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NCERT Solutions for Class 11 Physics Chapter 15 Waves

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Physics Waves Notes

Class 11 Physics Chapter 15 Waves:

Section NameTopic Name
15Waves
15.1Introduction
15.2Transverse and longitudinal waves
15.3Displacement relation in a progressive wave
15.4The speed of a travelling wave
15.5The principle of superposition of waves
15.6Reflection of waves
15.7Beats
15.8Doppler effect
Physics Waves Notes

Introduction

In this chapter we will see the importance of waves in our life.

We will also study about the different properties of waves, some terms related to waves and also about different types of waves.We will also learn how waves propagate.

For example: –

  1. Medium required by the waves to travel from one point to another:-
    • Consider a boy holding a thread and one end of thread is tied to the wall.
    • When a boy moves the thread, the thread moves in the form of a wave.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes
  • Similarly a boat sailing over the sea,the boat is able to move because of waves.
  • The ripples formed in a lake when we drop a stone in the lake.They are also waves.
  • Earthquakes are caused due to the waves under the surface of the earth.
  • The strings of the guitar when we play them are also waves again.
  • Music system which we use to hear songs.This is due to sound waves.
  • When 2 people talk they are able to hear each other because of the sound waves.

In the below Picture we can see waves need a medium to propagate.

Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes

Some type of waves can propagate from one point to another without any medium.

  1. Waves which are related to matter:-
  • There are some set of waves which are inside the matter.
  • For example: – whole of universe.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes

What is a wave?

A wave is s disturbance that propagates through spaceand time,usually with transference of energy.

For example: –

  • Consider the sound of the horn; this sound reaches our ear because of sound waves.
  • There is transfer of energy from one point to another with the help of particles in the medium.
  • These particles don’t move they just move around their mean position,but the energy is getting transferred from one particle to another and it keeps on transferring till it reaches the destination.
  • The movement of a particle is initiated by the disturbance.And this disturbance is transferred from one point to another through space and time.

Note:-Energy and not the matter is transferred from one point to another.

  1. When a source of energy causes vibration to travel through the medium a wave is created.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes

Types of Waves

  1. Mechanical waves
  2. Electromagnetic waves
  3. Matter waves

Mechanical waves: –

  1. The mechanical waves are governed by all the Newton’s laws of motion.
  2. Medium is needed for propagation of the wave.

For Example: – Water Waves, Sound Waves

Water waves: They are mechanical waves for which a medium is required to propagate.

Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes

Electromagnetic waves:-

  • Electromagnetic waves are related to electric and magnetic fields.
  • An electromagnetic wave, does not need a medium to propagate, it carries no mass,does carry energy.

Examples: – Satellite system, mobile phones,radio, music player, x-rays and microwave.

Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes

Transverse Waves

  • The transverse waves are those in which direction of disturbance or displacement in the medium is perpendicular to that of the propagation of wave.
  • The direction in which a wave propagates is perpendicular to the direction of disturbance.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes

Displacement in a progressive wave

  • Amplitude and phase together describe the complete displacement of the wave.
  • Displacement function is a periodic in space and time.
  • Displacement of the particles in a medium takes place along the y-axis.
  • Generally displacement is denoted as a function of X and T, but here it is denoted by y.
  • In case of transverse wave displacement is given as:
  • y(x,t) where x=propagation of the wave along x-axis, and particles oscillates along y-axis.
  • Therefore y(x,t)= A sin(kx – ωt + φ).This is the expression for displacement.
  • This expression is same as displacement equation which is used in oscillatory motion.
  • As cosine function;y(x,t)= B cos(kx – ωt + φ),As both sine and cosine function)y (x, t) = A sin (kx – ωt + φ) + B cos(kx – ωt + φ)

Mathematically:

  • Wave travelling along +X-axis: y(x, t) = a sin (kx – ωt + φ).
  • Consider y=asin(kx – ωt + φ)=> y/a=sin(kx – ωt + φ)
  • sin-1(y/a) = kx-ωt =>kx=sin-1(y/a) +ωt
  • x=(1/k)sin-1(y/a)+ (ωt/k)
  • Wave travelling along –X-axis: x=(1/k)sin-1(y/a)-(ωt/k)(only change in the sign ofωt)
  • Conclusion:-
  • As time t increases the value of x increases. This implies the x moves along x-axis.
  • As time t decreases the value of x decrease. This implies the x moves along (-)ive x-axis.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes

In case of longitudinal wave,

  • There are regions of compressions (particles are closely packed) and rarefactions (particles are far apart).
  • In compressions density of the wave medium is highest and in rarefactions density of the wave is lowest.
  • Consider when the particle is at rarefaction, in that region as particle gets more space as a result the particles oscillates to the maximum displacement.
  • Whereas in compressed region the particles oscillates very less as the space is not very much.
  • The peak or the maximum amplitude is the centre of two compressed regions.Because at the centre of the two compressed region the particle is most free to displace to maximum displaced position.

