Question

A 0.40 kg ball is suspended from a spring with spring constant 12 n/m . part a if the ball is pulled down 0.20 m from the equilibrium position and released, what is its maximum speed while it oscillates?

Answer 1

The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:

`E=K+U= (1)/(2)mv^2+ (1)/(2)kx^2`
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.

Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.

When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:

`E=U_(max)= (1)/(2)k(x_(max))^2`

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):

`E=K_(max)= (1)/(2)m(v_(max))^2`

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:

`U_(max)=K_(max)`

`(1)/(2)k(x_(max))^2 = (1)/(2)m(v_(max))^2`

`v_(max)= \sqrt{ (k x_(max)^2)/(m) }= \sqrt{ ((12 N/m)(0.20 m)^2)/(0.4 kg) }=1.1 m/s`