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Harmonic Motion
Chapter 13 4-8
Critical Concepts
• In the absence of friction, mechanical energy is
conserved
• Mechanical energy is the sum of the potential
and kinetic energy
• In oscillatory motion, the kinetic and potential
energy are “traded” back and forth
• When the kinetic energy is a maximum, the
potential energy is a minimum
• The total mechanical energy remains constant
Conservation of energy in mass-spring system
X0
We have learned that:
x  x0 cos t
v  x0 sin t
Substituting into K
and U, we get
The kinetic and
potential energy
are:
1 2
mv
2
1
U  Kx 2
2
K
E  K U
1
1
m 2 x02 sin 2 t  Kx02 cos 2 t
2
2
1
 Kx02 sin 2 t  cos 2 t
2
1
 Kx02  U MAX
2



The energy “sloshes” back and
forth from kinetic to potential
E  K U
1
1
m 2 x02 sin 2 t  Kx02 cos 2 t
2
2
1
 Kx02 sin 2 t  cos 2 t
2
1
 Kx02  U MAX
2



Energy in Mass-Spring System
Umax
Kmax
Total E is
CONSTANT.
Amplitude A can also be written x0, which is the turning point.
The pendulum has a similar energy relationship.
E
Turning points
Problem 13-28
Given spring constant K and
mass m, what is the period of
oscillation for Block 1 and Block
2?
Solution: The period is related to the angular frequency by: T  1  2
f

So, find the angular frequency and you have the period T.
The angular frequency is obtained from K and M:

The only tricky part is figuring out the spring constant. We
learned for springs in parallel that…
K effective  2k
What is the spring constant for Block 2? HINT: What is the total force?
K
M
Problem 13-45
A bullet of mass m and speed Vo is
embedded in a block of mass M
attached to spring K. If A is
measured, what was the initial
speed Vo? How long does it take
to compress the spring?
Strategy: First use conservation of momentum to find the speed of the
bullet-block just after collision. Use this to find the intial kinetic energy.
Use conservation of energy to find A. Then figure out the period of
oscillation, and take ¼ of that to find the time to compress the spring.
(You have to think about this part to see that it is ¼ of T that answers the
question.)
mv0  M  mV
E  K max 
Combining:
v0 
1
M TOTV 2
2
KM TOT
A
m
E  U max 

1
KA2
2
K
2

M TOT
T
Problem 13-33 (slightly modified)
A mass M is attached to a spring,
which drops a distance D. Then, it
oscillates with angular frequency .
What was D?
D
M
g
Strategy: From the angular frequency,
and the known mass, we can get the
spring constant K. From the spring force
law and K, we can get D.
Oscillations
K  M 2
Spring forces.
KD  Mg
Mg
g
D
 2
K
