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azamat
3 years ago
9

-g A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the

motion is 0.165 m. The maximum speed of the block is 3.90 m/s. What is the maximum magnitude of the acceleration of the block?
Physics
1 answer:
Nina [5.8K]3 years ago
5 0

Answer:

a= 92. 13 m/s²

Explanation:

Given that

Amplitude ,A= 0.165 m

The maximum speed ,V(max) = 3.9 m/s

We know that maximum velocity in the SHM  given as

V(max)  = ω A

ω=Angular speed

A=Amplitude

\omega =\dfrac{3.9}{0.165}\ rad/s

ω=23.63 rad/s

The maximum acceleration given as

a = ω² A

a= (23.63)² x 0.165 m/s²

a= 92. 13 m/s²

Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².

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Read 2 more answers
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