Question

An object with mass 2.3 kg is executing simple harmonic motion, attached to a spring with spring constant 330 N/m . When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.50 m/s . Part A Calculate the amplitude of the motion. Part B Calculate the maximum speed attained by the object.

Answer #1

Mass of the object = m = 2.3 kg

Spring constant = k = 330 N/m

Amplitude of the motion = A

Maximum speed of the object = V_{max}

Speed of the object at a distance of 0.02 m from equilibrium = V = 0.5 m/s

X = 0.02 m

The total energy of the system remains constant and is equal to the sum of potential energy of the spring and the kinetic energy of the object.

The total energy of the system is equal to the potential energy stored in the spring the object has maximum displacement(amplitude) or the kinetic energy of the object when it is at the equilibrium position.

kA^{2}/2 = kX^{2}/2 + mV^{2}/2

kA^{2} = kX^{2} = mV^{2}

(330)A^{2} = (330)(0.02)^{2} +
(2.3)(0.5)^{2}

A = 0.0463 m

mV_{max}^{2}/2 = kX^{2}/2 +
mV^{2}/2

mV_{max}^{2} = kX^{2} +
mV^{2}

(2.3)V_{max}^{2} = (330)(0.02)^{2} +
(2.3)(0.5)^{2}

V_{max} = 0.554 m/s

A) Amplitude of the motion = 0.0463 m

B) Maximum speed of the object = 0.554 m/s

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