1
oscillations_02
Oscillations
Time variations that repeat themselves at regular
intervals - periodic or cyclic behaviour
Examples: Pendulum (simple);
heart (more complicated)
Terminology:
Amplitude: max displacement from
equilibrium position [m]
Period: time for one cycle of motion [s]
SHM
Frequency: number of cycles per second
[s-1 = hertz (Hz)]
How can you determine the mass of a single E-coli bacterium or a
DNA molecule ?
CP458
CP Ch 14
2
Signal from ECG
voltage 
period T
time 
Period:
time for one cycle of motion [s]
Frequency: number of cycles per second [s-1 = Hz hertz]
1 kHz = 103 Hz
106 Hz = 1 MHz
1GHz = 109 Hz
CP445
3
Brightness
Example: oscillating stars
Time
CP445
4
oscillations_02:
MINDMAP SUMMARY
Reference frame (coordinate system, origin, equilibrium position),
displacement (extension, compression), applied force, restoring force,
gravitational force, net (resultant) force, Newton’s Second Law, Hooke’s
Law, spring constant (spring stiffness), equilibrium, velocity, acceleration,
work, kinetic energy, potential energy (reference point), gravitational
potential energy, elastic potential energy, total energy, conservation of
energy, ISEE, solve quadratic equations, SHM, period, frequency, angular
frequency, amplitude, sine function (cos, sin), phase, phase angle, radian,
SHM & circular motion
v 
dr
a 
dt
W 

dv
r2
F cos  dr
r1
x  A cos( t )
f 
1
F
dt
1
k x
F k x
T
1
 
2
f
1
2
k
2
1
2
  2 f 
2
2
T
k x m ax 
2
E  K Ue UG
a   x  A  cos(  t )    x
m
m v m ax 
Fe   k x
UG  m g h
2
v   A  sin( t )
T 
E  K Ue 
Ue 
 ma
1
2
k A
2
2
T  2
m
k
A  x m ax
5
Simple harmonic motion SHM
x=0
spring
Fe
Fe   k x
restoring force
x
+X
• object displaced, then released
• objects oscillates about equilibrium position
• motion is periodic

• displacement is a sinusoidal function of time (harmonic)
T = period = duration of one cycle of motion
f = frequency = # cycles per second
• restoring force always acts towards equilibrium position
• amplitude – max displacement from equilibrium position
CP447
6
By viewing the animation
You should have a better understanding of the
following terms
SHM – periodic motion
Equilibrium position
Displacement
Amplitude
Period
Frequency
7
Motion problems – need a frame of reference
origin 0
equilibrium position
displacement
x [m]
velocity
v [m.s-1]
- xmax
Vertical hung spring: gravity
determines the equilibrium position
– does not affect restoring force for
displacements from equilibrium
position – mass oscillates vertically
with SHM
Fe = - k y
x0
+ xmax
acceleration
a [m.s-2]
Force
Fe [N]
CP447
8
What is the connection between circular motion and SHM ?
What is the meaning of period T, frequency f, and angular frequency  ?
Connection SHM – uniform circular motion
9
1 revolution = 2 radians = 360o
 = d/dt
 = 2 / T
amplitude A
A
T=1/f
f=1/T
=2f
=2  / T
One cycle: period T [s]
Cycles in 1 s: frequency f [Hz]
Angular frequency  [rad.s-1]

Angles must be measured in- radians
CP453
10
+1
SHM & circular motion
0
x  cos( )

