Question

A 0.43-kg object mass attached to a spring whose spring constant is 561 N/m executes simple harmonic motion. If its maximum speed is 8 m/s, what is the amplitude of its oscillation (in m)? Round your answer to the nearest tenth.

Answer 1

Answer:

0.22 m

Explanation:

m = 0.43 kg, K = 561 N/m

Vmax = 8 m/s

Let the amplitude of the oscillations be A.

The formula for the angular frequency of oscillation sis given by

$\omega  = \sqrt{\frac{K}{m}}$

$\omega  = \sqrt{\frac{561}{0.43}}$

ω = 36.1 rad/s

The formula for the maximum velocity is given by

Vmax = ω x A

A = Vmax / ω

A = 8 / 36.1 = 0.22 m