Question

The spring shown in the 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 1

Amplitude will be equal to the position when the forces of spring and external forces are equal.

Let the amplitude be A

kA = F

A = F/k A = Amplitude

The time period is equal to $2\pi (Mk{)}^{1/2}$

The total energy at any instant is equal to$\left(k{A}^2\right)/2$

The energy stored in the block is the potential energy and is equal to$\left(k{x}^2\right)/2$

where x is the displacement from the mean position

At equilibrium x = 0 so

Ans (b) = 0

The kinetic energy is the total energy and is equal to $\left(k{A}^2\right)/2$

where A is F/k

Therefore kinetic energy is equal to $\left(k{F}^2\right)/(k^2\times 2)$