Conclusion:-

  • In case of longitudinal wave the particles will not oscillate to a very large distance. This displacement won’t represent the amplitude as it is not maximum possible displacement.

Amplitude is represented basically by the centre of the rarefaction region where the particle is most free to oscillate to its maximum displacement.

Physics Waves Notes
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes
  • Consider two points A and B on a wave.Their positions as well as their behaviour are same. Therefore points A and B are in phase.
  • Consider points A and C on a wave. They are not in phase with each other as their position is not same.
  • Similarly the points C and D are not in phase with each other as their positions are same but the behaviour is different. Thereforethey are not in phase with each other.
  • Consider the points F and G their positions are same but the behaviour is totally opposite. So F and G are out of phase.
  • Consider the points F and H;they are in phase with each other as their position is same as well as their behaviour.
Physics Waves Notes

Wavelength 

  • The term wavelength means length of the wave.
  • Wavelength is defined as the minimum distance between two consecutive points in the same phase of wave motion.
  • It is denoted by λ.
  • In case of transverse wave we use the term crest for thepeak of the maximum displacement.
  • The point of minimum displacement is known as trough.
  • In case of transverse wave wavelength is the distance between two consecutive crests or distance between two consecutive troughs.
  • In case of longitudinal wave wavelength is the distance between the two compressions or the distance between the two rarefactions provided the compressions or rarefactions are nearest.

Frequency of the ultrasonic sound, ν = 1000 kHz = 106 Hz

Speed of sound in water, vw = 1486 m/s

The wavelength of the transmitted sound is given as:

λr = 1486/106 =1.49 × 10–3 m

Problem:- A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of soundin the tissue in which the speed of sound is 1.7 km s–1? The operating frequency of the scanner is4.2 MHz

Answer: Speed of sound in the tissue, v = 1.7 km/s = 1.7 × 103 m/s.

Operatingfrequency of the scanner, ν = 4.2 MHz = 4.2 × 106 Hz.

The wavelength ofsound in the tissue is given as:

λ=ν/v

= (1.7×103)/ (4.2×106)

=4.1×10-4

Physics Waves Notes

Wave Number

Wave number describes the number of wavelengths per unit distance.

Denoted by ‘k’.

y(x,t)= a sin(kx – ωt + φ) assuming φ=0.

  • At initial time t=0:-
    • y(x,0)=asin kx (i)
    • When x=x+λ then y(x+λ,0)=a sink(x+λ)   (ii)
    • When x=x+2λ then y(x+2λ,0)=a sink(x+2λ)

Value of y is equal at all points because all the points’ λ, 2λ are in phase with each other. Therefore,

From(i) and (ii) asin kx= a sink(x+λ)  =asin(kx+k λ)

This is true if and only if: -k λ =2 πn, where n=1, 2, 3…

k=(2 π n)/ λ. This is the expression for wave number.

k is also known as propagation constantbecause it tells about the propagationof the wave.

Wave number is an indirect way of describing the propagation of wave.

Time Period, Frequency and Angular frequency

  1. Time Period of a wave: –
    1. Time Period of a wave is the time taken through one complete oscillation. It is denoted by’T’.
  2. Frequency of a wave:-
    1. Frequency of a wave is defined as number of oscillations per unit time.It is denoted by ν.
    2. ν=1/T.
  3. Angular frequency: –
    1. Angular frequency is defined as the frequency of the wave in terms of a circular motion.
    2. The term angular frequency is used only when there is an angle involved in the motion in that particular motion.
    3. It is denoted by ‘ω’.
    4. For example:-In linear motion angular frequency is not used but in case of wave the term angular frequency is used.
  • Relation of with ω frequency ν is given ω= 2πνorω= 2π/T.

