the mean position P.E. is O.K.E. is
` 1/2kx^2=1/2F^2/k`