Question

An object with mass 2.8 kg is executing simple harmonic motion, attached to a spring with spring constant 320 N/m . When the object is 0.021 m from its equilibrium position, it is moving with a speed of 0.65 m/s . Calculate the amplitude of the motion. Calculate the maximum speed attained by the object.

Answer 1

Mass of the object = m = 2.8 kg

Force constant of the spring = k = 320 N/m

Speed of the object when it 0.021 m from equilibrium position = V = 0.65 m/s

Position of the object = X = 0.021 m

Total energy of the system = E

E = 0.662 J

Amplitude of the motion = A

When the object is at maximum displacement the total energy of the system is equal to the potential energy in the spring.

A = 0.0643 m

Maximum speed of the object = V_max

When the object is at the equilibrium position the total energy of the system is equal to the kinetic energy of the object.

V_max = 0.688 m/s

a) Amplitude of the motion = 0.0643 m

b) Maximum speed attained by the object = 0.688 m/s