= 7.85 cm

(c) Now we relate T to ω by the relation

T = 2π/ω = (2 π)/ (3.0) s–1

Frequency, v = 1/T = 0.48 Hz.

The displacement y at x = 30.0 cm andtime t = 20 s is given by

y = (0.005 m) sin (80.0 × 0.3 – 3.0 × 20)

= (0.005 m) sin (–36 + 12π)

= (0.005 m) sin (1.699)

= (0.005 m) sin (970)

~ 5 mm

Problem:-

A transverse harmonic wave on a string is described by y(x,t) =3.0sin(36t+0.081x+(π/4))   Where x and y are in cm and t in s. The positive direction of x is from left to right. Is this a travelling wave or a stationary wave?

  • If it is travelling, what are the speed and direction of its propagation?
  • What are its amplitude and frequency?
  • What is the initial phase at the origin?
  • What is the least distance between two successive crests in the wave?

Answer:

  • Yes; Speed = 20 m/s, Direction = Right to left
  • 3 cm; 5.73 Hz,
  • (π/4)
  • 49 m

Explanation: The equation of a progressive wave travelling from right to left is given by the displacement

Function:y (x, t) = a sin (ωt + kx + Φ) … (i). The given equation is:

y(x, t) =3.0sin (36t+0.081x+(π/4))..(ii)

On comparing both the equations, we find that equation (ii) represents a travelling wave,

propagating fromright to left.

Now, using equations (i) and (ii), we can write:

ω = 36 rad/s and k = 0.018 m–1. We knowthat:

ν=ω/ (2π) and λ= (2π)/k

Also, v = νλ

Therefore, ν=(ω/ (2π)) x (2π/k) = ω/k

=36/ (0.018)=2000cm/s=20m/s

Hence, the speed of the given travelling wave is 20 m/s.

Amplitude of the given wave, a = 3 cm

Frequency of thegiven wave:

ν=ω/ (2π) = 36/ (2×3.14) =5.73Hz

On comparing equations (i) and (ii), we find that the initial phase angle, Φ= (π/4)

The distance between two successive crests or troughs is equal to the wavelength of the wave.

Wavelength is given by the relation:

k=(2π)/λ

Therefore,

λ= (2π)/k = (2×3.14)/ (0.018) = 348.89cm=3.49m

Physics Waves Notes

The speed of a travelling wave

  • To determine the speed of a travelling wave.
  • The propagation of the wave always takes place along the +(ive) x-axis.
  • It is denoted by v.
  • Consider transverse wave moving along (+) ive x axis.
    • Displacement equation y (x, t) = a sin (kx – ωt) assuming φ=0.
  • In case of transverse wave the y denotes the movement of particles along y-axis.x denotes the direction of wave propagation in terms of x axis.
  • y-axis denotes the displacement of the particles of the medium in case of transverse wave and in case of longitudinal wavex denotes the movement of the particles along x-axis.
  • In case of longitudinal wave the motion of particles takes place along x-axis and not along y-axis.
  • The form of equation remains the same for both transverse and longitudinal wave.
  • Only the direction of propagation i.e. the direction of the movement of the particles differs in case of transverse and longitudinal waves.
  • We have considered only the motion along the horizontal direction i.e. x-axis. i.e. motion of all the peaks and which is a motion of a straight line.
  • Displacement takes place along the straight line along x-axis.
  • y component does not play much role as it talks about the movement of particles of the medium.
  • When we are considering the peaks as a result phase does not have any role. That is phase is constant.
  • y/a=sin(kx – ωt) => sin-1 (y/a) = (kx- ωt )
    • => x=(1/k)sin-1 (y/a) +ωt/k (where (1/k)sin-1 (y/a)  is constant)
    • => Speed V=dx/dt =ω/k

(By differentiating with time,(1/k)sin-1 (y/a) = 0 as differentiating a constant term is 0.) 

  • Therefore Wave Speed V= ω/kwhereω=angular frequency and k=wave number.

Different Expressions for Wave Speed

  • Wave speed V= ω/k. By definition of ω= (2 π)/T where T is period of the wave.
  • Also k=(2 π)/λ .Therefore v=(2 π)/(T)/(2 π)/(λ) =λ/T
  • v= λv
    • As wavelength of a wave increases as a result frequency of the wave decreases as a result speed of wave is constant.
    • Wave speed is determined by the properties of the medium.

