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.
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 = Vmax
When the object is at the equilibrium position the total energy of the system is equal to the kinetic energy of the object.
Vmax = 0.688 m/s
a) Amplitude of the motion = 0.0643 m
b) Maximum speed attained by the object = 0.688 m/s
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