Question

You are watching an object that is moving in SHM. When the object is displaced 0.620 m to the right of its equilibrium position, it has a velocity of 2.10 m/s to the right and an acceleration of 8.60 m/s2to 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

in SHM,
f = -kx
=> ma = -kx
=> k = -ma/x
applying conservation of mechanical energy
1/2mv^2 - 1/2kx^2 = - 1/2kX^2
where, m = mass of the object
v = velocity of the object
k = spring constant
x = initial position of the shm
X = amplitude of shm
=> 1/2mv^2 - (1/2)(-ma/x)x^2 = - (1/2)(-ma/x)X^2
=> 1/2mv^2 + 1/2max = 1/2maX^2/x
=> v^2 + ax = aX^2/x
=> 2.1^2 + 8.6*0.62 = 8.6X^2/0.62
=> X = 0.838 m
distance travelled by object before it stops = 0.838 - 0.62
= 0.218 m