Question

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?

Answer 1

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)

I hope help you !!