Simple harmonic motion
Vibration / Oscillation

to-and-fro repeating
movement
1
Simple harmonic motion S.H.M.

A special kind of oscillation
X



O
Y
A
X, Y : extreme points
O: centre of oscillation / equilibrium position
A: amplitude
A special relation between
the displacement and
acceleration of the particle
2
Exploring the acceleration and displacement of
S.H.M.
a– x graph
1.5
0.3
acceleration
displacement
1
0.2
acceleration
0.5
0.1
displacement
0
0
0
45
90
135
180
225
270
315
360
405
- 0.5
- 0.4
- 0.3
- 0.2
- 0.1
0
0.1
0.2
0.3
0.4
time
- 0.1
- 0.5
- 0.2
-1
- 0.3
- 1.5
3
0.5
Exploring the acceleration and displacement
of S.H.M.
a – x graph
a
a ∝ -x
x

x
Definition of Simple harmonic motion (S.H.M.)
An oscillation is said to be an S.H.M if
(1)
the magnitude of acceleration is directly proportional to
distance from a fixed point (centre of oscillation), and
4
Definition of Simple harmonic motion (S.H.M.)
An oscillation is said to be an S.H.M if
(1)
the magnitude of acceleration is directly proportional to
distance from a fixed point, and
(2)
the acceleration is always directed towards that point.
a(-)
a(-)
a(-)
a(+)
a(+)
a(+)
a = 3k a = 2k a = k
a = 0 a = k a = 2k a = 3k
x
-3
-2
-1
0
1
2
3
Note:
For S.H.M.,
direction of acceleration and displacement is always opposite to
each other.
5
For the projection P’
• Moves from X’ to O’ to Y’ and
returns through O’ to X’ as P
w
2
rw cos q
completes each revolution.
P
Displacement
a = rw2
rw2 sin q
• Displacement from O
q
• x = r cos q = r cos w t
O
Acceleration
Acceleration of P’ = component
r
of acceleration of P along the
x x-axis
O’
P’ X’
• a = -rw2 cos q (-ve means
x
directed towards O)
∴ a = -w2 x
q = wt
The motion of P’ is simple
harmonic.
Equations of S.H.M.
Y’
6
Equations of S.H.M.
rw
q
w
Period
The period of oscillation of P’
= time for P to make one
revolution
T = 2p / angular speed
∴ T = 2p/w
P
rw sin q
O
rw cos q
q
r
x
Y’
O’
P’ X’
x
Velocity
Velocity of P’
= component of velocity of P
along the x-axis
v = -rw sin q = -rw sin w t
7
Equations of S.H.M.
rw
q
w
P
rw sin q
O
rw cos q
q
A
r
Y’
O’
P’ X’
x
Motion of P’
 Amplitude of oscillation
= Radius of circle
⇒A = r
 Displacement x:
x = A cos q
 Velocity v:
v = -wA sin q
x
 Acceleration a:
a = -w2A cos q
8
Relation between the amplitude
of oscillation A and x, w, and v:
v
x
 sin q  
, cosq 
Aw
A
2
2
 v  x
 
   1
 Aw   A 
 v 2  w 2 x 2  A2w 2

 v 2  w 2 A2  x 2

Note: Maximum speed = wA at x = 0
(at centre of oscillation / equilibrium position).
9
Example 1
A particle moving with S.H.M. has velocities of 4 cm s-1 and
3 cm s-1 at distances of 3 cm and 4 cm respectively from its
equilibrium. Find
(a)
the amplitude of the oscillation
-1
4
ms
 Solution:
By v2 = w2(A2 – x2)
O
when
x = 3 cm, v = 4 cm s-1,
3 ms-1
-1
x = 4 cm, v = 3 cm s .
42 = w2(A2 – 32) --- (1)
32 = w2(A2 – 42) --- (2)
(1)/(2):
16/9 = (A2 – 9) / (A2 – 16)
9A2 – 81 = 16 A2 - 256
A2 = 25
A = 5 cm
∴ amplitude = 5 cm
10
(b)
the period,
(c)
the velocity of the particle as it passes through
the equilibrium position.

