Topic 1 shm

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Topic 1 Simple Harmonic Motion (SHM)UEEP1033 Oscillations and Waves

Topic 1:Simple Harmonic Motion

(SHM)

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Simple pendulum Flat disc

SHM Systems

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Mass-Spring Mass at the centre of a light string

SHM Systems

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Frictionless U-tube of liquidOpen flask of volume V

and neck of length l

SHM Systems

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Hydrometer LC circuit

SHM Systems

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• When the above systems are slightly disturbed from the equilibrium or rest position, they will oscillate with SHM

• A small displacement x from its equilibrium position sets up a restoring force F, which is proportional to x acting in a direction towards the equilibrium position

F = – sx (Hooke’s Law of Elasticity)

• s = proportional constant (called the stiffness)• negative sign shows that the force is acting against the direction

of increasing displacement and back towards the equilibrium position

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• By Newton’s Law, xmF

sxxm

0 sxxm 2

2

dt

xdx

0 xm

sx

222

fTML

MLT

m

s

• Dimensions of

• T = period of oscillation• f = frequency of oscillation

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• Since the behaviors of x with time has a sinusoidal dependence, so it more appropriate to consider the angular frequency = 2f

2

21

s

m

fT

m

s 2

02 xx

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Displacement in Simple Harmonic Motion (SHM)

• The behaviour of SHM is expressed in terms of its:

: displacement x from equilibrium

: velocity

: acceleration at any given timex

x

xtAx

tAx

tAx

22 cos

sin

cos

A = constant with the same dimensions as x

• One possible solution: satisfies tAx cos 02 xx

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xtBx

tBx

tBx

22 sin

cos

sin

• another solution: tBx sin

• Complete or general solution for

xtBtAx

tBtAx22 )sincos(

sincos

02 xx

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• Rewrite:

where is a constant angle sinaA cosaB

22

222222 )cos(sin

BAa

aaBA

)sin(

sincoscossin

tax

tatax

where a is the amplitude of the displacement and is the phase constant

• Limiting values of sin(t+) are 1, so the system will oscillate between the values of x = a

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• Displacement x can be represented by projection of a radius vector of constant length a rotates in anticlockwise direction with constant angular velocity

• Phase constant defines the position in the cycle of oscillation at time t = 0

• takes the values between 0 and 2 radian

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Sinusoidal displacement of simple harmonic oscillator with time, showing variation of starting point in cycle in terms of phase angle

[source: H.J.Pain, The Physics of Vibrations and Waves 6th Ed. Fig. 1.2]

Displacement In Simple Harmonic Motion (SHM)

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Velocity and Acceleration in SHM

• Displacement

• Velocity

• Acceleration

• Maximum value of velocity a (velocity amplitude)

• Maximum value of acceleration a2 (acceleration amplitude)

)sin( tax

)cos( taxdt

dx

)sin(22

2

taxdt

xd

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• Velocity leads the displacement by /2• Velocity is maximum when displacement is zero• Velocity is zero when displacement is maximum

• Acceleration is anti-phase ( rad) with respect to displacement• Acceleration is maximum positive when displacement is maximum

negative and vice versa

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Variation with time of displacement, velocity and acceleration in SHM

[source: H.J.Pain, The Physics of Vibrations and Waves6th Ed. Fig. 1.3]

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• Two oscillators having the same frequency and amplitude may be consider in terms of their phase difference 1 - 2

• When two systems are diametrically opposed, the system are anti-phase:

phase difference 1 - 2 = n rad (n is odd integer)

• When two systems are exactly equal values of displacement, velocity and acceleration, the system are in phase:

phase difference 1 - 2 = 2n rad ( n is any integer)


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txx m cosBlue curve:

txx m cos'Red curve: Differ in amplitude only

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txx m 'cosRed curve:

Differ in period only

T = T/2

= 2

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)4/cos( txx mRed curve: Differ in phase only

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Example 1.1

a) What are the angular frequency, the frequency, the period , the amplitude of the resulting motion?

b) What is the amplitude of the oscillation?c) What is the maximum speed of the oscillating block, and

where is the block when it occurs?d) What is the amplitude of the maximum acceleration of the

block?e) What is the phase constant for the motion?f) What is the displacement function x(t) for the spring-block

system?

A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 N/m. The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0.

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Example 1.1

The block-spring system

The block moves in SHM once it has been pulled to the side and released

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Example 1.1 - Solution

(a) What are the angular frequency, the frequency, the period of the resulting motion?

