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

2. Simple harmonic motion S.H.M.
A special kind of oscillation O X 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

3. Exploring the acceleration and displacement of S.H.M.
a– x graph
displacement
acceleration
time
displacement
acceleration

4. Exploring the acceleration and displacement of S.H.M.
a – x graph a x  
a ∝ -x
a ∝ 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

5. Note: 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. x
a(-) a(-) a(-) a(+) a(+) a(+)
a = 3k a = 2k a = k a = a = k a = 2k a = 3k
Note:
For S.H.M.,
direction of acceleration and displacement is always opposite to each other.

6. Equations of S.H.M. 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 q = r cos w t
Acceleration
Acceleration of P’ = component of acceleration of P along the x-axis
a = -rw2 cos q (-ve means directed towards O)
∴ a = -w2 x
The motion of P’ is simple harmonic. w
rw2 sin q
rw2 cos q P
a = rw2 q O r x
Y’ O’
P’ X’ x
q = wt

7. Equations of S.H.M. rw cos q rw w Period q
rw sin q
rw cos q q rw w
Period
The period of oscillation of P’
= time for P to make one revolution
T = 2p / angular speed
∴ T = 2p/w P q O
Velocity
Velocity of P’
= component of velocity of P along the x-axis
v = -rw sin q = -rw sin w t r x
Y’ O’
P’ X’ x

8. Equations of S.H.M. rw cos q rw w q Motion of P’
rw sin q
rw cos q q rw w
Motion of P’
Amplitude of oscillation = Radius of circle
⇒ A = r
Displacement x:
x = A cos q
Velocity v:
v = -wA sin q
Acceleration a:
a = -w2A cos q P q A O r x
Y’ O’
P’ X’ x

9. Relation between the amplitude of oscillation A and x, w, and v:
Note: Maximum speed = wA at x = 0
(at centre of oscillation / equilibrium position).

10. Example 1 A particle moving with S. H. M
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
4 ms-1
Solution:
By v2 = w2(A2 – x2)
when x = 3 cm, v = 4 cm s-1,
x = 4 cm, v = 3 cm s-1.
42 = w2(A2 – 32) --- (1)
32 = w2(A2 – 42) --- (2)
(1)/(2):
16/9 = (A2 – 9) / (A2 – 16)
9A2 – 81 = 16 A
A2 = 25
A = 5 cm
∴ amplitude = 5 cm O
3 ms-1

11. (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

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 T/4 T/2 3T/4 T T/4 T/2 3T/4 T A wA
w2A x v a t
w2A x y v a A wA w
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.

14. Phase difference of x-t, v-t and a-t graphs
w2A x y v a A wA w
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)

15. 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) x
K.E. 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
∵ w2 = k/m
∴ P.E. = ½ mw2x2
x = A
x = -A
Centre of oscillation ix x
P.E. A -A
P.E. is maximum at extreme points.
(Spring is most stretched.)
P.E. is minimum when x = 0.
(Spring is not stretched)

17. Total energy = K.E. + P.E. (constant) A -A Energy K.E. P.E. ½ mw2A2A -A
Energy
K.E.
P.E.
½ mw2A2
Total energy x

18. Energy and time From equation of S.H.M. K.E. = P.E. =
Total energy = K.E. + P.E.
(constant)

19. Total energy
Energy
P.E. = ½ mw2A2cos2wt
T/ T/2 3T/4 T Time
½ mw2A2
K.E. = ½ mw2A2sin2wt
t = 0
t = T/2
Centre of oscillation 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 the extension
By Newton’s second law
T = -ma
kx = -ma
a = -(k/m)x
which is in the form of a = -w2x
Hence, the motion of the mass is simple harmonic, and
w2 = k/m
Period of oscillation
Natural length (l)
Extension (x) T
Centre of oscillation ix

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
Extension at equilibrium
Displacement from equilibrium x e mg T’
Centre of oscillation
In equilibrium
Spring unstretched
Displaced from equilibrium
which is in the form of a = -w2x
Hence, the motion of the mass is simple harmonic
and w2 = k/m.
Period of oscillation

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
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 T2 m
Line of best fit

24. ∴ ∴ or T2 slope = y-intercept = x ∴ effective mass ms y-intercept m
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 = 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
F1 = k1x
F2 = k2x F
where k = k1 + k2

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
F1 = k1x1
F2 = k2 x2 F
∵ x = x1 + x2 or
where

27. Case 2: Springs in series
Note: the equivalent force constant of the
system is k where
F1 = k1x1
When the mass is set into vibration, the oscillation is simple harmonic.
Period of oscillation
F2 = k2 x2 F .
where

28. Case 3: The mass is connected by two springs on both sides
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
where k = k1 + k2.

29. Simple pendulum Resolve tangentially (perpendicular to the string)
mg sin q = -ma
where a is the acceleration along the arc
If q is small (i.e. <10o), sin q ≈ q and x ≈ lq ,
mg sin q = -ma becomes
mg q = -ma
a = -g(x/l) = -(g/l)x
which is in the form of a = -w2x
Hence, the motion of the bob is simple harmonic and w2 = g/l
Period of oscillation T q l T x P O
mg sin q
mg cos q mg

30. (a) maximum magnitude of velocity = wA = p(5) = 5p cm s-1
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:
(a) maximum magnitude of velocity
= wA = p(5) = 5p cm s-1
(b) maximum magnitude of velocity
= w2A = 5p2 cm s-2