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S.H.M. P.1
PHYSICS
Simple Harmonic Motion
1. Mechanical Oscillations
- A system can be set into ( vibration ) oscillation if it has
(i) elasticity ( or springiness ) for storing P.E., and
(ii) mass ( or inertia ) for possessing K.E.
- An oscillation represents the continual interchange of P.E. and K.E.
- Example : Consider the case " the body on the lower end of a spiral spring is
pulled down and released.
i) OA = OB = amplitude of the oscillation
ii) it is a periodic motion in which a body continually retraces its path at equal time interval
iii) its motion is isochronous. i.e. the period is independent of the amplitude.
2. Simple Harmonic Motion ( S.H.M. )
a) Definition
- Motion of a body is said to be simple harmonic if its acceleration is proportional and opposite to
its displacement from a fixed point.
a -x
a = - ( constant ) x
d 2x
dt 2
x  displaceme nt
a
b) Important points of S.H.M.
- T does not depend on amplitude. ( isochronous )
- Equation of motion is
a = - ( constant ) x
- Any periodic motion which is not simple harmonic can be analysed into its simple harmonic
components.
3. Graphical representation of S.H.M.
- Consider a point P moves round a circle of radius A and centre O.
S.H.M. P.2
w = uniform angular velocity
v = constant speed = Aw
a) Motion of N ( x-component )
- N is the projection of P on x-axis.
i) displacement
x = A cos (  +  )
........... ( 1 )
ii) acceleration
a=
(v2) / A =
A w2
ax =  Aw2 cos(   )
( towards the centre )
...........( 2 )
From (1), (2), we have
ax =  w 2 x
- It implies that N has S.H.M.
- it can also be obtained by direct differentiation
iii) Period
2A
v
2
T
w
T
In this S.H.M. , T is constant since w is a constant.
iv) velocity
x = A cos (  +  )
dx
d
Vx 
  A sin(    )
dt
dt
Vx   Aw sin(    )
  Aw(1  cos (   ))
2
(V x 
dx
)
dt
1
2
x
  Aw 1  ( ) 2
A
 w A2  x 2
- +/- means that at a certain position, the velocity of N has two values. They are equal but
opposite.
S.H.M. P.3
v) Critical Conditions
1. acceleration
ax = - w2 x
max.
ax = +/- w2 A
at x = +/- A ( limiting points or extreme points.)
ax = 0
at
x = 0 ( at equilibrium point or mid-point )
2. velocity
vx= +/- w
max.
A2 - x 2
vx = +/- w A
at x = 0
vx = 0
at x = +/- A
vi) Phase difference
- the graph of
x vs t, v vs t, a vs t
can be plotted as below.
- displacement has a phase lead of 90 w.r.t. velocity. or
velocity has a phase lag of 90 w.r.t. displacement.
b) Motion of M ( y-component )
- M is the projection of P on y-axis.
S.H.M. P.4
y = A sin (  +  )
i) displacement
Similarly, ii) velocity
vy = +/- w
A2  y 2
ay = - w2 y
iii) acceleration
iv) period
T = 2 / w
c) Solution of differential equation a = - w2 x
x = A cos (  +  )
x = A sin (  +  )
Either
or
A = amplitude
 = phase angle
 = wt
is used to represent the solution of this differential equation.
Example.
1. Given x = xo , v = vo at t = 0 . Determine  and A.
 x  A cos( wt   )
 x0  A cos  ................(1)
( x  x0 at
t  0)
sin ce v x   Aw sin( wt   )
v0   Aw sin  ............(2)
(2)
(1)
v0
wx0
  tan 1 ( v0
(v x  v0
at
t  0)
tan   
wx0 )
(1) 2
cos 2   x0
( 2) 2
sin 2   v0
2
2
A 2 ......................(3)
A 2 w 2 ....................(4)
(3)  (4)
1  x0
2
A 2  v0
A  x0  v0
2
2
2
A2 w2
w2
- If you use the solution x = A sin (  +  ), A will be found unchanged but the phase angle 
has a difference 90˚
S.H.M. P.5
2. Find the time taken for a particle moving in S.H.M. from
period of oscillation is 12s.
method 1
sin  
method 2
since x  A cos( wt   )
A
  A cos 
2
1
cos  
2
  60
1
 A  A cos( wt  60)
2
1
2
  cos( wt 
)
2
6
2
120  wt 
6
2 2
2