-1
0
2
4
6
Displacement is sinusoidal function of time
x  x m ax
t 

cos  2 
  x m ax cos  2  f t   x m ax cos   t 
T 

uniform circular motion
v
radius A, angular frequency 
  2 f 
2
T
 t
x
x component is SHM
X
x  A cos   t 
x m ax  A
CP453
11
Simple Harmonic Motion
Displacement is a sinusoidal function of time
displacement
T
x max
T
x  x m ax
amplitude
time
T

t 

cos  2    x m ax cos  2  f t   x m ax cos   t 
T 

By how much does phase change over one period?
CP451
12
x
Simple Harmonic Motion
k
force F   k x  m a
m
+X
x=0
displacement
x  x m ax co s   t 
velocity = dx/dt
v   x m ax  sin   t    
acceleration = dv/dt
a   x m ax  co s   t     x
2
( x m ax  x )
2
2
2
angular frequency, frequency, period
 
k
m
 f 
1
k
2
m
 T  2
m
k
CP457
13
a   x
Simple harmonic motion
2
acceleration is  rad (180) out of phase with displacement
v  x m ax  sin
2
2
2
2
 t 
 x m ax  (1  c os
2
  (x
2
2
2
 x )
2
m ax
 t )
E  K Ue 
1
2
mv 
2
v 
2
v  
( x m ax  x )
2
2
2
1
1
2
k x m ax 
2
1
kx 
2
2
k
x

m
v  
2
x
2
m ax
2
x m ax  x
2
k x m ax
1
2
m v m ax  constant
2
2

2
x0
Ue  0
CP457
14
SHM
position x
10
0
-10
0
10
20
30
40
50
60
70
80
90
100
0
10
20
30
40
50
60
70
80
90
100
0
10
20
30
40
50
time t
60
70
80
90
100
velocity v
5
0
acceleration a
-5
1
0
-1
Describe the phase relationships between displacement, velocity and acceleration?
What are the key points on these graphs (zeros and maximums)?
345678
CP459
15
Problem solving strategy: I S E E
Identity:
What is the question asking (target variables) ?
What type of problem, relevant concepts, approach ?
Set up:
Diagrams
Equations
Data (units)
Physical principals
PRACTICE ONLY
MAKES PERMANENT
Execute: Answer question
Rearrange equations then substitute numbers
Evaluate:
Check your answer – look at limiting cases
sensible ?
units ?
significant figures ?
16
Problem 1
If a body oscillates in SHM according to the equation
x  5.0 cos(0.40 t  0.10) m
where each term is in SI units. What are
(a) the amplitude?
(b) the angular frequencies, frequency and period?
(c) the initial phase at t = 0 ?
(d) the displacement at t = 2.0 s ?
use the ISEE method
9 10 11
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Solution 1
x  5.0 cos(0.40 t  0.10) m
Identify / Setup
SHM
x  x m ax cos(  t   )
x m ax  5.0 m
Execute
  0.40 rad.s
T
-1
(a)
amplitude
(b)
angular frequency  = 0.40 rad.s-1
frequency
period
Execute
  2 f 
  0.10 rad
A = xmax = 5. 0 m
f =  / 2 = 0.40 / (2) Hz = 0.064 Hz
T = 1 / f = 1 / 0.064 s = 16 s
(c)
initial phase angle  = 0.10 rad
(d)
t = 2.0 s
OK
2
x = 5 cos[(0.4)(2) + 0.1] m = 3.1 m
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Problem 2
An object is hung from a light vertical helical spring that
subsequently stretches 20 mm. The body is then displaced and
set into SHM.
Determine the frequency at which it oscillates.
use the ISEE method
12 13 14
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Solution 2
Identify / Setup
k = ? N.m-1
SHM
k
  2 f 

m
f 
1
x = 20 mm
k
2
m
m
F mg kx
x = 20 mm = 2010-3 m
Execute
mg kx
k 
mg
x
f 
1
2
Execute
mg
mx
OK

