Question

A 0.34-kg mass at the end of a spring oscillates 2.3 times per second with an amplitude of 0.20 m.

Determine the magnitude of the velocity when it passes the equilibrium point.

Determine the magnitude of the velocity when it is 0.10m from equilibrium.

Determine the total energy of the system.

Answer 1

Mass of the object = m = 0.34 kg

Frequency of the oscillation = f = 2.3 Hz

Amplitude of oscillation = A = 0.2 m

Angular frequency of oscillation =

= 2f

= 2(2.3)

= 14.45 rad/s

Velocity when it passes the equilibrium point = V₁

Magnitude of velocity in a SHM can be given by,

where X is the displacement of the object from the equilibrium point.

Displacement of the object at equilibrium point = X₁ = 0

V₁ = 2.89 m/s

Velocity when it is 0.1 m from the equilibrium point = V₂

Displacement of the object 0.1 m from the equilibrium point = X₂ = 0.1 m

V₂ = 2.502 m/s

Total energy of the system = E

The total energy of the system is equal to the maximum kinetic energy of the mass.

Maximum kinetic energy of the mass occurs at maximum velocity which is at the equilibrium point.

E = mV₁²/2

E = (0.34)(2.89)²/2

E = 1.42 J

a) Magnitude of velocity when it passes the equilibrium point = 2.89 m/s

b) Magnitude of velocity when it is 0.1 m from equilibrium = 2.502 m/s

c) Total energy of the system = 1.42 J