Question

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.

Answer 1

The amplitude of the motion is 0.0286 m.

Step-by-step explanation:

Given;

mass of the object, m = 0.2 kg

spring constant, k = 120 N/m

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

The amplitude A of the motion is given by;

`V_m = \omega A\\\\`

where;

ω is the angular velocity given as;

`\omega = \sqrt{(k)/(m) }\\\\\omega = \sqrt{(120)/(0.2) }\\\\\omega =24.5 \ rad/s`

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

`V_m = \omega A\\\\A = (V_m)/(\omega)\\\\A = (0.7)/(24.5)\\\\A = 0.0286 \ m`

Therefore, the amplitude of the motion is 0.0286 m.