Simple harmonic motion

- Vibration / Oscillation
- to-and-fro repeating movement

Simple harmonic motion S.H.M.

- A special kind of oscillation

- X, Y extreme points
- O centre of oscillation / equilibrium position
- A amplitude

A special relation between the displacement and

acceleration of the particle

Exploring the acceleration and displacement of

S.H.M.

Exploring the acceleration and displacement of

S.H.M.

a x graph

?

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

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.

Note For S.H.M., direction of acceleration and

displacement is always opposite to each other.

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

P

q

O

r

x

Y O

P X

q wt

Equations of S.H.M.

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

Equations of S.H.M.

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

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).

Example 1A 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 A2 - 256
- A2 25
- A 5 cm
- ? amplitude 5 cm

O

3 ms-1

(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

Isochronous oscillations

- Definition period of oscillation is independent

of its amplitude. - Examples Masses on springs and simple pendulums

Phase difference of x-t, v-t and a-t graphs

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.

Phase difference of x-t, v-t and a-t graphs

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)

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)

Note 1. K.E. is maximum when x 0 (equilibrium

position) 2. K.E. is minimum at extreme points

(speed 0)

- Potential energy
- P.E. ½ kx2
- ? w2 k/m
- ? P.E. ½ mw2x2

- P.E. is maximum at extreme points.
- (Spring is most stretched.)
- P.E. is minimum when x 0.
- (Spring is not stretched)

Total energy K.E. P.E.

(constant)

x

Energy and time

From equation of S.H.M.

K.E.

P.E.

Total energy K.E. P.E.

(constant)

(No Transcript)

Examples of S.H.M.Mass on spring horizontal

oscillationHookes law F kx where k is the

force constant and x is the extension

- By Newtons 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

Mass on spring vertical oscillation

- At equilibrium,
- T mg
- ke mg
- Displaced from equilibrium,
- T mg -ma
- k(e x) mg -ma

- which is in the form of a -w2x
- Hence, the motion of the mass is simple harmonic
- and w2 k/m.
- Period of oscillation

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.

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.

? effective mass ms

In theory, effective mass of a spring is about ?

of the mass of string. Usually, we would neglect

the effective mass for simplicity.

Combined Springs OscillationCase 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

F2 k2x

F1 k1x

F

where k k1 k2

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

Case 2 Springs in series

F1 k1x1

When the mass is set into vibration, the

oscillation is simple harmonic. Period of

oscillation

F2 k2 x2

F

.

Case 3 The mass is connected by two springs on

both sides

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

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. lt10o), 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

A

q

l

T

P

x

O

mg sin q

mg cos q

mg

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