t
3
T
6
T
t   2s
6
1
1
A A
2
2
  30
time required 
(1/2)A to (-1/2)A. Given that the
60
T
360
T 6
 2s
4. Application to include the simple pendulum and loaded spring.
-
it states that the tension in the string is direct proportional to the extension.
F = kx
k increase stiff spring
k decrease less stiff spring
5. Application to include the simple pendulum and loading spring
a) Mass attached to a horizontal spring
ma = -kx
a = -(k/m)x
hence, we have
w2 = k/m
T = 2π/w
T  2
-
(for S.H.M.)
m
k
what is the relation between the period and mass of the object as well as the spring constant?
S.H.M. P.6
b) Loaded Spring
consider the body at the equilibrium state
mg = k1
………………....(1)
Now consider the body have a displacement x from the equilibrium position
By Newton’s 2nd law,
ma  (k ( x  l )  mg )
ma  kx  kl  mg
...........(2)
From (1) and (2), we have
ma   kx
k
x
m
k
w2 
( forS .H .M )
m
2
m
T
 2
w
k
a
Experimentally, the graph of T vs m does not pass through the origin as we might expect
from the equation. It is due to the mass of spring itself have been neglected.
c) Simple pendulum
-
i) Period of oscillation
By Newton's 2nd law,
F   mg sin 
ma   mg sin 
a   g sin 
a   g
x
 g

(when θ is small, sin θ = θ)
S.H.M. P.7
w2 
T
g

2
w
T  2

g
(condition : θ must be small)
ii) measurement of g
- measure T of a simple pendulum for different values of l and plotting a graph of
l Vs T2.
6. Energy of S.H.M.
- If there is no work done against resistive forces ( i.e. undamped ), the total energy is constant.
a) Kinetic energy ( Ek )
A2  x 2
v= +/- w
1
Ek = mv 2
2
Ek 
1
mw2 ( A 2  X 2 )
2
(K.E. at displacement x)
b) Potential Energy ( Ep )
total energy = constant
P.E. gained = K.E. lost
= work done
dE p   Fdx
x
E p   Fdx
0
x
   madx
0
x
 m  ( w 2 x)dx
0
Ep 
1
mw2 x 2
2
c) Total energy ( Et )
Et = Ek + Ep
(P.E. at x = 0 is chosen to be zero)
S.H.M. P.8
1
1
mw 2 ( A 2  X 2 )  mw 2 x 2
2
2
1
 mw 2 A 2
2

- Consider the position where Ek = Ep .
1
1
mw2 ( A 2  X 2 )  mw2 X 2
2
2
2
2
2
A X X
X 
A
2
- Consider the graph of Energy Vs Time ( if x = A at t = 0 )
7. Other example of S.H.M.
a) Oscillations of a liquid in a U-tube
- Consider a U-tube which contains a length l of liquid of density and has an internal area of
cross-section A. One surface is depressed a distance x and then released.
By Newton’s
nd
2 law,
ma  (2 xA ) g
column = ℓA)
(m = mass of whole liquid
S.H.M. P.9
2 xAg
A
2g

x
a

w2  2g
T

2

 2
w
2g
- the amplitude of the oscillation depends on the initial depression
S.H.M. P.10
b) The floating rod
- Consider a rod of mass m and of cross-sectional area A floating in a liquid of density . l is the
length of rod below the surface . The rod is depressed a distance x and then released.
By Newton's second law,
ma  
mgx

