1) The spring constant k is found to be 200 N/m by using Hooke's law and the force of gravity on the 0.6 kg block. 2) The amplitude A of the oscillation is 0.02 m, which is the distance the block is compressed below the equilibrium position. 3) The maximum speed vmaximum is found to be 0.365 m/s using the conservation of energy and setting the kinetic and potential energy equal at the endpoints of oscillation. 4) The position as a function of time x(t) is given by x(t) = 0.02 cos (18.3T) where T is the period of 0.344 s.

1. Physics 101 Learning Object 1
Simple Harmonic Motion
Question
A relaxed Hooke’s Law spring is shown in fig. 1. When a 0.6 kg block sits on top of the spring,
the spring is compressed by 0.03 m as shown in fig. 2. In fig. 3, the block is then pushed down
0.02 m and then released from rest and is sent into oscillation.
a) Find the spring constant, k. Use g= 10 m/s2.
b) Find the amplitude of the oscillation, A or xmaximum.
c) Find the maximum speed of the box during the oscillation, vmaximum.
d) Find the equation for the position as a function of time.
0.03 m
0.02 m
x = 0
x
0.6 kg
0.6 kg
Relaxed
T = 0

2. Physics 101 Learning Object 1
Simple Harmonic Motion (Answer Key)
Answer Key
a) To find k, refer to fig. 2. When the block sits on top of the spring, the
forces are balanced and the spring is at equilibrium.
1. Use Fspring = kx
2. Because it is in equilibrium Fspring = mg
3. Therefore, mg = kx
x being the distance compressed
4. Solve for k k = mg = (0.6 kg)(10 m/s) = 200 N
(0.03 m) m
b) To find A or xmaximum, refer to fig. 2 and fig. 3. Recall that the amplitude is the distance
between the end point and the equilibrium, where the net force on the box is equal to zero,
shown in fig. 2. As the box is pushed down 0.02 m from the equilibrium point and then released,
the distance pushed down becomes the end point. Therefore,
A = xmaximum= 0.02 m.
c) To find vmaximum, refer to fig. 3. Here we can use the equation for the Conservation of Energy.
1. The total mechanical energy is the E = constant
same throughout the oscillation.
Fspring
mg

3. 2. At the end point, Eend = 1/2 kA 2
3. At equilibrium, Eequilibrium = 1/2 mv2
4. By setting Eend = Eequilibrium, we can
solve for v Eend = Eequilibrium
1/2 kA2 = 1/2 mv2
kA2 = mv2
√ (kA2/m) = v
v = √ ((200 Nm-1)(0.02m)2/ (0.6 kg)) = 0.365 m
s
d) To find the equation for the position as a function of time, we must recognize that the
equation is either a function of sine or cosine. By plotting a graph, we can see that in fig. 3,
when the spring is at rest, T = 0 s and A = 0.02 m below the equilibrium, x = 0 m.
1. Therefore, the equation must be a cosine function.
x(t) = _____ cos (____)
2. The function fluctuates from positive to negative according to the amplitude, A. Therefore,
x(t) = A cos (____)
3. What goes inside the cosine function is a constant, z, proportional to time, T.
x(t) = A cos ( z · T)
4. To find T, use the period equation.
T = 2π (√(m/k) )
= 2π (√(0.6 kg) / (200 N/m))
= 0.344 s
0
x
T

4. 5. We can then solve for the unknown variable, z.
z · T = 2π
z = 2π / T
= 2π / (0.344 s)
= 18.3
6. Our final equation for the position as a function of time is then,
x(t) = A cos ( z · T)
x(t) = 0.02 cos ( 18.3 T )