CHAPTER 10
Elasticity and Oscillations
Oscillations
 Oscillation
or vibration = back-and-forth
motion.
 Repeats itself, hence it is periodic.
 Vibrations common:
Vocal cords when singing/speaking
String/rubber band,
 Spring
 Simple Pendulum – Grandfather clock
An object in stable equilibrium:
• If displaced slightly from equilibrium
point, will oscillate about the
equilibrium point.
• It will experience a restoring force.
• The restoring force - directly
proportional to displacement.
• The restoring force is opposite to
displacement. F = -kx
Types of Equilibrium
Simple Harmonic Motion
Simple Harmonic Motion (SHM)
Periodic motion in which
Magnitude of the restoring force is
proportional to displacement.
 Restoring force is opposite to the
displacement.
F(x)
Spring
(Elastic) force: F = -kx
 So spring motion is SHM.
x
Springs
 Hooke’s
Law: The force exerted by a
spring is proportional to the distance by
which the spring is stretched or compressed
from its relaxed position.
FX = -k x
where x is the displacement from the
relaxed position and k is the constant of
proportionality.
relaxed position
(Equilibrium Position)
FX = 0
x
x=0
Springs
FX = -kx
relaxed
position
Compressed
FX > 0
x<0
x
x=0
relaxed position
Stretched
FX < 0
x>0
x=0
x
x=A F=-kA v=0 a=-kA/m
x=0 F=0 v=-vo a=0
x=-A F=kA v=0 a=kA/m
x=0 F=0 v=+vo
a=0
x=A F=-kA v=0 a=-kA/m
x
+A
t
-A
• AMPLITUDE (x = A) = Maximum displacement
from equilibrium.
• PERIOD (T) = Time to make one complete
cycle.
• FREQUENCY (f) = # of cycles per second.
Total energy of the spring-mass system:
E = U(spring) + K(mass) = ½kx2 + ½mv2
x=A F=-kA v=0 a=-kA/m
U=½kA2 K=0
x=0 F=0 v=-vo a=0
U=0 K=½mvo2
x=-A F=kA v=0 a=kA/m
U =½kA2 K=0
x=0 F=0 v=+vo
U=0 K=½mvo2
a=0
Total Energy
• At equilibrium point, E1 = U + K = ½mvo2 + 0
• At maximum displacement points,
E2 = U + K = ½kA2 + 0
• At any other position where displacement is x
and the velocity is vx;
E3 = ½kx2 + ½mvx2
• Conservation of energy: E1 = E2 = E3
½kA2 = ½mvo2 = ½kx2 + ½mvx2
Total Energy
E1 = E2 = E3
½kA2 = ½mvo2 = ½kx2 + ½mvx2
½mvo2 = ½kA2
OR vo = (k/m) A
Energy in SHM
 A mass
is attached to a spring and set to
motion. The maximum displacement is x = A
Energy = U + K = constant!
= ½ k x2 + ½ m v2
At maximum displacement x = A, v = 0
Total Energy = ½ k A2 + 0
At zero displacement x = 0
Total Energy = 0 + ½ mvm2
Conservation of energy: ½ k A2 = ½ m vm2
Hence vo = (k/m) A
Analogy w/ marble in bowl
Energy in SHM
½ k A2 = ½ m vm2
US
vo = (k/m) A
x
0
m
x=0
x
Sinusoidal Nature of SHM
Sinusoidal equation = equation involving sine
x
and cosine
.
t
v
t
a
t
Example #1
A mass on a spring oscillates back &
forth with simple harmonic motion of
amplitude A. At what points during
its oscillation is the magnitude of the
acceleration of the block biggest?
1. When x = +A or -A
2. When x = 0 (i.e. zero displacement)
3. The acceleration of the mass is
constant.
Example #2
A mass on a spring oscillates back &
forth with simple harmonic motion of
amplitude A. At what points during
its oscillation is the magnitude of the
potential energy (U) of the spring
biggest?
1. When x = +A or -A
2. When x = 0 (i.e. zero displacement)
3. Potential energy of the spring is
constant.
Example #3

