Physics 115 Lecture 3
Oscillations and energy
January 29, 2016
1
Position of a Simple Harmonic
Oscillator as a function of time
x = A cos( )
Plug in  =  t + 
x(t) = A cos(t + )
y = Asin( )
Plug in  =  t + 
y(t) = A sin(t + )
Simple Harmonic Motion
x(t) = A cos( t + )
phase angle
time
angular speed
amplitude
displacement as a function of time
Simple Harmonic Motion
y(t) = A sin( t + )
phase angle
time
angular speed
amplitude
displacement as a function of time
Simple Harmonic Motion
Period = Time (in seconds) for one complete cycle.
For a spring-and-mass system:
m
T  2
k
T = period (in seconds)
m = mass (in kg)
k = spring constant (in N/m)
Once the period has been calculated, two other
important quantities can be calculated:
Frequency f  1 T  f is in Hz if T is in seconds
Angular speed   360 T   is in deg/s if
T is in seconds
Find the frequency at which this system oscillates:
k = 16 N/m
2 kg
40 mm
A. 2.2 Hz
B. 8.0 Hz
C. 0.040 Hz
D. 0.450 Hz
Response
Counter
0%
A.
0%
B.
0%
C.
0%
D.
180
Find the frequency at which this system oscillates:
k = 16 N/m
2 kg
40 mm
A. 2.2 Hz
m
2 kg
T  2
 2
 2.22 s
B. 8.0 Hz
k
16 N/m
1
1
C. 0.040 Hz
f  
 0.450 Hz
T 2.22 s
D. 0.450 Hz
Simple Harmonic Motion
For a pendulum:
L
T  2
g
T = period (in seconds)
L = length of pendulum (in m)
g = accel of gravity (in m/s2)
What length pendulum will have a period of 10.0 s?
Assume g = 9.80 m/s2.
A. 10.0 m
B. 980 m
C. 0.629 m
D. 24.8 m
Response
Counter
0%
A.
0%
B.
0%
C.
0%
D.
180
What length pendulum will have a period of 10.0 s?
Assume g = 9.80 m/s2.
A. 10.0 m
B. 980 m
C. 0.629 m
D. 24.8 m
L
2
2 L
T  2
 T  4
g
g
9.80 m/s  10.0 s 

gT
L
 24.8 m
2
2
4
4
2
2
2
Kinetic Energy
Any mass that is in motion has kinetic energy.
The amount of kinetic energy it has is given by:
KE  mv
1
2
2
m = the mass of the moving object, in kilograms.
v = the speed of the object, in meters per second.
KE = kinetic energy, in joules (J).
Potential Energy
Any spring that is stretched or compressed
has potential energy. The amount of potential
energy it has is given by:
PE  k x
1
2
2
k = the spring constant, in newtons per meter.
x = amount of stretch or compression of spring,
in meters.
PE = potential energy, in joules.
Potential Energy
Any pendulum that is pulled
away from its equilibrium position
has potential energy. The amount
of potential energy it has is given by:
PE  m g h
m = the mass of the pendulum bob, in kg.
g = acceleration of gravity, 9.80 m/s2.
h = vertical distance the bob is raised above its
lowest position, in meters.
PE = potential energy, in joules.
Conservation of Energy


Energy for mass on a
spring applet
Energy for pendulum
applet
THE ENERGY OF SIMPLE HARMONIC MOTION
KINETIC ENERGY
POTENTIAL ENERGY
16
ENERGY (in mJ)
14
12
10
8
6
4
2
0
0
1
2
3
TIME (in seconds)
4
5
Total Energy: E = KE + PE
2
2
1
1
E  2 mv  2 k x
CONSTANT (assuming there is no friction)
2
max
E  mv
because x = 0 when v = vmax
2
max
E kx
because v = 0 when x = xmax
1
2
1
2
E kA
1
2
2
where A = xmax = amplitude
If the amplitude of an oscillator is tripled, what
happens to the energy stored in the oscillation?
A. It is tripled.
B. It increases by a
factor of 9.
C. It stays the same.
D. It is cut to a third.
Response
Counter
0%
A.
0%
B.
0%
C.
0%
D.
120
If the amplitude of an oscillator is tripled, what
happens to the energy stored in the oscillation?
A. It is tripled.
B. It increases by a
factor of 9.
C. It stays the same.
D. It is cut to a third.
Recall that E  12 k A2 . Because
E increases with A2 , E will increase
by a factor of nine. Or try a ratio:
2
new
2
old
Enew 12 knew A
 1
Eold
2 kold A

1
2
knew  3 Aold
1
2
2
old
kold A

2
 9
Brief review



A restoring force produces
oscillation
The position of an oscillator can be
mathematically described by a sine
function or a cosine function
Oscillators store energy, which
converts back and forth between
kinetic and potential energies
Power
The rate at which energy is
transferred from one system to
another is called the power
energy joule
P

 watt (W)
time second
A weightlifter hoists 337 lb (1500 N) a
distance of 2.1 m, requiring 3150 J of work. If
he does this in 1.80 s, the required power is:
3150 J
1 hp
P
 1750 W 
 2.3 hp
1.80 s
746 W
A loudspeaker generates 4.8 W of acoustic power.
How much acoustic energy does it generate in 60 s?
A. 12.5 J
B. 48 J
C. 288 J
D. 1380 J
Response
Counter
0%
A.
0%
B.
0%
C.
0%
D.
120
A loudspeaker generates 4.8 W of acoustic power.
How much acoustic energy does it generate in 60 s?
E
P
t
A. 12.5 J
B. 48 J
J

C. 288 J P t  E   4.8  60 s
s

D. 1380 J

 288 J