The block-spring system forms a linear SHM

angular frequency: rad/s78.9kg0.68

N/m65

m

k

frequency: Hz56.12

rad/s78.9

2

f

period: s64.01

f

T

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(b) What is the amplitude of the oscillation?


The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0

The amplitude xm = 11 cm

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(c) What is the maximum speed of the oscillating block, and where is the block when it occurs?

The maximum speed occurs when the oscillating block is moving through the origin, i.e. at x = 0

Maximum speed = velocity amplitude vm = xm

vm = xm = (0.11 m) (9.78 rad/s) = 1.1 m/s

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(d) What is the amplitude of the maximum acceleration of the block?

amplitude of maximum acceleration = xm 2

am = xm 2 = (0.11 m) (9.78 rad/s)2 = 11 m/s

The maximum acceleration occurs when the block is at the ends of its path

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(e) What is the phase constant for the motion?

At time t = 0, the block is located at x = xm

)cos( txx m

0

1cos)0cos(

mm xx

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(f) What is the displacement function x(t) for the spring-block system?

)8.9cos(.11.0)(

])0)rad/s8.9cos([()m11.0()(

ttx

ttx

)cos()( txtx m


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Energy of a SHM

• An exchange between kinetic and potential energy

• In an ideal case the total energy remains constant but this is never realized in practice

• If no energy is dissipated:

1. Etotal = KE + PE = KEmax = PEmax

2. Amplitude a remains constant

• When energy is lost the amplitude gradually decays

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Energy of a SHM

• Potential energy PE is found by summing all the small elements of work sx dx (force sx times distance dx) done by the system against the restoring force over the range 0 to x

• where x = 0 gives zero potential energy

2

0 2

1sxdxsxPE

x

• Kinetic energy 2

2

1xmKE

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Energy of a SHM

• Total Energy:

22

2

1

2

1sxxmE

0)( xsxxmdt

dE

• Since total energy E is constant

0 sxxm

PEKEE

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Energy of a SHM

• Maximum potential energy occurs at thereforeax

2max 2

1saPE

22

max

max222

max

2max

2

1

)(cos2

1

2

1

maKE

tma

xmKE

• Maximum kinetic energy is

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Energy of a SHM

The total energy at any instant of time t or position of x is:

2

22

2222

22

2

12

1

)(sin)(cos2

12

1

2

1

saE

ma

ttma

sxxmE

)cos( taxdt

dx

)sin( tax

2ms

slide-9

= PEmax

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En

erg

y

At x = 0the energy is all kinetic

At x = athe energy is all potential

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Potential energy PE(t), kinetic energy K(t) and total energy E as function of time t for a linear harmonic oscillator

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A 0.5 kg cart connected to a light spring for which the force constant is 20 N m-1 oscillates on a horizontal, frictionless air track.

(a) Calculate the total energy of the system and the maximum speed of the cart if the amplitude of the motion is 3 cm.(b) What is the velocity of the cart when the position is 2 cm.(c) Compute the kinetic and potential energy of the system when the position is 2.00 cm.

Example 1.2

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Example 1.2 - Solutions

(a) Total energy =

J 009.003.0202

1

2

1 22 kA

1-2 s m 19.0009.0)5.0(2

1 xx Maximum speed

1-

22

s m 141.0

009.002.0202

15.0

2

1

x

x

(b)

J 005.0141.05.05.0 2

J 004.002.0205.0 2

(c) Kinetic energy =

Potential energy =

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SHM in an Electrical System

• An inductor L is connected across the plates of a capacitor C

• The force equation becomes the voltage equation

dt

dIL q/C

_

+

_

+

I

0C

qqL

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• In the absence of resistance, the energy of the electrical system remains constant

• Exchanged between magnetic field energy stored in the inductor and electric field energy stored between the plates of the capacitor

• The voltage across the inductor is

where I is the current flowing and q is the charge on the capacitor

2

2

dt

qdL

dt

dILV

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• The voltage across the capacitor is q/C

• Kirchhoff’s law

LC

qq

C

qqL

1

0

0

2

2

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• Energy stored in the magnetic field

• Electrical energy stored in the capacitor

22

0 2

1

2

1qLLILIdIE

Idtdt

dILdtVIE

I

L

L

C

qCV

22

1 22


2q

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Force or Voltage Energy

Mechanical

Force

Electrical

Voltage

0 sxxm

0C

qqL

22

2

1

2

1sxxmE

C

qqLE

22

2

1

2

1

Comparison between Mechanical and Electrical Oscillators

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Superposition of Two Simple Harmonic Vibrations in One Dimension