(b) Put A = 5 cm into (1)
42 = w2(52 – 32)
w2 = 1⇒ w = 1 rad s-1
T = 2p/w = 2p s
(c) at equilibrium position, x = 0
By v2 = w2(A2 – x2)
v2 = 12(52 – 02)
v = 5 cm s-1
11
Isochronous oscillations
Definition: period of oscillation is
independent of its amplitude.
 Examples: Masses on springs and simple
pendulums

12
Phase difference of x-t, v-t and a-t graphs
x
v
x  A cosw t
A
t
0 T/4 T/2 3T/4 T
y
v  wA sin w t
wA
t
0 T/4 T/2 3T/4 T
w
x
v
A wA w2A
a
a  w 2 A cosw t
w2A
t
0 T/4 T/2 3T/4 T
Vectors x, v and a rotate with the
same angular velocity w.
Their projections on the y-axis give
the above x-t, v-t and a-t graphs.
a
13
Phase difference of x-t, v-t and a-t graphs
y
w
x
v
A wA w2A
a
Note:
1 a leads v by 90o or T/4.
(v lags a by 90o or T/4)
2 v leads x by 90o or T/4.
(x lags v by 90o or T/4)
3 a leads x by 180oor T/2.
(a and x are out of phase or
antiphase)
14
Energy of S.H.M.
(Energy and displacement)



From equation of S.H.M.
v2 = w2(A2 – x2),
∴ K.E. = ½ mv2 = ½ mw2(A2 – x2)
K.E.
x
-A
0
A
Note:
1.
K.E. is maximum when x = 0 (equilibrium position)
2.
K.E. is minimum at extreme points (speed = 0)
15




Potential energy
P.E. = ½ kx2
∵ w2 = k/m
∴ P.E. = ½ mw2x2
Centre of oscillation
ix
P.E.
x = -A
x=A
x
-A


0
A
P.E. is maximum at extreme points.
(Spring is most stretched.)
P.E. is minimum when x = 0.
(Spring is not stretched)
16
Total energy = K.E. + P.E.


1
1
1
2
2
2
2 2
 mw A  x  mw x  mw 2 A2 (constant)
2
2
2
Energy
½ mw2A2
Total energy
P.E.
K.E.
-A
0
A
x
17
Energy and time
From equation of S.H.M.
x  A cosw t and v  wA sin w t
1
1
1
2
2


mv

m

w
A
sin
w
t

mw 2 A 2 sin 2 w t
K.E. =
2
2
2
1
1
1
2
2 2
2


m
w
x

m
w
A
cos
w
t

mw 2 A 2 cos 2 w t
P.E. =
2
2
2
Total energy = K.E. + P.E.

1
1
1
mw 2 A 2 sin 2 w t  mw 2 A 2 cos 2 w t  mw 2 A 2 sin 2 w t  cos 2 w t
2
2
2
1
 mw 2 A2
(constant)
2

18

Energy
Total energy
½ mw2A2
P.E. = ½ mw2A2cos2wt
K.E. = ½ mw2A2sin2wt
0
T/4
T/2 3T/4
T
Time
Centre of oscillation
ix
t = T/2
t=0
19
Examples of S.H.M.
Mass on spring – horizontal oscillation
Hooke’s law: F = kx where k is the force constant and x is
the extension
 By Newton’s second law
Centre of oscillation
T = -ma
kx = -ma
a = -(k/m)x
ix T
 which is in the form of a = -w2x
 Hence, the motion of the mass
is simple harmonic, and
Natural Extension
w2 = k/m
length (l)
(x)
 Period of oscillation
2p
m

 2p
w
k
20
Mass on spring – vertical oscillation

Natural
length (l)
Extension at
e
equilibrium
Displacement x
from
equilibrium

T’
T
Centre of
oscillation
mg
mg
Displaced
In
Spring
from
unstretched equilibrium
equilibrium
At equilibrium,
T’ = mg
ke = mg
Displaced from equilibrium,
T – mg = -ma
k(e + x) – mg = -ma
mg
k(
 x)  mg  ma
k
k
a x
m


which is in the form of a = -w2x
Hence, the motion of the mass
is simple harmonic
and w2 = k/m.
Period of oscillation
2p
m

