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Theory · Physics Circle Master Book

Wave Motion

01
CONCEPT

Wave Basics & Properties of Sound

In one line
A wave carries energy and shape through a medium; the particles only vibrate about their mean positions.
WHAT IS A WAVE

17.1.1 Characteristics of wave motion

  1. Particles vibrate about their mean position; they do not travel with the wave.
  2. Energy moves forward with no net transport of the medium.
  3. Particle velocity changes from point to point; wave speed in a medium is constant.
  4. Wave speed depends on the medium, not on frequency, wavelength or intensity.
  5. A medium needs elasticity, inertia and low friction.
TYPES

17.1.2 Classifying waves

  1. Mechanical waves need a medium: string waves, sound, seismic waves.
  2. Non-mechanical waves need no medium: light, radio waves, X-rays.
  3. Sound cannot travel through vacuum.
Transverse wave on a stringvparticles vibrate up and downLongitudinal wave in airCRCRCRCvPparticles vibrate along v
Transverse wave: crests and troughs. Longitudinal wave: compressions (C) and rarefactions (R)
TransverseLongitudinal
vibration perpendicular to wavevibration along the wave
crests and troughscompressions and rarefactions
needs rigidityneeds volume elasticity
solids, liquid surfacesolids, liquids, gases
can be polarisedcannot be polarised
string of sitar, tabla skinsound in air, air in organ pipes
  1. Sound in air is longitudinal; light is transverse.
  2. Progressive waves carry energy forward; stationary waves do not.
  3. Audible range: 20 Hz to 20 kHz. Below it infrasonic; above it ultrasonic.
  4. Ultrasonic uses: SONAR, ultrasonography.
TERMS

17.1.3 Amplitude, wavelength, frequency, time period

λAcresttrough
Wavelength is the distance between two successive points in the same phase
  1. Amplitude a: largest displacement from the mean position.
  2. Wavelength λ: distance travelled in one time period; crest to next crest.
  3. Crest to neighbouring trough = half a wavelength.
  4. Frequency n: vibrations per second, in hertz.
T = 1n
v = nλ
  1. The frequency is fixed by the source and stays the same in every medium.
  2. Going to a new medium: v and λ change together, λ ∝ v.
  3. x vibrations of a fork: wave travels distance xλ.
  4. N waves counted in time t: n = Nt.
  5. A wave front is a surface of equal phase: spherical from a point source, plane far from it.
SphericalPlane
Rays are perpendicular to wave fronts
MUSICAL SOUND

17.1.4 Pitch, loudness and quality

  1. Pitch depends on frequency: high frequency sounds shrill.
  2. Female voice has higher pitch than male voice; a mosquito higher than a lion.
  3. Loudness is the sensation of intensity; it also depends on the ear.
  4. Quality (timbre) depends on the overtones present: it separates two sounds of equal pitch and intensity.
  5. Percussion instruments are struck: tabla, drum, dholak. Flute is a wind instrument; sitar and violin are string instruments.
INTENSITY

17.1.5 Energy flow per unit area per second

I = 2π2n2a2ρv
  1. At fixed medium: I ∝ a2n2.
  2. From a point source, the energy spreads over a sphere.
I = P4πr2
  1. So I ∝ 1r2 and a ∝ 1r.
  2. Two instruments of equal energy: a1n1 = a2n2.
SOUND LEVEL

17.1.6 The decibel scale

L = 10 log10 II0
  1. I0 = 10−12 W m−2 is the threshold of hearing.
  2. Every extra 10 dB multiplies the intensity by 10.
  3. Level difference of 30 dB: intensity ratio 1000.
  4. Doubling the intensity adds about 3 dB.
JEE · NEET

Corner

Shock waves and Mach number
  1. A body faster than sound (supersonic) leaves a cone-shaped shock wave behind it.
Mach number = speed of sourcespeed of sound
Reverberation and Sabine’s law
  1. Reverberation: sound persists in a hall after the source stops.
  2. V = volume of the hall, A = total absorption of its surfaces.
T = 0.16 VA
  1. All dimensions doubled: V becomes 8 times and A becomes 4 times, so T doubles.
Practise PYQs on this topicWave Basics & Properties of Sound · 43 questions with solutions
02
CONCEPT

Speed of Waves on a String

In one line
Speed of a wave on a string depends only on the tension and the mass per unit length.
THE FORMULA

17.2.1 Transverse wave on a stretched string

v = √Tμ
  1. T = tension, μ = mass per unit length.
  2. μ = πr2ρ for a wire of radius r and density ρ.
  3. Stress S = Tarea, so:
v = √Sρ
  1. Thicker wire, same tension and material: v ∝ 1r.
  2. Frequency does not change the speed; it only changes the wavelength.
RATIOS

17.2.2 Quick changes

  1. v ∝ √T, so T = k v2 : to make v thrice, T must be 9 times.
  2. Small changes: percentage change in v is half the percentage change in T.
Δvv = 12 ΔTT
  1. Large changes: use the ratio, never the half rule. 40% more v needs T × 1.96.
  2. Stretched elastically: T = YA × strain, so
v = √Y × strainρ
TIME OF TRAVEL

17.2.3 Pulse on a string

t = lv
  1. Two wires joined: add the times, t = t1 + t2; the tension is common, and v differs in each.
LOAD ON THE STRING