g
x

g
w2 

2
T
w
a
T  2
( forS .H .M )

g
c) Combined Spring Oscillation.
S.H.M. P.11
 (k1  k 2 ) x  ma
k1  k 2
x
m
w 2  ( k1  k 2 ) / m
a
T  2 (
1
m
)2
k1  k 2
k1  k 2  k
If
T  2
m
2K
1
1
m 1
 2 ( ) 2 ( ) 2
k
2
T
 0
2
 T0
(T0 is the period of a single spring oscillation)
By Newton's third law,
k1 x1  k 2 ( x  x1 )
x1 
k2 x
k1  k 2
.........................(1)
From Newton's second law, the equation of motion of m is
(1) → (2)
ma  k 2 ( x  x1 )
k x
ma  k 2 ( x  2 )
k1  k 2
k1 k 2
a
x
m ( k1  k 2 )
w 2  k1 k 2 m ( k1  k 2 )
T  2
If
T  2
 2 2
m ( k1  k 2 )
k1 k 2
k1  k 2  k
2m
k
m
 2T0  T0
k
S.H.M. P.12
 (k1  k 2 ) x  ma
( k1  k 2 )
x
m
k  k2
w2  1
m
m
T  2
k1  k 2
a
If
T  2
k1  k 2  k
T
m
 0  T0
2k
2
d) Oscillation inside the earth
GM ' m

 mx
x2
GM '
  2  x.........................(1)
x
x3
M '  3 M ............................(2)
R
sub. (2) into (1),
GM
x   3 x...........................(3)
R
GM
 g  2 ............................(4)
R
sub. (4) into (3),
g
x   x
R
g
w 2  ( forS .H .M )
R
2
T
w
1
R
T  2 ( ) 2
g
e)
Torsion Pendulum
Torque = -kθ (k = torsion constant
I  k
I  k
= torque per unit angular
displacement)
(I = moment of inertia)
d 2 
( 
)
dt
k
I
k
w 2  ( forS .H .M )
I
2
T
w
I
T  2
k
   
S.H.M. P.13
f)
Heavy piston inside Frictionless cylinder
— The heavy piston is pressed little and then released. (Ignore atmospheric pressure since
the piston is heavy.)
P0 A  mg. . . . . . . . . . . . . .1). (
By Boyle’s law,
By Newton’s 2nd law,
P0 hA  P(h  x) A
P0 h
P 
.................(2)
(h  x)
 ( P  P0 ) A  mx
P0 h
 1) A  mx
hx
mg h