 2p
21
w
k
Effective mass of spring



Not only the mass oscillates when it
is released, but also the spring itself.
The period of oscillation is affected
by the mass of the spring.
Hence, the equation T  2p
m
k
should be rewritten as T  2p
m  ms
k
where ms is the effective mass
of the spring.
22
Measurement of effective mass of spring


To find the effective mass, we can do an experiment by
using different masses m and measure the
corresponding periods T.
Use the results to plot a graph of T2 against m which is a
straight line but it does not pass through the origin.
T2
x
Line of
best fit
x
x
x
x
m
23
∴
∴
m  ms
k
m  ms
T 2  4p 2
k
T  2p
2
2
4
p
4
p
or T 2 
m
ms
k
k
T2
slope =
x
x
x
4p 2
k
y-intercept =
x
x
m
4p 2
ms
k
∴ effective mass ms
k

 y-intercept
2
4p
In theory, effective mass of a spring is about ⅓ of the mass of string.
Usually, we would neglect the effective mass for simplicity.
24
Combined Springs Oscillation
Case 1: Springs in parallel

F1 = k1x
F2 = k2x

F


Let x be the common extension
of the spring.
∵ the springs are in parallel,
∴ upward force F = F1 + F2
F = k1x + k2x = (k1 + k2)x
Note: k1 + k2 is the equivalent
force constant of the system.
When the mass is set into
vibration, the oscillation is simple
harmonic.
Period of oscillation
m
T  2p
k
where k = k1 + k2
25
Case 2: Springs in series

F1 = k1x1



F2 = k2 x2

Let x1 and x2 be the extensions of the first
and the second spring respectively.
The total extension x = x1 + x2
∵ the springs are in series,
∴ upward force F = F1 = F2
F = k 1 x1 = k2 x2
 x1 
F
F
1
1

k1 k 2
and x2 
x = x1 + x2
F F
x 
k1 k2
or F  kx where

x
∵
F
k1
F
k2
1 1
1
 
k k1 k 2
26
Case 2: Springs in series
Note: the equivalent force constant of the
F1 = k1x1
system is k where
1 1
1
 
k k1 k 2
When the mass is set into vibration, the
oscillation is simple harmonic.
F2 = k2 x2
F
Period of oscillation T  2p
.
where
m
k
1 1
1
 
k k1 k 2
27
Case 3: The mass is connected by two springs
on both sides
Equilibrium position
Force
Force
constant k1
constant k2
T1
T2
Under
extension
x
Suppose the springs are
initially unstretched.
Under
compression
When the mass is displaced to the right by x,
1st spring is extended but 2nd spring is compressed.
Resultant force on the bob F = T1 + T2
∴ F = k1x + k2x = (k1 + k2)x
Note: k1 + k2 is the equivalent force constant of the system.
The oscillation is simple harmonic.
Period of oscillation
T  2p
m
k
where k = k1 + k2.
28
Simple pendulum
Resolve tangentially (perpendicular to the
string)
mg sin q = -ma
q
where a is the acceleration along the arc
l
If q is small (i.e. <10o), sin q ≈ q and x ≈ lq ,
T
mg sin q = -ma becomes
mg q = -ma
P
x
a = -g(x/l) = -(g/l)x
O
which is in the form of a = -w2x
mg sin q
 Hence, the motion of the bob is simple
2 = g/l
harmonic
and
w
mg cos q
2p
l
mg
 Period of oscillation T 
 2p
w
g
A

29
A simple pendulum has a period of 2 s and an
amplitude of swing 5 cm.
Calculate the maximum magnitudes of
(a)
velocity, and
(b)
acceleration of the bob.

Solution:
By T 


2p
w
2
2p
w
 w  p rad s 1
(a) maximum magnitude of velocity
= wA = p(5) = 5p cm s-1
(b) maximum magnitude of velocity
= w2A = 5p2 cm s-2
30