375823749-1-9-Simple-Harmonic-Motion.ppt

1
Simple harmonic motion
Vibration / Oscillation
 to-and-fro repeating
movement

2
Simple harmonic motion S.H.M.
 A special kind of oscillation
X Y
O
 X, Y : extreme points
 O: centre of oscillation / equilibrium position
 A: amplitude
A
A special relation between
A special relation between
the displacement and
the displacement and
acceleration of the particle
acceleration of the particle

3
Exploring the acceleration and displacement of
S.H.M.
- 1 .
5
- 1
- 0 .
5
0
0 .
5
1
1 .
5
0 4 5 9 0 1 3 5 1 8 0 2 2 5 2 7 0 3 1 5 3 6 0 4 0 5
time
displacement
acceleration
a
a–
– x
x graph
graph
- 0 .
3
- 0 .
2
- 0 .
1
0
0 .
1
0 .
2
0 .
3
- 0 .
5 - 0 .
4 - 0 .
3 - 0 .
2 - 0 .
1 0 0 .
1 0 .
2 0 .
3 0 .
4 0 .
5
displacement
acceleration

4
Exploring the acceleration and displacement
of S.H.M.
a – x graph a
x
a
a ∝
∝ x
x

a
a ∝
∝ -x
-x
Definition of Simple harmonic motion (S.H.M.)
An oscillation is said to be an S.H.M if
(1)
(1) the
the magnitude
magnitude of acceleration is directly proportional to
of acceleration is directly proportional to
distance from a fixed point (centre of oscillation), and
distance from a fixed point (centre of oscillation), and


5
-3 -2 -1 0 1 2 3
-3 -2 -1 0 1 2 3
x
a
a(-)
(-) a
a(-)
(-) a
a(-)
(-) a
a(+)
(+) a
a(+)
(+) a
a(+)
(+)
a
a = 3
= 3k
k a
a = 2
= 2k
k a
a =
= k
k a
a = 0
= 0 a
a =
= k
k a
a = 2
= 2k
k a
a = 3
= 3k
k
(1)
(1) the magnitude of acceleration is directly proportional to
the magnitude of acceleration is directly proportional to
distance from a fixed point, and
distance from a fixed point, and
(2)
(2) the acceleration is always
the acceleration is always directed towards that point
directed towards that point.
.
Note:
For S.H.M.,
direction of acceleration
acceleration and displacement
displacement is always opposite
to each other.

6
Equations of S.H.M.


P
O
r
a
a =
= r
r
2
2
r2
sin 
r2
cos 
Y’ O’ P’
P’ X’
x
x
For the projection P’
• Moves from X’ to O’ to Y’ and
returns through O’ to X’ as P
completes each revolution.
Displacement
• Displacement from O
• x = r cos  = r cos  t
Acceleration
Acceleration of P’ = component
of acceleration of P along the
x-axis
• a = -r2
cos  (-ve means dire
cted towards O)
∴ a
a = -
= -
2
2
x
x
The motion of P’ is simple har
monic.
 = t

7
Equations of S.H.M.


P
O
r
Y’ O’ P’
P’ X’
x
x
Period
The period of oscillation of P’
= time for P to make one
revolution
T = 2 / angular speed
∴ T = 2
T = 2
/
/

Velocity
Velocity of P’
= component of velocity of P al
ong the x-axis
v
v = -
= -r
r
 sin
sin 
 = -
= -r
r
 sin
sin 
t
t
r
rsin

rcos



8
Equations of S.H.M.


P
O
r
Y’ O’ P’
P’ X’
x
x
Motion of P’
 Amplitude of oscillation
= Radius of circle
⇒
⇒ A
A =
= r
r
 Displacement x:
x
x = A cos
= A cos 

 Velocity v:
v
v = -
= -
A
A sin
sin 

 Acceleration a:
a
a = -
= -
2
2
A
A cos
cos 

r
rsin

rcos


A

9
Relation between the amplitude
of oscillation A and x, , and v:
 
2
2
2
2
2
2
2
2
2
2
2
1
cos
,
sin
x
A
v
A
x
v
A
x
A
v
A
x
A
v

































Note: Maximum speed = A at x = 0
(at centre of oscillation / equilibrium positio
n).

