Simple Harmonic Motion
A trampoline exerts a restoring force on the
jumper that is directly proportional to the
displacement of the mat. (Fs=kx). Such
restoring forces provide the driving forces
necessary for objects that oscillate with
simple harmonic motion.
Simple Harmonic Motion
Simple harmonic motion Back and forth motion that
is caused by a force that is directly proportional to the
displacement. (F=kx) The displacement centers around
an equilibrium position.
1
f 
T
Amplitude
A
Period, T, is the time
for one complete
oscillation. (seconds)
Frequency, f, is the
number of complete
oscillations per
second. Hertz (s-1)
Example The suspended mass makes 30
complete oscillations in 15 s. What is the
period and frequency of the motion?
15 s
T
 0.50 s
30 cycles
x
F
Period: T = 0.5 s
1
1
f  
T 0.5 s
Frequency: f = 2 Hz
Example
A load of 50 N attached to a spring hanging vertically stretches
the spring 5.0 cm. The spring is now placed horizontally on a
table and stretched 11.0 cm. What force is required to stretch
the spring this amount?
Fs  kx
Fs  kx
50  k (0.05)
k  1000 N/m
Fs 
Fs  (1000)(0.11)
110 N
Position graph of a spring
When you graph the motion of a spring, it is
a sine function
Position time graph for SHM
The amplitude of the curve is the max
displacement ,x
Equilibrium Line
Period, T, is the time for one revolution or in the case of springs the time for
ONE COMPLETE oscillation (One crest and trough). Oscillations could also
be called vibrations and cycles. In the wave above we have 1.75 cycles or
waves or vibrations or oscillations.
SHM and Uniform Circular Motion
Springs (and waves) behave very similar to objects that move in
circles.
If you trace out the position of an object on the outside of a
circle, you get a sine function (remember the unit circle??)
http://www.animations.physics.unsw.edu.a
u//jw/SHM.htm#projection
xspring  Awave  rcircle
Period of a spring
F  ma  kx
For a vibrating body with an elastic restoring force:
Recall that F = ma = kx:
m
T  2
k
 2 r 
m

 T   kx
r
m4 2 r
 kr
2
T
2
T kr  m 4 2 r
2
4 2 m
T 
k
2
The frequency f and the period T can be found if the spring constant k and
mass m of the vibrating body are known.
Example
A 200 g mass is attached to a spring and executes
simple harmonic motion with a period of 0.25 s If the
total energy of the system is 2.0 J, find the (a) force
constant of the spring (b) the amplitude of the motion
m
0.200
Ts  2
 0.25  2
k
k
SPE  1 kx 2  2  1 kA2
2
2
k
A
126.3 N/m
0.18 m
Velocity in SHM
v (-)
v (+)
m
x = -A
x=0
x = +A
• Velocity is positive when moving to the right
and negative when moving to the left.
• It is zero at the end points and a maximum
at the midpoint in either direction (+ or -).
Acceleration in SHM
+a
-a
-x
+x
m
x = -A
x=0
x = +A
• Acceleration is in the direction of the
restoring force. (a is positive when x is
negative, and negative when x is positive.)
• Acceleration is a maximum at the end points
and it is zero at the midpoint of oscillation.
Example A 2-kg mass hangs at the end of a
spring whose constant is k = 400 N/m. The
mass is displaced a distance of 12 cm and
released. What is the acceleration at the
instant the displacement is x = +7 cm?
kx
a
m
(400 N/m)(+0.07 m)
a
2 kg
a = -14.0 m/s2
a
m
Note: When the displacement is +7 cm
(downward), the acceleration is -14.0 m/s2
(upward) independent of motion direction.
+x
Example: What is the maximum acceleration
for the 2-kg mass in the previous problem? (A
= 12 cm, k = 400 N/m)
The maximum acceleration occurs when the restoring
force is a maximum; i.e., when the stretch or
compression of the spring is largest.
F = ma = -kx
xmax =  A
kA 400 N(  0.12 m)
a

m
2 kg
m
Maximum Acceleration:
amax = ± 24.0 m/s2
+x
Graph of Force vs. displacement
Remember that the slope of a force vs. displacement graph is equal to k
Force vs. Displacement
y = 120x + 1E-14
R² = 1
80
70
Force(Newtons)
60
50
40
30
20
k =120 N/m
10
0
0
0.1
0.2
0.3
0.4
Displacement(Meters)
0.5
0.6
0.7
Graph of Force vs. displacement
And that the area is equal to the spring potential energy
1 2
SPE  kx
2
Example
A slingshot consists of a light leather cup, containing a stone, that
is pulled back against 2 rubber bands. It takes a force of 30 N to
stretch the bands 1.0 cm (a) What is the potential energy stored
in the bands when a 50.0 g stone is placed in the cup and pulled
back 0.20 m from the equilibrium position? (b) With what speed
does it leave the slingshot?
a ) Fs  kx 30  k (0.01) k  3000 N/m
SPE  1 kx 2  0.5(k )(.20) 2 
2
b) EB  E A
SPE  KE
SPE  1 mv 2  1 (0.050)v 2
2
2
v  49 m/s
60 J
Pendulums
Pendulums, like springs, oscillate
back and forth exhibiting
simple harmonic behavior.
http://www.walterfendt.de/ph14e/pendulum.htm
The restoring force is the force that brings the pendulum back to its equilibrium
Pendulums
Consider the FBD for a pendulum. Here we have the
weight and tension. Even though the weight isn’t at an
angle let’s draw an axis along the tension.
q
q
mgcosq
mgsinq
mg sin q  Restoring Force
mg sin q  kx
Pendulums
mg sin q  Restoring Force
mg sin q  kx
s s
q 
R L
s  qL  Amplitude
mg sin q  kqL
sin q  q , if q  small
mg  kl
m l

k g
m
Tspring  2
k
What is x? It is the
amplitude! In the picture to
the right, it represents the
chord from where it was
released to the bottom of
the swing (equilibrium
position).
l
T pendulum  2
g
Example What must be the length of a
simple pendulum for a clock which has a period
of two seconds (tick-tock)?
L
T  2
g
2
2 L
T  4
;
g
(2) 2 (10)
L
2
4
L
T 2g
L=
2
4
L = 1.01 m
Example
A visitor to a lighthouse wishes to determine the
height of the tower. She ties a spool of thread
to a small rock to make a simple pendulum,
which she hangs down the center of a spiral
staircase of the tower. The period of oscillation
is 9.40 s. What is the height of the tower?
l
TP  2
 l  height
g
2
2
2
T
g
4

l
9.4
(10)
2
P
TP 
l 


2
2
g
4
4( )
L = Height = 22.4 m