Case-1: Vibrations having equal frequencies

)cos( 111 tax

)cos( 222 tax

cos2

)sin()cos(

212

22

12

22

221

2

aaaaR

aaaR

Displacement of first motion:

where = 2 - 1 is constant

Resulting displacement amplitude R:

Displacement of second:

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each representing SHM along the x axis at angular frequency to give a resulting SHM displacement : x = R cos(t + )

--- here shown for t = 0

Addition of Vectors

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• The phase constant of R is

• Resulting simple harmonic motion has displacement

• An oscillation of the same frequency but having an amplitude R and a phase constant

2211

2211

coscos

sinsintan

aa

aa

)cos( tRx



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Displacement of first motion:

Displacement of second:

where 2 > 1

Resulting displacement:

tax 11 sintax 22 sin

2

)(cos

2

)(sin2

)sin(sin

1221

2121

ttax

ttaxxx


Case-2: Vibrations having different frequencies

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2

)(cos

2

)(sin2 1221 tt

ax

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• Resulting oscillation at the average frequency (2+ 1)/2

• Resulting amplitude of 2a which modulates• Amplitude varies between 2a and zero under the influence of the

cosine term of a much lower frequency (2- 1)/2

• Acoustically this growth and decay of the amplitude is registered as ‘beat’ of strong reinforcement when two sounds of almost equal frequency are heard

• The frequency of the ‘beat’ is (2- 1)


Case-2: Vibrations having different frequencies

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• One along the x-axis, the other along the perpendicular y-axis

• Displacement,

• Eliminating t, get a general equation for an ellipse

)sin( 11 tax

)sin( 22 tay

)(sin)cos(2

122

1221

22

2

21

2

aa

xy

a

y

a

x

Superposition of Two Perpendicular Simple Harmonic Vibrations Having Equal Frequency

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Expanding the arguments of the sines:

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• Some special case:

• An ellipse:

If a1 = a2 = a, a circle x2 + y2 = a2

• A straight line:

• A straight line:

212

122

2

21

2

a

y

a

x

... ,4 ,2 ,012 xa

ay

1

2

... ,5 ,3 ,12 xa

ay

1

2


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• If the amplitudes of vibrations are a and b respectively, the Lissajous figure will always be contained within the rectangle of sides 2a and 2b

• The axial frequencies bear the simple ratio

Superposition of Two Perpendicular Simple Harmonic Vibrations Having Different Frequency

• When the frequencies of the two perpendicular simple harmonic vibrations are not equal, the resulting motion becomes more complicated

• The patterns which are traced are called Lissajous figures

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Simple Lissajous figures produced by perpendicular simple harmonic motions of different angular frequencies

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Superposition of a Large Number nof Simple Harmonic Vibrations

• n simple harmonic vibrations of equal amplitude a and equal successive phase difference

• Resultant magnitude R and phase difference

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Vector superposition of a large number n of simple harmonic vibrations of equal amplitude a and equal successive phase difference

resultant amplitude:

its phase difference with respect to the first

component acos t:

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Geometrically, each length:

where r is the radius of the circle enclosing the (incomplete) polygon

From the isosceles triangle OAC the magnitude of the resultant:

and its phase angle is seen to be:

In the isosceles triangle OAC:

In the isosceles triangle OAB:

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i.e. half the phase difference between the first and the last contributions

The resultant:

magnitude of the resultantis not constant but depends on the value of

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When n is very large and is very small and the polygon becomes an arc of the circle centre O, of length na = A, with R as the chord

Then:

In this limit:

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The pattern is symmetric about the value = 0 and is zero whenever sin

= 0 except at 0 i.e. when sin / 1

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(b) When = 0, = 0 and the resultant of the n vectors is the straight line of length A

(c) As increases A becomes the arc of a circle until at = /2 the first and last contributions are out of phase (2 = ) and the arc A has become a semicircle of which the diameter is the resultant R

(d) A further increase in increases and curls the constant length A into the circumference of a circle ( = ) with a zero resultant

(e) At = 3/2, the length A is now 3/2 times the circumference of a circle whose diameter is the amplitude of the first minimum.