10
Example 1
Example 1
A particle moving with S.H.M. has velocities of 4 cm s
A particle moving with S.H.M. has velocities of 4 cm s-1
-1
and
and
3 cm s
3 cm s-1
-1
at distances of 3 cm and 4 cm respectively from its
at distances of 3 cm and 4 cm respectively from its
equilibrium. Find
equilibrium. Find
(a)
(a) the amplitude of the oscillation
the amplitude of the oscillation
 Solution:
By v2
= 2
(A2
– x2
)
when x = 3 cm, v = 4 cm s-1
,
x = 4 cm, v = 3 cm s-1
.
42
= 2
(A2
– 32
) --- (1)
32
= 2
(A2
– 42
) --- (2)
(1)/(2):
16/9 = (A2
– 9) / (A2
– 16)
9A2
– 81 = 16 A2
- 256
A2
= 25
A = 5 cm
∴ amplitude = 5 cm
4 ms-1
3 ms-1
O

11
(b)
(b) the period,
the period,
(c)
(c) the velocity of the particle as it passes through
the velocity of the particle as it passes through
the equilibrium position.
the equilibrium position.
 (b) Put A = 5 cm into (1)
42
= 2
(52
– 32
)
2
= 1⇒  = 1 rad s-1
T = 2/ = 2 s
(c) at equilibrium position, x = 0
By v2
= 2
(A2
– x2
)
v2
= 12
(52
– 02
)
v = 5 cm s-1

12
Isochronous oscillations
 Definition: period of oscillation is
independent of its amplitude.
 Examples: Masses on springs and simple
pendulums

13
Phase difference of x-t, v-t and a-t graphs
0 T/4 T/2 3T/4 T 0 T/4 T/2 3T/4 T 0 T/4 T/2 3T/4 T
A A 2
A
x v a
t t t
t
A
x 
cos
 t
A
a 
 cos
2


t
A
v 
 sin


2
A
x
y
v
a
A A

Vectors x, v and a rotate with the sa
me angular velocity .
Their projections on the y-axis give
the above x-t, v-t and a-t graphs.

14
Phase difference of x-t, v-t and a-t graphs
2
A
x
y
v
a
A 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 180o
or T/2.
(a and x are out of phase or ant
iphase)

15
Energy of S.H.M.
(Energy and displacement)
 From equation of S.H.M.
 v2
= 2
(A2
– x2
),
 ∴ K.E. = ½ mv2
= ½ m2
(A2
–
x2
)
x
K.E.
0 A
-A
Note:
1. K.E. is maximum when x = 0 (equilibrium position)
2. K.E. is minimum at extreme points (speed = 0)

16
 Potential energy
 P.E. = ½ kx2
 ∵ 2
= k/m
 ∴ P.E. = ½ m2
x2
 P.E. is maximum at extreme points.
(Spring is most stretched.)
 P.E. is minimum when x = 0.
(Spring is not stretched)
x
P.E.
0 A
-A
x = A
x = -A
Centre of oscillation
ix

17
Total energy = K.E. + P.E.
x
Total energy
0 A
-A
Energy
K.E.
P.E.
½ m2
A2
  2
2
2
2
2
2
2
2
1
2
1
2
1
A
m
x
m
x
A
m 

 


 (constant)

18
Energy and time
t
A
v
and
t
A
x 

 sin
cos 


  t
A
m
t
A
m
mv 


 2
2
2
2
2
sin
2
1
sin
2
1
2
1



  t
A
m
t
A
m
x
m 



 2
2
2
2
2
2
2
cos
2
1
cos
2
1
2
1


2
2
2
1
A
m

From equation of S.H.M.
K.E. =
P.E. =
Total energy = K.E. + P.E.
(constant)
 
t
t
A
m
t
A
m
t
A
m 





 2
2
2
2
2
2
2
2
2
2
cos
sin
2
1
cos
2
1
sin
2
1





19
Total energy
Energy
P.E. = ½ m2
A2
cos2
t
0 T/4 T/2 3T/4 T Time
½ m2
A2
K.E. = ½ m2
A2
sin2
t
t = 0
t = T/2
ix

20
Examples of S.H.M.
Mass on spring – horizontal oscillation
Hooke’s law: F = kx where k is the force constant and x is t
he extension  By Newton’s second law
T = -ma
kx = -ma
a = -(k/m)x
 which is in the form of a = -2
x
 Hence, the motion of the mass i
s simple harmonic, and
2
= k/m
 Period of oscillation
Natural
length (l)
Extension
(x)
T
ix
k
m