1
2
g
x

1
2
9.8
20  10
3
H z  3.5 H z
f = ? Hz
20
Problem 3 What are all the values at times t = T/4, T/2, 3T/4, T ?
-A
0
+A
t
x
V
a
KE
PE
0
A
0
- 2 A
0
½ k A2
T/4
T/2
3T / 4
T
A  x max
Problem 4
A spring is hanging from a support without any object attached to it and its length is
500 mm. An object of mass 250 g is attached to the end of the spring. The length
of the spring is now 850 mm.
(a) What is the spring constant?
The spring is pulled down 120 mm and then released from rest.
(b) What is the displacement amplitude?
(c) What are the natural frequency of oscillation and period of
motion?
(d) Describe the motion on the object attached to the end of the
spring.
Another object of mass 250 g is attached to the end of the spring.
(e) Assuming the spring is in its new equilibrium position, what is the length of
the spring?
(f) If the object is set vibrating, what is the ratio of the periods of oscillation for
the two situations?
use the ISEE method
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Solution 4
Identify / Setup
L0
L0 = 500 mm = 0.500 m
L1 = 850 mm = 0.850 m
m1 = 250 g = 0.250 kg
ymax = 120 mm = 0.120 m
k = ? N.m-1
f1 = ? Hz T1 = ? s
m2 = 0.500 kg
L2 = ? m
T2 / T1 = ?
L1
L2
equilibrium position
y=0
F = k (L1 – L0)
SH M
Fe   k x
F = FG
x  x m ax cos(  t )
  2 f 
2
T
T  2
m
k
m1
a=0
FG = m g

k
m
m1
ymax
m2
23
Execute
(a)
Object at end of spring – stationary
F = FG  k (L1 - L0) = m g
k = m g /(L1 – L0)
k = (0.250)(9.8) / (0.850 – 0.500) N.m-1
k = 7.00 N.m-1
2
  2 f 
k

T
f1 
T1 
1
k
2
m
1
f1

1

(b) (c) (d)
Object vibrates up and down with SHM
about the equilibrium position with a
displacement amplitude
A = ymax = 0.120 m
m
1
7
2
0.25
H z  0.842
s  1.19
Hz
Evaluate
s
0.842
(e)
Again F = m g = k y  y = m2 g / k
m2 = 0.500 kg k = 7 N.m-1
y = (0.5)(9.8) / 7 m = 0.700 m L2 = L0 + y = (0.500 + 0.700) m = 1.20 m
(f)
T1  2 
m1
k1
T2  2 
m2
k2
k  k 1  k 2  T 2 / T1 
m2
m1

0.50
0.25
 1.4
24
Problem 5
A 100 g block is placed on top of a 200 g block. The coefficient of static friction
between the blocks is 0.20. The lower block is now moved back and forth
horizontally in SHM with an amplitude of 60 mm.
(a)
Keeping the amplitude constant, what is the highest frequency for which the
upper block will not slip relative to the lower block?
Suppose the lower block is moved vertically in SHM rather than horizontally. The
frequency is held constant at 2.0 Hz while the amplitude is gradually increased.
(b)
Determine the amplitude at which the upper block will no longer maintain
contact with the lower block.
use the ISEE method
25
Solution 5
Identify / Setup
FN
m1 = 0.1 kg
m2 = 0.2 kg
1
 = 0.20
2
A2 = 60 mm = 0.06 m
m1
Ff =  N
=mg
FN = m g
FG
max freq f = ? Hz
SHM
amax = A 2 = A(2  f)2 = 4 2 f 2 A
FN
1
2
a1y = 0
m1
FG
Execute
(a)
max frictional force between blocks
Ff =  m1 g
max acceleration of block 1
a1max = Ff / m1 =  g
max acceleration of block 2
a2max = a1max =  g
SHM
a2max = 4 2 f2 A
 g = 4 2 f2 A
f = [ g / (4 2 A)] = [(0.20)(9.8)/{(4)(2)(0.06)}] Hz = 0.91 Hz
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27
Execute
(b)
max acceleration of block 1 (free fall)
a1max = g
max acceleration of block 2
a2max = a1max = g
SHM
a2max = 4 2 f2 A
g=42f2A
f = 2 Hz
A = g / (4 2 f 2) = (9.8) / {(4)(2)(22)} m = 0.062 m
Evaluate