Dimensional Analysis to show how the speed is related to mass per unit length and Tension

  • μ = [M]/[L] = [ML-1] (i)
  • T=F=ma =[M][LT-2] = [MLT-2] (ii)
  • Dividing equation (i) by (ii) :- [ML-1]/[MLT-2] = L-2T-2 =[T/L]2 =[TL-1]2 =1/v2
  • Therefore μ/T = 1/v2
  • v=C√T/ μ where C=dimensionless constant
  • Conclusion: v depends on properties of the medium and not on frequency of the wave.

Problem:- A steel wire 0.72 m long hasa mass of 5.0 ×10–3 kg. If the wire is under a tension of 60 N, what is the speed oftransverse waves on the wire?

Answer: -Mass per unit length of the wire,0.72 m

μ = 5.0×10−3 kg

= 6.9 ×10–3 kg m–1

Tension, T = 60 N

The speed of wave on the wire is given by,v=C√T/ μ

√ (60)/ (6.9×10-3kgm-1) = 93ms-1

Problem:- A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Answer:-

Mass of the string, M = 2.50 kg

Tension in the string, T = 200 N

Length of the string, l = 20.0 m

Mass per unit length,μ=M/l=2.50/20=0.125kgm-1

The velocity (v) of the transverse wave in the string is given by the relation: v=√T/μ

=√200/0.125=√1600=40m/s

Therefore, time taken by the disturbance to reach the other end, t =l/v=20/40=0.50s

Problem:- A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1.

Answer:-

Length of the steel wire, l = 12 m

Mass of the steel wire, m = 2.10 kg

Velocity of the transverse wave, v = 343 m/s

Mass per unit length,μ=m/l= 2.10/12 =0.175kgm-1

For tension T, velocity of the transverse wave can be obtained using the relation:

v= √T/μ

Therefore T=v2μ

= (343)2 × 0.175 = 20588.575 ≈ 2.06 × 104 N

Speed of a longitudinal wave in a stretched string

Longitudinal wave speed determined by:

  • Density – Longitudinal wave is formed due to compressions (particles very close to each other) and rarefactions (particles are far from each other).
  • At certain places it is very dense and at certain places it is very less dense.So density plays a very important role.
  • It is denoted by ρ.
  • Bulk modulus– Bulk modulus tells how does the volume of a medium changes when the pressure on it changes.
    • If we change the pressure of compressions orrarefactions then the volume of the medium changes.
    • It is denoted by B.

Dimensional Analysis to show how the speed is related to density and bulk modulus

  • ρ =mass/volume= [ML-3]
  • B = – (Change in pressure (ΔP))/Change in volume(ΔV/V))
  • ΔV/V is a dimensionless quantity as they are 2 similar quantities.
  • ΔP =F/A = ma/A =[M][LT-2]/[L2] = [ML-1T-1]
  • Dividing ρ/B= [ML-3]/[ML-1T-2] = [L-2T2]=[TL-1]2 =1/v2
  • Therefore ρ/B = 1/v2 or v2=B/ ρ
  • v= C √ (B/ ρ) where C=dimensionless constant
  • In case of fluids :- v= C √ (B/ ρ)
  • In case of solids :- v= C √ (Y/ ρ)

Problem:- Estimate the speed ofsound in air at standard temperature andpressure. The mass of 1 mole of air is29.0 ×10–3 kg.

Answer:-

We know that 1 mole of any gasoccupies 22.4 litres at STP. Therefore, density

of air at STP is:ρo = (mass of one mole of air)/ (volume of onemole of air at STP)

29.0 x 103 kg/22.4×103 m3

= 1.29 kg m–3

According to Newton’s formula for the speedof sound in a medium, we get for the speed of

sound in air at STP,

v = [(1.01×105Nm-2)/ (1.29kgm-3)]1/2

=280ms-1

The principle of superposition of waves

  • Principle of superposition of waves describes how the individual waveforms can be algebraically added to determine the net waveform.
  • Waveform tells about the overall motion of the wave.It does not tell about individual particles of the wave.
  • Suppose we have 2 waves and
  • Example of superposition of waves is Reflection of waves.
  • Mathematically: –

Case1:-

  • Consider 2waves which are in phase with each other. They have the same amplitude, same angular frequency, and same angular wave number.
  • If wave 1 is represented by y1(x, t) =a sin (kx – ωt).
  • Wave 2 is also represented by y2(x, t) =a sin (kx – ωt).
  • By the principle of superposition the resultant wave (2a sin (kx – ωt)) will also be in phase with both the individual waves but the amplitude of the resultant wave will be more.

