A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic motion that occurs has a maximum speed of 0.70 m/s. Determine the amplitude A of the motion.

The amplitude of the motion is 0.0286 m.

Explanation:

Given;

mass of the object, m = 0.2 kg

spring constant, k = 120 N/m

maximum speed of the simple harmonic motion, = 0.70 m/s

The amplitude A of the motion is given by;

where;

ω is the angular velocity given as;

Now, substitute the value of angular velocity and solve the amplitude;

Therefore, the amplitude of the motion is 0.0286 m.

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