A mass on a spring oscillates back &
forth with simple harmonic motion of
amplitude A. If the mass takes 2
seconds to move from x = 0 to x = A,
what is the period of the oscillation?
(A) 0.5s (B) 1s (C) 2s (D) 4s (E) 8s
x=0
x=A
Example #4
A mass on a spring oscillates back & forth with
simple harmonic motion of amplitude A. A plot
of displacement (x) versus time (t) is shown
below. At what points during its oscillation is
the total energy (K+U) of the mass and spring a
maximum? (Ignore gravity).
1. When x = +A or -A (i.e. max displacement)
2. When x = 0 (i.e. zero displacement)
3. The energy of the system is constant.
x
+A
t
-A
Example #5
A mass on a spring oscillates back & forth with
simple harmonic motion of amplitude A. A plot
of displacement (x) versus time (t) is shown
below. At what points during its oscillation is
the speed of the block biggest?
1. When x = +A or -A (i.e. max displacement)
2. When x = 0 (i.e. zero displacement)
3. The speed of the mass is constant
x
+A
t
-A
Example #6
A spring oscillates back and forth on a frictionless
horizontal surface. A camera takes pictures of the
position every 1/10th of a second. Which plot best
shows the positions of the mass.
(A)
EndPoint
Equilibrium
EndPoint
(B)
EndPoint
Equilibrium
EndPoint
(C)
EndPoint
Equilibrium
EndPoint
What does moving in a circle have to
do with moving back & forth in a
straight line ??
vo
m
A

y
x
m
x=-A
x=0
x=A
• Mass m moving in uniform circular motion, with
constant speed vo [Angular velocity  = vo/A].
• x-component of position of the mass is same as
SHM of the spring-mass system.
vo
• Circular motion can be
used to describe SHM
of the spring-mass
system.
m
A

• Just pay attention to
the x-component of the
circular motion.
x =Acos  = Acos t
y
x
m
x=-A
x=0
x=A
• Total energy:
E = = ½mvo2 = ½kA2
OR vo = (k/m) A
vo = A  = 2A/T = 2Af
vo = A  = (k/m) A
OR  = (k/m)
vo
m
A y

x
x=-A
m
x=0
x=A
• Also, vo = 2A/T = (k/m) A
OR T = 2 (m/k)
When amplitude (A) is doubled, maximum velocity (vo) will ……
In the final analysis, period (T) depends only on m and k.
When amplitude (A) is doubled, period (T) will ……
x
+A
¼T
½T
¾T T
-A
Displacement x = Acos t = Acos (2/T)t
Velocity
v = -[A]sint
Acceleration
a = -[A2]cost
t
y
vo
vy = v0cos

vx = -v0sin


x
v
A
m
x=-A
x=0
x=A
0
¼T ½T ¾T T
-A
vo = A
vx = -v0sin = (-A)sin = (-A)sint
t
x=A F=-kA v=0 a=-kA/m
x=0 F=0 v=-vo a=0
x=-A F=kA v=0 a=kA/m
x=0 F=0 v=+vo
a=0
x=A F=-kA v=0 a=-kA/m
x
Displacement x = [A]cos t
+A
t
A
v
Velocity v = -[A]sint
t
A2
a
Acceleration a = -[A2]cost
t
Simple Harmonic Motion:
x(t) = [A]cos(t)
v(t) = -[A]sin(t)
a(t) = -[A2]cos(t)
xmax = A
vmax = vo = A
amax = A2
P10.25/28
The period of oscillation of an object in
an ideal spring-and-mass system is
0.50 s and the amplitude is 5.0 cm.
What is the speed at the equilibrium
point?
T = 0.5 s, A = 5.0 cm, vo = ?
vo = A,  = 2/T, vo = (2/T)A
P10.42/49
A body is suspended vertically from an ideal spring
of spring constant 2.5 N/m. The spring is initially
in its relaxed position. The body is then released
and oscillates about its equilibrium position. The
motion is described by
y = (4.0 cm)sin[(0.70 rad/s)t]
What is the maximum kinetic energy of the body?
Kmax = ½ mvo2
y =Asin t and  = (k/m)
A = 4.0 cm,  = 0.7 rad/s, m = k/2
vo = A,  = 2/T, vo = (2/T)A
 Simple
Harmonic Oscillator
=2f =2/T
x(t) = [A] cos(t)
v(t) = -[A] sin(t)
a(t) = -[A2] cos(t)
 For
a Spring F = kx
 amax = (k/m) A
 A2 = (k/m) A
 = sqrt(k/m)
=
k
m
T = 2
m
k
Vertical Mass and Spring
 If
we include gravity, there are two forces
acting on mass. With mass, new equilibrium
position has spring stretched d
SF = kd – mg = 0
d = mg/k
Let this point be y=0
SF = k(d-y) – mg
= -k y
Same as horizontal! SHO
New equilibrium position y = d
The Simple Pendulum
Simple pendulum – small object (pendulum bob)
attached to the end of a light inextensible string of
length L and swung through small displacements.
 Motion of the bob back and forth along arc length (s)
is a simple harmonic motion.
 Restoring force bringing bob back to eqlbm position
F = -mgsin