Case2:-

  • Consider when the two waves are completely out of phase.i.e. φ = π
  • If wave 1 is represented by y1(x, t) =a sin (kx – ωt).
  • Wave 2 is represented by y2(x, t) =a sin (kx – ωt+ π).
  • =>y2=asin(π-(-kx+ ωt) =>y2=-a sin(kx- ωt)
  • Therefore by superposition principle y=y1+y2=0

Reflection at rigid boundary

  • Consider a string which is fixed to the wall at one end. When an incident wave hits a wall,it will exerta force on the wall.
  • By Newton’s third law, the wall exerts an equal and opposite force of equal magnitude on the string.
  • Since the wall is rigid wall won’t move, therefore no wave is generated at the boundary.This implies the amplitude at the boundary is 0.
  • As both the reflected wave and incident wave are completely out of phase at the boundary.Therefore φ=π.
  • Therefore, yi(x, t) = a sin (kx – ωt),
  • yr(x, t) = a sin (kx + ωt + π) = – a sin (kx + ωt)
  • By superposition principle y= y+ yr =0
  • Conclusion: –
    • The reflection at the rigid body will take place with a phase reversal of π or 180.

Standing (Stationary) Waves

  • A stationary wave is a wave which is not moving,i.e. it is at rest.
  • When two waves with the same frequency,wavelengthand amplitude travelling in opposite directions will interfere they produce a standing wave.
  • Explanation:-
    • Consider Ist wave in the figure and suppose we have a rigid wall which does not move. When anincident wave hits the rigid wall it reflects back with a phase difference of π.
    • Consider IInd wave in the figure, when the reflected wave travels towards the left there is another incident wave which is coming towards right.
    • The incident wave is continuously coming come from left to right and the reflected wave will keep continuing from right to left.
    • At some instant of time there will be two waves one going towards right and one going towards left as a result these two waves will overlap and form a standing wave.
  • Mathematically:
  • Wave travelling towards left yl(x,t) =a sin(kx– ωt) and towards right yr(x,t) =a sin (kx + ωt)
  • The principle of superposition gives, for the combined wave
  • y (x, t) = yl(x, t) + yr(x, t)= a sin (kx – ωt) + a sin (kx + ωt)
  • y(x,t)= (2a sin kx) cos ωt (By calculating and simplifying)
  • The above equation represents the standing wave expression.
  • Amplitude = 2a sin kx.
    • The amplitude is dependent on the position of the particle.
    • The cos ωt represents the time dependent variation or the phase of the standing wave.

Difference between the travelling wave and stationary wave

Travelling Wave(Progressive Wave)Stationary Wave (Standing wave)
Waveform moves. Movement of the waveform is always indicated by the movement of the peaks of the wave.Waveform doesn’t move.Peaks don’t move.
Wave amplitude is same for all the elements in the medium. Denoted by ‘A’.Wave amplitude is different for different elements.Denoted by asinkx.
Amplitude is not dependent on the position of the elements of the medium.Amplitude is dependent on the position of the elements of the medium.
y(x,t)=asin(kx– ωt + φ )y(x,t)=2asin(kx)cos(ωt)

Nodes and Antinodes of Standing Wave

  • The amplitude of a standing wave doesn’t remain the same throughout the wave.
  • It keeps on changing as it is a function of x.
  • At certain positions the value of amplitude is maximum and at certain positions the value of amplitude is 0.
  • Nodes: – Nodes represent the positions of zero amplitude.
  • Antinodes: – Antinodes represent the positions of maximum amplitude.

Nodes:-

  • At nodes, amplitude is 0.
  • In case of the standing wave amplitude is given as :- 2asinkx
  • => 2asinkx = 0 ,=>sinkx = 0,=>sinkx =sin n π => kx=n π
  • The value of x represents position of nodes where amplitude is 0.
  • x=(nπ)/k … equation(i)
  • From the definition of k=(2π)/λ … equation(ii)
  • The position of nodes is represented by: –x=(n λ)/2from(i) and (ii),where n=1, 2, 3…
  • Note: –Half a wavelength (λ/2) separates two consecutive nodes.