2
2



21
Mass on spring – vertical oscillation
 At equilibrium,
T’ = mg
ke = mg
 Displaced from equilibrium,
T – mg = -ma
k(e + x) – mg = -ma
Natural
length (l)
T
T
Extension at
equilibrium
Displacement
from
equilibrium
x
e
mg
mg
T’
T’
mg
mg
Centre of
oscillation
In
equilibrium
Spring unstr
etched
Displaced
from
equilibrium which is in the form of a = -2
x
 Hence, the motion of the mass i
s simple harmonic
and 2
= k/m.
ma
mg
x
k
mg
k 


 )
(
x
m
k
a 

k
m



2
2



22
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 k
m
T 
2

k
m
m
T s

 
2
should be rewritten as
where ms is the effective mass
of the spring.

23
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.
x
x
x
x
x
T2
m
Line of
best fit

24
k
m
m
T s

 2
2
4 s
m
k
m
k
T
2
2
2 4
4 



∴ or
k
m
m
T s

 
2
∴
x
x
x
x
x
T2
m
k
2
4
slope =
s
m
k
2
4
y-intercept =
∴ effective mass ms

 2
4
k
y-intercept
In theory, effective mass of a spring is about ⅓ of the mass of
string.
Usually, we would neglect the effective mass for simplicity.

25
Combined Springs Oscillation
Case 1: Springs in parallel
 Let x be the common extension of
the spring.
∵ the springs are in parallel,
∴ upward force F
F = F
F1
1 + F
F2
2
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.
F
F
F2 =k2x
where k = k1 + k2
k
m
T 
2

F1 =k1x

26
Case 2: Springs in series
 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 = k1x1 = k2x2
F
F
F
F1
1 =
= k
k1
1x
x1
1
F2 = k2 x2
2
2
1
1
k
F
x
and
k
F
x 


 ∵ x = x1 + x2
2
1 k
F
k
F
x 

2
1
1
1
k
k
x
F


 kx
F 
2
1
1
1
1
k
k
k


or where

27
Case 2: Springs in series
F
F
F
F1
1 =
= k
k1
1x
x1
1
F2 = k2 x2
Note: the equivalent force constant of the
system is k where
2
1
1
1
1
k
k
k


k
m
T 
2

When the mass is set into vibration, the
oscillation is simple harmonic.
Period of oscillation
2
1
1
1
1
k
k
k


where
.

28
Case 3: The mass is connected by two springs
on both sides
T1
Equilibrium position
T2
Under
compression
Under
extension
Force
constant k1
Force
constant k2
x
Suppose the springs are
initially unstretched.
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
k
m
T 
2
 where k = k1
+ k2
.

29
Simple pendulum
 Resolve tangentially (perpendicular to the string)
mg sin  = -ma
where a is the acceleration along the arc
If  is small (i.e. <10o
), sin  ≈  and x ≈ l ,
mg sin  = -ma becomes
mg  = -ma
a = -g(x/l) = -(g/l)x
which is in the form of a = -2
x
 Hence, the motion of the bob is simple harmonic an
d 2
= g/l
 Period of oscillation T
O
P
A
mg
mg cos
mg sin
l

x
T
g
l



2
2



30
A simple pendulum has a period of 2 s and an
A simple pendulum has a period of 2 s and an
amplitude of swing 5 cm.
amplitude of swing 5 cm.
Calculate the maximum magnitudes of
Calculate the maximum magnitudes of
(a)
(a) velocity, and
velocity, and
(b)
(b) acceleration of the bob.
acceleration of the bob.
 Solution:
 (a) maximum magnitude of velocity
= A = (5) = 5 cm s-1
 (b) maximum magnitude of velocity
= 2
A = 5
cm s-2
1
s
rad
2
2
2
By 




 





T

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