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Serjik [45]
1 year ago
10

when a body moves in a simple harmonic motion, its acceleration at the ends of its path is: group of answer choices zero less th

an at equilibrium more than g maximum
Physics
1 answer:
Hoochie [10]1 year ago
7 0

When a body moves in a simple harmonic motion, its acceleration at the end of its path is more than that of the acceleration at equilibrium.

The acceleration of the body moving in Simple Harmonic Motion(SHM) is given by,

Acceleration = ω^2 A

Here w is the angular velocity of the body and A is the amplitude.

A is the maximum distance of the body from the equilibrium position.

When the body is at an equilibrium position, its amplitude is equal to zero.

Hence, acceleration = ω^2 x 0 = 0

When the body is at the end of its path, its amplitude is equal to A.

Hence, acceleration = ω^2 x A

Therefore,

when a body moves in a simple harmonic motion, its acceleration at the end of its path is more than that of the acceleration at equilibrium.

To know more about "Simple Harmonic Motion", refer to the following link:

brainly.com/question/20885248?referrer=searchResults

#SPJ4

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The solution to the question above is explained below:

Explanation:

For which solid is the lumped system analysis more likely to be applicable?

<u>Answer</u>

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<em>Question :Why?</em>

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<u>Further explanations:</u>

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3 years ago
A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wal
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2.59 m

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