Example 12-4 SHM III: Kinetic and Potential Energy
As in Example 12-3, an object of mass 0.80 kg is attached to an ideal horizontal spring of spring constant 1.8 * 102 N>m.
The object is initially at rest at equilibrium. You then start the object oscillating with amplitude 2.0 * 1022 m. (a) How
much work do you have to do on the object to set it into oscillation? (b) What is the speed of the object when the spring is
compressed by 1.0 * 1022 m? (c) How far is the object from equilibrium when the kinetic energy of the object equals the
potential energy in the spring?
Set Up
We’ll use energy ideas to answer these questions. Equation 12-21 will let us find the
­(constant) total mechanical energy E of the
system from the given spring constant and
amplitude. Equation 6-19 will let us find the
spring potential energy for any displacement
x; we can then find the kinetic energy K using
Equation 12-21, and from that find the speed
for that value of x.
Solve
(a) Initially the system has zero kinetic energy
and zero potential energy. You can start the
oscillation by pulling the object so that you
stretch the spring by a distance A = 2.0 * 102 m
from equilibrium. The work you do goes into
the potential energy of the spring.
Total mechanical energy of a mass
oscillating on an ideal spring:
1
(12-21)
E = K + Uspring = kA2
2
Spring potential energy:
1
Uspring = kx 2
2
k
m
(6-19)
(Work you do) = (Change of
potential energy of the spring as
you stretch it from x = 0 to x = A)
x=0
Initial potential energy
1
= k102 2 = 0
2
1
Final potential energy = kA2
2
So the work you do is
1
W = kA2 - 0
2
1
= 11.8 * 102 N>m2 12.0 * 10-2 m2 2
2
= 3.6 * 1022 N # m = 3.6 * 1022 J
(b) When the spring is compressed by
1.0 * 1022 m, the displacement is x =
21.0 * 1022 m. The first step in determining
the block’s speed at this value of x is to
determine its kinetic energy.
x =A
= 2.0 × 10–2 m
From (a), the total mechanical
energy is
E =
x=0
1 2
kA = 3.6 * 10-2 J
2
From Equation 6-19, the
spring potential energy for
x = 21.0 * 1022 m is
1
Uspring = kx 2
2
1
= 11.8 * 102 N>m2 1 -1.0 * 10-2 m2 2
2
= 9.0 * 1023 J
x = –1.0 × 10–2 m
So from Equation 12-21, the kinetic energy at this value of x is
K = E 2 Uspring = 3.6 * 1022 J 2 9.0 * 1023 J
= 2.7 * 1022 J
Given the kinetic energy K of the block and its
mass m = 0.80 kg, calculate the speed of the
block.
Solve for the speed v:
Kinetic energy is K =
v2 =
v =
1
mv 2, so
2
2K
m
212.7 * 10-2 J2
2K
=
= 0.26 m>s
Bm
C
0.80 kg
(Recall from Chapter 6 that if the kinetic energy is in joules and the
mass is in kilograms, the speed is in meters per second.)
The velocity of the block could be +0.26 m>s or -0.26 m>s,
­depending on what direction it is moving as it passes through
this point.
(c) Use the same ideas as in part (b) to solve
for the value of x at which the kinetic energy K
equals the spring potential energy Uspring.
From above, the total mechanical energy is
1
E = kA2 = 3.6 * 10-2 J
2
The spring potential energy is
1
Uspring = kx 2
2
and the kinetic energy is
K = E 2 Uspring
At the value of x for which K = Uspring,
1 2
1
1
kA - kx 2 = kx 2
2
2
2
Multiply both sides by 2>k and solve for x:
A2 - x 2 = x 2 so 2x 2 = A2 and x 2 =
A2
2
A2
A
= {
B2
12
2.0 * 10-2 m
= {
= { 1.4 * 10-2 m
12
Note that x can be either positive (the spring is stretched by
1.4 * 1022 m) or negative (the spring is compressed by
1.4 * 1022 m).
x = {
Reflect
The result for part (b) would have been very difficult to find without using the energy approach. You would have needed
to use the equations from Section 12-3 to solve for the time t at which the block passes through x = 21.0 * 1022 m, then
use this value of t to find the velocity of the block at that time. The energy approach makes this much easier.
Note that the points x = {A> 12 = {0.71A where the kinetic and potential energies are equal are shown in
Figure 12-10 (see parts (b), (d), (f), and (h) of that figure). These points are not halfway between the equilibrium position (x = 0) and the extremes of the motion (x = { A); they are actually closer to the extremes. Can you show that at
x = {A>2, the energy is 75% kinetic and 25% potential? (Hint: See the solution to part (b).)