Question

Energy in simple harmonic motion

A 2.90 kg object oscillates with simple harmonic motion on a spring of force constant 600 N/m. The maximum speed is 0.800 m/s.

A) What is the total energy of the object and the spring?

B) What is the maximum amplitude of the oscillation?

Answer 1

Given that

The mass of the object (m) =2.90kg

The spring force constant (k) =600N/m

The maximum speed is(v_max) = 0.800 m/s

a)

The total energy of the object and the spring is given by

TE =(1/2)mv_max²+(1/2)kx²

We know that when the velocity is maximum then the potential energy is zero then total energy is

TE =E =(1/2)mv_max² =0.5*2.90*(0.800)² =0.928J

b)

There will be maximum amplitude , when the kinetic energy is minimum then

The maximum amplitude of the oscillation is given by

TE =E =(1/2)kx²

where x is the amplitude then

E =(1/2)kx²

x =Sqrt(2E/k) =Sqrt(2*0.928J/600N/m) =0.0556m