Question

# You are watching an object that is moving in SHM. When the object is displaced 0.580...

You are watching an object that is moving in SHM. When the object is displaced 0.580 m to the right of its equilibrium position, it has a velocity of 2.50 m/sto the right and an acceleration of 8.30 m/s2 to the left.

How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

Solution-:

Given data ,

Velocity = 2.50 m/s (towards right ) Acceleration = 8.30 m/s2 (towards left)

x = 0 580 meter

Force = Mass * Acceleration = Ma

Kx = Ma

So,

0.580 K = 8.3 M

or K/M = 8.3 / 0.580

= 14.31

Fron Using Energy Conservation ,

Total mechanical energy = constant

Then Energy conservation between point x= 0.580 and the extreme point i.e x = A at this point the velocity becomes is 0 .

So ,

1/2 M (2.5)^2 + 1/2 K(0.580)^2  = 1/2 KA^2

3.125 M = 0.5K(A^2 - 0.3364 )

So,

6.25 = (K/M) (A^2 - 0.3364) (K/M) = 14.31 from previous equation is ,

6.25 = 14.31(A^2 - 0.3364)

So find A ,

A^2-0.3364 = 6.25 / 14.31

A^2-0.3364 = 0.43675

A^2 = 0.4367 + 0.3364

A^2 = 0.7731 m

A = 0.879 m

Then , A - x

0.879 - 0.580 = 0.299 meter (Answer)

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