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Triss [41]
3 years ago
13

A 0.43-kg object mass attached to a spring whose spring constant is 561 N/m executes simple harmonic motion. If its maximum spee

d is 8 m/s, what is the amplitude of its oscillation (in m)? Round your answer to the nearest tenth.
Physics
1 answer:
VLD [36.1K]3 years ago
4 0

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

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