Simple harmonic motion

Question

A 200g object is attached to the end of a spring which rests on a smooth horizontal surfac while in equilbrium the object is pulled by 8cm and released if the maximum speed the object exhibit is 40cm/s
A) find spring constant
B) find the speed of the object at the moment when it is 6cm away from equilbrium position

Answer 1

(A)

 

Equate the elastic potential energy of the spring to the kinetic energy of the object at maximum speed, and solve for k.

 

 kx²/2 = mv_max²/2

 

k = mv_max²/x²

 

Plug in

 

m = 0.200 kg

v_max = 0.40 m/s

x = 0.08 m

 

and get k in newtons/meter.

 

(B)

 

Set total energy at 0.06 meters equal to mv_max²/2, and solve for v.

 

mv²/2 + k(0.06 meters)²/2 = mv_max²/2

 

mv² + k(0.06)² = mv_max²

 

v = √[v_max² - (k/m)(0.06)²]

 

You know v_max and m, and you have k from part (A).  Plug in the numbers and get v in m/s.