(
 1) A  mx
From (1),
A hx
1
x   g (
 1)..................(3)
x
1
h
x
x
x
I
since (1  ) 1 = 1  ...........(4)
for
h
h
h
g
sub. (4) into (3), x   x
h
g
w 2  ( forS.H .M .)
h
2
h
T
 2
(for small displacements
w
g
only)
From (2),
8.
(
Damped Oscillations
a) Introduction
- we observed that the amplitude of the oscillation of a pendulum gradually decreases
to zero due to the air resistance. The motion is therefore not a perfect S.H.M. and is
said to be damped. Its energy becomes internal energy of the surrounding of air.
-
Damping is the process whereby energy of the oscillating system is lost.
Hence a damped oscillation has its amplitude decreases.
S.H.M. P.14
b) Amplitude of damped oscillations
(i) Undamped oscillations are said to be free oscillation
(ii) Slightly damped oscillation
-
It has oscillations but the amplitude of oscillation decays exponentially with
time.
(iii) Critically damped oscillations
- It has no real oscillation. The time taken for thc displacement to become zero is
a minimum.
-
Displacement dies to zero, but never becomes negative.
(iv) Heavily damped oscillations
- No oscillation occurs and the system returns very slowly to its equilibrium
position.
9.
Examples of damped oscillations
a) Shock Absorber
- It critically damps the suspension of the vehicle and so resist the setting up of
vibrations which could make control difficult or cause damage.
- The viscous force exerted by the liquid contributes this resistive force.
S.H.M. P.15
b) Electrical Meters
- they are critically damped (i.e. dead-beat) so that the pointer moves quickly to the
correct position without oscillation.
c) Acoustic Examples
(i) A percussive musical instrument (e.g. a drum) gives out a note whose intensity
decreases with time.
(slightly damped oscillations due to air resistance)
(ii) The paper cone of a loud speaker vibrates, but is heavily damped so as to lose energy
(as sound wave energy) to the surrounding air.
d) Other damping examples
(i) Damping due to the eddy current produced in the aluminum plate
(ii) Damping is due to the viscosity of the liquid
10. Forced vibration and Resonance
a) Forced vibration
(i) A body is said to be forced vibration (oscillation )if there is an external periodic
driving force acting on it.
(ii) In the early stages, beats will occur between the forced and natural vibration, giving
rise to transient oscillations.
(This stage is usually ignored since its time interval is small.)
(iii) And after a short time, the body will oscillate steadily with the same frequency as
the driving one but the amplitude is different.
b) Resonance
— Resonance occurs when a body is acted upon by a periodic driving force and the forcing
frequency is equal to its natural frequency.
S.H.M. P.16
— When resonance occurs, the forced vibration has it amplitude to be maximum i.e. the
driving oscillator transfers its energy most easily in this case.
c) The response of damped systems to varying driving frequency
— Maximum amplitude is attained when resonance occurs at which driving frequency =
natural frequency.
11. Examples on forced oscillation and resonance
a) Barton’s Pendulum
1)
2)
3)
4)
5)
6)
forcing oscillator – heavy bob
forced oscillator – paper cones
forced oscillation – the paper cones vibrate with the same frequency as
the
driver pendulum.
Resonance – the paper cone pendulum whose length equals that of the
driver
has the greatest amplitude. It is because its natural
frequency is the same as
that of the driving pendulum and resonance occurs.
Plastic ring - it is used to reduce the damping effect caused by the air resistance so
that the resonance effect is more dominant.
(amplitude of light damping oscillation is greater than that of heavy damping.)
Observation on phase difference
— the resonant pendulum has a phase lag of 90° (or T/4) w.r.t. the driver
pendulum.
— The shorter pendulum are nearly in phase with the driver.
— The longer pendulum are almost half a period behind the driver pendulum.
S.H.M. P.17
b) Hacksaw blade oscillator
forcing oscillator – driver pendulum with heavy bob
forced oscillator – hacksaw blade
resonance – adjust the position of the sliding mass to vary the natural frequency of
the hacksaw blade. Resonance occurs when the natural frequency is equal to the
driving frequency.
4. Pin – to indicate the amplitude of the oscillation
5. Postcard – to vary the degree of damping
1.
2.
3.
12. Classified Examples of resonance
a) Mechanics
(i) Barton’s pendulum
(ii) Hacksaw blade oscillator
(iii) Wind-induced oscillation – destruction of Tacoma bridge
(iv) Spring mass
(America 1940)
b) Acoustic
(i) resonance tube – resonance occurs in the vibrations of columns and hence musical
sound can be produced.
(ii) Kundt’s tube – resonance occurs when the frequency of the loudspeaker is equal to
natural frequency of the glass tube.
** It can be used compare the velocity of sound in different gases.
Gas 1 : v1  f1
Gas 2 : v2  f2
v1 1

v 2 2
S.H.M. P.18
c) Electricity (e.g. tuning a radio)
— the natural frequency of the radio is adjusted to that of the incoming
electromagnetic wave by changing its capacitance.
d) Atomic – maximum absorption of infra-red waves by NaCl crystal occurs when their
frequency equals that of the vibration of the Na+ and Cl- ions.