Nodes and Antinodes: system closed at both ends

  • System closed at both ends means both the ends are rigid boundaries.
  • Whenever there is rigid body there is no displacement at the boundary. This implies at boundary amplitude is always 0. Nodes are formed at boundary.
  • Standing waves on a string of length L fixed at both ends have restricted wavelength.
  • This means wave will vibrate for certain specific values of wavelength.
  • At both ends,nodes will be formed.=>Amplitude=0.
  • Expression for node x =(nλ)/2.This value is true when x is 0 and L.
  • When x=L:- L=(nλ)/2 =>λ=(2L)/n ; n=1,2,3,4,…..
  • λ cannot take any value but it can take values which satisfy λ=(2L)/n this expression.
  • That is why we can say that the standing wave on a string which is tied on both ends has the restricted wavelength.
  • As wavelength is restricted therefore wavenumber is also restricted.
  • ν =v/λ (relation between wavelength and frequency)
  • Corresponding frequencies which a standing wave can have is given as: –ν= (vn)/2Lwhere v= speed of the travelling wave.
  • These frequencies are known as natural frequency or modes of oscillations.

Modes of Oscillations:-

  • ν= (vn)/2L where v=speed of the travelling wave, L=length of the string,  n=any natural number.
  • First Harmonic:-
    • For n=1, mode of oscillation is known as Fundamental mode.
    • Therefore ν1=v/(2L).This is the lowest possible value of frequency.
    • Therefore ν1is the lowest possible mode of the frequency.
    • 2 nodes at the ends and 1 antinode.
  • Second Harmonic:-
    • For n=2, ν2=(2v)/ (2L) =v/L
    • This is second harmonic mode of oscillation.
    • 3 nodes at the ends and 2 antinodes.
  • ThirdHarmonic:-
    • For n=3,ν= (3v)/ (2L).
    • This is third harmonic mode of oscillation.
    • 4 nodes and 3 antinodes.

Problem:- Find the frequency of note emitted (fundamental note) by a string 1m long and stretched by a load of 20 kg, if this string weighs 4.9 g. Given, g = 980 cm s–2?

Answer:-

L = 100 cm T = 20 kg = 20 × 1000 × 980 dyne

m= 4.9/100 = 0.049 g cm-1

Now the frequency of fundamental note produced,

ν= (1/2L) √ (T/m)

ν = 1/ (2×100) √(20x1000x980)/(0.049)

=100Hz

Problem:- A pipe 20 cm long is closed at one end, which harmonic mode of the pipeis resonantly excited by a 430 Hzsource? Will this same source can bein resonance with the pipe, if both ends are open? Speed of sound = 340ms–1?

Answer:-

The frequency of nth mode of vibration of a pipe closed at one endis given by

νn=(2n 1)ν/4L

ν=340ms-1; L=20cm=0.2m;νn=430Hz.

Therefore 430= ((2n-1) x 340)/ (4×0.2)

=>n=1

Therefore, first mode of vibration of the pipe is excited, for open pipe sincen must be an integer, the same source cannot be in resonance with thepipe with both ends open.

Nodes and Antinodes: system closed at one end

  • For a system which is closed at one end,only one node is formed at the closed end.
  • Consider there is one fixed end(x=0) and one open end(x=L), and a string is attached between these two ends.
  • At x=L, antinodes will be formed.This means amplitude will be maximum at this end.
  • Condition for formation of antinodes is x= (n+ (1/2)) (λ/2).=>L=(n+ (1/2)) (λ/2)
  • =>λ=(2L)/(n+(1/2)).This expression shows that the values of wavelength is restricted. n=0, 1, 2, 3…
  • Corresponding frequencies will be ν= v/(2L)(n+1/2) (By using ν=v/λ);n=0,1,2,3, …

Modes of oscillations:-

  • Fundamental frequency:- Also known as First Harmonic:-It corresponds to lowest possible value for n. That is n=0.
  • The expression for fundamental frequency is ν0 = v/(2L)x1/2 = v/(4L)
  • Odd Harmonics
    • n=1 ; ν =v/(2L)(1+(1/2)) = (3v)/(4L) =3ν0
    • n=2; ν =v/(2L)(2+(1/2))= (5v)/(4L)=5ν0
    • n=3;ν =v/(2L)(3+(1/2))=(7v)/(4L)=7ν0
  • For a system which is closed at one end and open at another end will get one fundamental frequency and all other odd harmonics.