FT
m
x
s
mgsin 
m
mg
Types of Equilibrium
Physics 101: Lecture 19, Pg 37
The Simple Pendulum
 Restoring
force bringing bob back to eqlbm
position = F = -mgsin
 For small angles, sin   = s/L  x/L
F = -mg  = -mgx/L, i.e. F  x
Compare with F = -kx, means k  mg/L
OR k/m = g/L . But k/m = 2
So 2 = g/L
Since T = 2(m/k) = 2(L/g)
Period of simple pendulum T = 2(L/g)

FT
L
x
s
mgsin 
m
mg
The Simple Pendulum
Period of simple pendulum T = 2(L/g)
What should the length of a pendulum clock be for it
to measure time accurately?
You need period T = 1s = 2(L/g). [L ~ 25 cm]
What will happen
(a)If L is increased? [ie if L > 25 cm]
(b) If L is decreased? [ie if L < 25 cm]
(c) If g is decreased? (eg clock taken to the moon).
#1
A grandfather clock is running too fast.
To fix it, should the pendulum be
lengthened or shortened?
(A) lengthened
(B) shortened
(C) need to know the period
T = 2(L/g)
#2
A simple pendulum has a period of
oscillation of one second here on Earth.
On Mars (where g < 9.8 m/s2), the
pendulum will have a period of
oscillation
(A) greater than one second.
(B) equal to one second.
(C) less than one second.
T = 2(L/g)
Two simple pendulums A and B
Same lengths, Mass of A is twice the mass of B.
Equal vibrational amplitudes.
Periods TA and TB
Energies EA and EB
Choose the correct statement:
A) TA = TB and EA > EB
B) TA > TB and EA > EB
C) TA > TB and EA < EB
D)TA = TB and EA < EB
A mass attached to a spring oscillates with a
simple harmonic motion. If it takes 28 seconds
to make 40 cycles, what is the frequency of the
oscillation?
A.
B.
C.
D.
E.
7.0 Hz
1,120 Hz
0.70 Hz
2.80 Hz
1.43 Hz
1
2
3
4
5
6
7
8
9
10
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12
13
14
15
16
17
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19
20
21
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23
24
25
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28
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31
32
0%
33
34
0%
35
36
0%
37
38
0%
39
0%40
41
42
43
44
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46
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48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
A.
B.
C.
D.
E.
A mass on a spring oscillates with SHM. It
takes 0.82 sec to move from x = 0 to x =A.
What is the period of the oscillation?
20% 20% 20% 20%
20%
A.
B.
C.
D.
E.
1.22 s
0.21 s
0.41 s
1.64 s
3.28 s
1
2
3
4
5
6
7
8
9
10
11
12
13
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15
16
17
18
19
20
21
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55
56
57
58
59
60
61
62
63
64
65
A.
B.
C.
D.
E.
A mass on a spring oscillates with simple
harmonic motion. If the amplitude (A) is
increased by a factor 3, the maximum kinetic
energy of the mass will increase by a factor
A.
B.
C.
D.
E.
F.
1 G.2
1
1/3
1/6
1/9
3
6
93 4
5
6
7
8
9
10
11
31
21
22
23
24
25
26
27
28
29
30
41
42
43
44
45
46
47
48
49
50
61
62
63
64
65
12
13
14
15
32
0%
51
52
33
0%
53
34
35
36
0%
0%
54
55
56
A.
B.
C.
16
D.
17
19
20
37
38
39
0%
0%
57
58
59
40
0%
60
E.
18
F.
G.
A spring-mass system is in simple harmonic motion
and the velocity of the mass is plotted against time
as shown below. When is the potential energy of the
50%
spring at its minimum?
48%
A. 2s, 6s
B. 0s, 4s, 8s
C. 0s, 2s, 4s, 6s, 8s
D. 1s, 2s, 3s
E. The potential energy is constant throughout
A.
B.
0%
0%
C.
D.
3%
E.