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Rotating-Vector Representation of SHM

Source: A.P. French, “Vibrations and Waves” MIT Introduction Physics Series, CRC Press

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• Describing SHM by regarding it as the projection of uniform circular motion of a disk of radius R rotates about a vertical axis at the rate of angular frequency rad/s

Small block

Shadow of the block P

The shadow performs SHM with period 2/ and amplitude A along a horizontal line on the screen

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SHM as the geometrical projection of uniform circular motion

Horizontal axis Ox = the line along which actual oscillation takes places

Rotating vector OP projected onto a diameter of the circle

Instantaneous position of the point P is defined by the constant length A and the variable angle

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• Take counterclockwise direction as positive

Actual value of : = t + is the value of at t = 0

• Displacement x of the actual motion:

x = A cos = A cos (t + )

• Orthogonal oscillation along y axis perpendicular to the real physical axis Ox of the actual motion:

y = A sin (t + ) -- physical unreal component of the motion

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Cartesian and polar representations of a rotating vector

vector OP has the plane polar coordinates (r, )

Cartesian coordinates (x, y)

x = r cos y = r sin

Rotating Vectors and Complex Numbers

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Representation of a Vector in the Complex Plane

z = a + jb

j 2 = 1

jz = ja + j2b

jz = b + ja

Multiplication of z by j is equivalent to a 90° rotation

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• Sinusoidal signals are characterised by their magnitude, their frequency and their phase

• If the frequency is fixed (e.g. at the frequency of the AC supply) and we are interested in only magnitude and phase

• In such cases we often use phasor diagrams which represent magnitude and phase within a single diagram

Phasor Diagrams

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• Phasor diagram is a graphical method that helps in the understanding waves and oscillations, and also helps with calculations, such as wave addition

• imagine a rigid rotor or vector moving around in circles around the origin, as illustrated in the diagram 1

• the projection or shadow of the tip of the vector moves back and forth exactly like the oscillation

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diagram 1 – Phasor Diagramthe projection or shadow of the tip of the vector moves back and forth exactly like

the oscillation

The projection that the vector makes on the horizontal or x-axis:

x = A cos (ωt + ϕ)

• A phasor is a complex number that represents the amplitude and phase of a sinusoid

• Phasors provide a simple means of analyzing linear system excited by sinusoidal sources

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Representations of Complex Numbers

Rectangular form: z = x + jy

Re

x

Im

y

jyz

r

Real part of z

)Re(zx

Imaginary part of z

)Im(zy

1j

12 j

123 jjj

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• Complex number z can be represented in three ways:

z = x + jyRectangular form:

z = re jExponential form:

e j = cos + j sin

22 yxr x

y1tan

x = r cos y = r sin

1j

z = rPolar form:

jrerjrrjyxz sincos

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Some Basic Properties of Complex Numbers

e ±j = cos ± j sin

Real part: cos = Re (e j)

Imaginary part: sin = Im (e j)

x(t) = Xm cos (t + )

Example:

x(t) = Re(X e jt)

where

X is the phasor representation of the sinusoid v(t)

Phasor is a complex representation of the magnitude and phase of a sinusoid

To obtain the sinusoid corresponding to a given phasor X, multiply the phasor by the time factor e jwt and take the real part

X = Xm e j = Xm

= Re(Xm e j(t + )) = Re(Xm e j e jt)

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x(t) = Re(X e jt) To obtain the sinusoid corresponding to a given phasor X, multiply the phasor by the time factor e jt and take the real part

x(t) = Xm cos (t + ) = Re(Xm e j(t + )) = Re(Xm e j e jt)

• To get the phasor corresponding to a sinusoid, we first express the sinusoid in the cosine form so that the sinusoid can be written as the real part of a complex number

• Then we take out the time factor e jwt, and whatever is left is the phasor corresponding to the sinusoid

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• Arithmetic• Operations

Complex Conjugate

222yxzz* ;

1 where

sincos*

rjj

θ) j θr(rerjyxz j

z1 = x1 + jy1 = r1 1 z2 = x2 + jy2 = r2 2

rz

rz

r

r

z

z

rrzz

yyjxxzz

yyjxxzz

1

1

)()(

)()(

2

1

2

1

2121

212121

212121

Addition:

Division:

Reciprocal:

Multiplication:

Subtraction:

Square Root:

–

1– 2

1+ 2

/2

–

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Xm

Xm

Xmx(t) = Re(X e jt)

Xe jt = Xm e j(t +)Plot of the sinor on the complex plane

As time increases, the sinor rotates on a circle of radius of Xm at an angular velocity in the counterclockwise direction

x(t) is the projection of the sinor Xejwt on the real axis

the value of sinor at time t = 0 is the phasor X of the sinusoid x(t)

X = Xm e j = Xm

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• Phasor diagrams can be used to represent the addition of signals. This gives both the magnitude and phase of the resultant signal

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• Phasor diagrams can also be used to show the subtraction of signals

Topic 1 shm

Science

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