Therefore nx550=1100 => n=2. It is the second harmonic.

  1. Fundamental Frequency: – ν1= (V)/ (4L)

Odd harmonics :- (3V)/ (4L), (5V)/ (4L), (7V)/ (4L) ….

=>ν1Y =275Hz. Where νY = Fundamental Frequency

=>nν1 = 1100 => n=1100/275=4

This should correspond to 4th harmonic but 4th harmonic cannot be present as only odd harmonics are present.

Therefore the same source cannot be resonated if the pipe is closed at one end.

Beats

Beats is the phenomenon caused by two sound waves of nearly same frequencies and amplitudes travelling in the same direction.

For example:-

  • Tuning of musical instruments like piano, harmonium etc. Before we start playing on these musical instruments they are set against the standard frequency. If it is not set a striking noise will keep on coming till it is set.

Mathematically

  • Consider only the time dependent and not the position dependent part of the wave.
  • s1=a cos ω1t and s2=a cos ω2t; where amplitude and phase of the waves are same,but the frequencies are varying. Also considering ω1> ω2.
  • When these 2 waves superimpose s= s1+ s2=a[cos ω1t + cos ω2t]
  • By simplifying , 2a (cos(ω1 – ω2)/2)t cos(ω1 + ω2)/2)t)
  • =>ω1 – ω2 is very small as ω1> ω2.Let (ω1 – ω)=ωb
  • =>ω1 + ωis very large. Let (ω1 + ω)=ωa
  • s= 2a cos ωbt cos ωat
  • cosωat will vary rapidly with time and 2acosωbt will change slowly with time.
  • Therefore we can say 2acosωbt = constant. As a result 2acosωbt = amplitude as it has small angular variation.

Problem:- Two sitar strings A and Bplaying the note ‘Dha’ are slightly out oftune and produce beats of frequency 5 Hz.The tension of the string B is slightlyincreased and the beat frequency is foundto decrease to 3 Hz. What is the originalfrequency of B if the frequency of A is 427 Hz?

Answer:-  Increase in the tension of a stringincreases its frequency. If the original frequencyof B (νB) were greater than that of A (νA), furtherincrease in νB should have resulted in anincrease in the beat frequency. But the beatfrequency is found to decrease. This shows thatνB< νA. Since νA – νB = 5 Hz, and νA = 427 Hz, weget νB = 422 Hz.

Doppler’s Effect

  • Doppler Effect is the phenomenon of motion-related frequency change.
  • Consider if a truck is coming from very far off location as it approaches near our house, the sound increases and when it passes our house the sound will be maximum. And when it goes away from our house sound decreases.
  • This effect is known as Doppler Effect.
  • A person who is observing is known as Observer and object from where the sound wave is getting generated it is known as Source.
  • When the observer and source come nearer to each other as a result waves get compressed. Therefore wavelength decreases and frequency increases.
  • Case 1:- stationary observer and moving source
  • Let the source is located at a distance L from the observer.
  • At any time t1, the source is at position P1.
  • Time taken by the wave to reach observer =L/v where v=speed of the sound wave.
  • After some time source moves to position P0 in time T0.
  • Distance between P1 and P0 =vsTo where vs is the velocity of the source.
  • Let t2be the time taken by the second wave to reach the observer
    • Total time taken by the for the second wave to be sent to the observer = To +( L+vsTo)/v
    • Total time taken by the for the third wave to be sent to the observer=2To +( L+2vsTo)/v
    • Therefore for nth point tn+1 =nTo +( L+nvsTo)/v
    • =>In time tn+1the observer captures n waves.
  • Total time taken by the waves to travel Time period T= (tn+1 – t1)/n
  • =To +(vsTo)/v =>T=To(1+vs/v)
  • Or v= 1/T
  • =>v = v0(1+vs/v)-1
  • By using binomial Theorem,v= v(1- vs/v)
  • If the source is moving towards the observer the expression will become v= v(1+ vs/v)
  • Case 2:- moving observer and stationary source
  • As the source is not moving therefore vs is replaced by -v0.
  • Therefore v= v(1+ v0/v)

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