Answer:
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, [tex]V_m[/tex] = 0.70 m/s
The amplitude A of the motion is given by;
[tex]V_m = \omega A\\\\[/tex]
where;
ω is the angular velocity given as;
[tex]\omega = \sqrt{\frac{k}{m} }\\\\\omega = \sqrt{\frac{120}{0.2} }\\\\\omega =24.5 \ rad/s[/tex]
Now, substitute the value of angular velocity and solve the amplitude;
[tex]V_m = \omega A\\\\A = \frac{V_m}{\omega}\\\\A = \frac{0.7}{24.5}\\\\A = 0.0286 \ m[/tex]
Therefore, the amplitude of the motion is 0.0286 m.
