Simple harmonic motion

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Simple harmonic motion

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

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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
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Exploring the acceleration and displacement of
S.H.M.
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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
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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.
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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
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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
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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
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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).
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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
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(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

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Isochronous oscillations

• Definition period of oscillation is independent
  of its amplitude.
• Examples Masses on springs and simple pendulums

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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.
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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)
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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)
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• 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)

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

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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

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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.
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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.

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? 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.
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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
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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

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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
.
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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
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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
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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