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You are watching an object that is moving in SHM. When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left. Part A How much farther from this point will the object move before it stops momentarily and then starts to move back to the left

Answer :

maxwellekoh

Answer:

2.95m

Explanation:

The farthest distance the object can move is the radius of the circle of which the Simple harmonic motion is assumed to be a part

But V = w× r; where V is velocity,

w is angular velocity and r is radius.

Also,

a= w2r; where a is linear acceleration

but a = v× r ; by comparing both equations

Hence r = a/v =8.6/2.45 =3.51m

But the horizontal distance of the motion is given by:

X = rcosx ; where x is the angle

X is the distance covered.

We know that the maximum value of cos x is 1 which is 0°

When the object moves in a fashion directly parallel to an horizontal distance, maximum distance would be reached and hence:

X = r=3.51m

Meaning the object needs to travel 3.51-0.56=2.95m further.

Note: the acceleration of the motion is constant whether it is swinging towards the left or right.

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