LECTURE 2
1
Oscillations
Time variations that repeat themselves at regular intervals periodic or cyclic behaviour
Examples: Pendulum (simple); heart (more complicated)
Terminology:
Period: time for one cycle of motion [s]
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 ?
CP466
CP Ch 14
2
Signal from ECG
period T
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
CP453
Brightness
Example: oscillating stars
3
Time
CP453
4
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
CP455
5
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
+ xmax
acceleration
a [m.s-2]
Force
Fe [N]
CP455
6
Connection SHM – uniform circular motion
ω = dθ/dt
ω = 2π / T
amplitude A
A
T=1/f
f=1/T
ω=2πf
ω=2 π / T
One cycle: period T [s]
Cycles in 1 s: frequency f [Hz]
θ
- -1
]
angular frequencyω [rad.s
CP461
7
SHM & circular motion
Displacement is sinusoidal function of time
t

x = xmax cos  2π  = xmax cos ( 2π f t ) = xmax cos (ω t )
 T
uniform circular motion
radius A, angular frequency ω
G
v
A ≡ xmax
θ =ωt
x
2π
ω = 2π f =
T
x component is SHM:
x = A cos (ω t )
xmax ≡ A
CP461
8
Simple harmonic motion
displacement
T
amplitude
x max
T
time
T
Displacement is a sinusoidal function of time

t
x = x max cos2π  = x max cos(2πft ) = x max cos(ωt )
 T
By how much does phase change over one period?
CP458
9
Simple harmonic motion
x = xmax cos (ω t )
v = − xmaxω sin (ω t )
a = − xmaxω 2 cos (ω t )
a = −ω x
2
equation of motion
substitute
→ω =
xmax ≡ A
m a = −k x
k
1
→ f =
m
2π
(restoring force)
k
m
→ T = 2π
m
k
oscillation frequency and period
CP459
10
SHM
a = −ω 2 x
acceleration is π rad (180°) out
of phase with displacement
2
v 2 = xmax
ω 2 sin 2 (ω t )
At extremes of oscillations, v = 0
When passing through equilibrium,
v is a maximum
2
= xmax
ω 2 (1 − cos2 (ω t ))
2
= ω 2 ( xmax
− x2 )
2
− x2 )
v = ±ω ( xmax
CP459
11
Simple harmonic motion
Hecht: figure 10.25
CP459
12
S HM
pos ition
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
tim e t
60
70
80
90
100
veloc ity
v
5
0
ac c eleration
a
-5
1
0
-1
How do you describe the phase relationships between displacement, velocity and
acceleration?
CP459
13
Problem 2.1
If a body oscillates in SHM according to the equation
x = 5.0cos(0.40 t + 0.10) m
where each term is in SI units. What are
(a) the amplitude
(b) the frequency and period
(c) the initial phase at t = 0?
(d) the displacement at t = 2.0 s?
[Ans: 5.0 m, 0.064 Hz, 16 s, 0.10 rad, 3.1 m]
14
Solution 2.1
Identify / Setup
SHM
xmax
x = 5.0cos(0.40 t + 0.10) m
2π
ω = 2π f =
x = xmax cos(ω t + φ )
T
= 5.0 m
ω = 0.40 rad.s-1
φ = 0.10 rad
Execute
(a)
amplitude
A = xmax = 5. 0 m
(b)
frequency
f = ω / 2π = (0.40 / (2π)) Hz = 0.064 Hz
period
T = 1 / f = (1 / 0.064) s = 16 s
(c)
initial phase angle φ = 0.10 rad
(d)
t = 2.0 s
Execute
OK
x = 5 cos{(0.4)(2) + 0.1} m = 3.1 m
15
Problem 2.2
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.
[Ans: ω = 22 rad.s-1; f = 3.5 Hz]
16
Solution 2.2
Identify / Setup
k = ? N.m-1
SHM
ω = 2π f =
k
m
1 k
2π m
F =mg =k x
⇒
x = 20 mm
f =
m
x = 20 mm = 20×10-3 m
Execute
mg
mg =k x k =
x
1 mg
1
=
f =
2π m x 2π
Execute
OK
1
g
=
x 2π
9.8
Hz = 3.5 Hz
−3
20 × 10
f = ? Hz