Question

A 0.59 kg object connected to a light spring with a force constant of 19.2 N/m oscillates on a frictionless horizontal surface. If the spring is compressed 4.0 cm and released from rest.

(a) Determine the maximum speed of the object.
cm/s
(b) Determine the speed of the object when the spring is compressed 1.5 cm.
cm/s
(c) Determine the speed of the object when the spring is stretched 1.5 cm.
cm/s
(d) For what value of x does the speed equal one-half the maximum speed?
cm

Answer 1

(a) Use conservation of energy:

SPEmax = 1/2*k*xmax^2
KEmax = 1/2*m*vmax^2

Because energy is conserved:
SPEmax = KEmax

Thus:
k*xmax^2 = m*vmax^2

vmax = xmax*sqrt(k/m)
22.81 cm/s

(B) Call the 1.5 cm compression point, point B. Use conservation of energy to equate total energy at B to SPEmax.

SPEmax = KEb + SPEb

KEb = 1/2*m*vb^2
SPEb = 1/2*k*xb^2

1/2*k*xmax^2 = 1/2*m*vb^2 + 1/2*k*xb^2

Solve for vb:
vb = sqrt(k/m*(xmax^2 - xb^2))
= 21.15 cm/s

(C) Because of symmetry of the simple harmonic motion, the velocity at the compression you mention in part B is equal and opposite to this velocity in part C. Because we are only interested in the speed, forget about the sign. Thus vc = vb.
-----------------
(D) Call point D, this point of interest.

vd = vmax/2

From part A, vmax = xmax*sqrt(k/m), therefore:
vd = 1/2*xmax*sqrt(k/m)

Conservation of energy from SPEmax point to point D:
1/2*k*xmax^2 = 1/2*k*xd^2 + 1/2*m*vd^2

solve for xd:
vd =