17.2.4 Hanging mass, in air and in a liquid

MvT = MgMσT′ = Mg(1 − σρ)
Immersing the load reduces the tension by the buoyant force
v = √Mgμ
  1. ρ is the density of the load, σ the density of the liquid.
v′ = v √1 − σρ
HANGING ROPE

17.2.5 Tension varies along the rope

yyv
Speed grows as the square root of the height from the free end
  1. At height y from the lower end, tension = weight of the rope below it.
v = √gy
  1. At the top of a rope of length L: v = √(gL). At the middle: v = √gL2.
  1. Time for a pulse to climb the whole rope does not depend on its mass:
t = 2√Lg
ROPE WITH A BLOCK

17.2.6 Rope of mass m1 with block m2 at the end

  1. The pulse is made at the bottom and read at the top.
  2. Frequency stays constant, so λ ∝ v ∝ √T.
λtopλbottom = √m1 + m2m2
JEE · NEET

Corner

Heated wire between rigid supports
  1. A temperature change Δθ produces a thermal stress in the clamped wire.
v = √YαΔθρ
Accelerating support
  1. A ball hangs by a light string in a car with acceleration a: tension uses the effective g.
T = M√g2 + a2
Practise PYQs on this topicSpeed of Waves · 40 questions with solutions
03
CONCEPT

Speed of Sound

In one line
Sound speed is set by the elasticity and density of the medium; in a gas it changes with temperature, not with pressure.
THE FORMULA

17.3.1 Speed in any medium

v = √Eρ
MediumElasticity ESpeed
solid rodYoung’s modulus Y√Yρ
liquid, gasbulk modulus B√Bρ
  1. Solids are most elastic: vsolid > vliquid > vgas.
  2. Roughly: steel 5000, water 1500, air 330 m/s.
  3. Compressibility = 1B.
NEWTON AND LAPLACE

17.3.2 Sound in a gas

  1. Newton took the process as isothermal: B = P.
v = √Pρ ≈ 280 m/s
  1. This is too low. Laplace said compressions are fast, so the process is adiabatic: B = γP.
v = √γPρ = √γRTM
  1. For air γ = 1.41, so v ≈ 332 m/s.
  2. Two gases at the same T: v ∝ √γM.
  3. Same γ, for example two monoatomic gases: v ∝ 1√M.
  4. Compared with rms speed:
vsoundvrms = √γ3
WHAT CHANGES v

17.3.3 Factors for a gas

  1. Pressure: no effect at constant temperature, because P and ρ rise together.
  2. Temperature: v ∝ √T, with T in kelvin.
v2v1 = √T2T1
vt = v0 + 0.61 t
  1. The second form is for small t in °C; v0 = 332 m/s.
  2. To double the speed, T in kelvin must become 4 times: 0 °C becomes 819 °C.
  3. Density: at the same P, v ∝ 1√ρ.
  4. Humidity: moist air is less dense, so sound is faster in humid air.
  5. A rigid box heated: ρ is fixed and v ∝ √P.
  6. A fixed fork in warmer air: n stays the same and λ rises with v.
WIND

17.3.4 Moving medium

Ssound, vwind, wθ
Only the wind component along the sound counts
v′ = v + w cos θ
  1. All frequencies travel with the same speed in a given medium.
  2. Phase, loudness and pitch have no effect on v.
Practise PYQs on this topicSpeed of Waves · 40 questions with solutions
04
CONCEPT

Progressive Wave Equation

In one line
One equation, y = a sin(ωt − kx), gives the amplitude, frequency, wavelength, speed and direction of a wave. Bring it to the standard form and compare.
THE EQUATION

17.4.1 Plane progressive wave

xyOPxv
P repeats the motion of O after a time xv
y = a sin 2π(tT − xλ)
y = a sin(ωt − kx)
y = a sin 2πλ(vt − x)
  1. Sign between t and x is minus: wave moves along +x.
  2. Sign is plus: wave moves along −x.
  3. A starting phase φ0 can be added inside the bracket.
READ FROM IT

17.4.2 Compare with the standard form

QuantityFormula
angular frequencyω = 2πn
wave numberk = 2πλ
wave speedv = ωk = nλ
frequencyn = ω2π
v = coefficient of tcoefficient of x
  1. Keep the coefficients of t and x outside the brackets before reading.
  2. For y = a sin(At − Bx): v = AB.
  3. Check the units asked: x in cm gives v in cm/s.
WRITING AND TESTING

17.4.3 What counts as a travelling wave

xyt = 0time tvt
The shape moves without change: y = f(x − vt)
  1. Any function of (x − vt) or (x + vt) is a travelling wave.
  2. A form with x2 and t2 separately, such as sin(4x2 − 9t2), is not.
  3. To write an equation: put a, ω = 2πn, k = 2πλ, and pick the sign from the direction.
  4. Fix the phase from the given point at t = 0.
PARTICLE VELOCITY

17.4.4 Do not mix it with wave speed

vp = aω cos(ωt − kx)
vp,max = aω
amax = aω2
vp,maxv = ak = 2πaλ
  1. Wave speed v is constant; particle velocity changes with time.
  2. Maximum particle speed = wave speed when λ = 2πa.
  3. Maximum particle speed = m × wave speed when λ = 2πam.
PHASE DIFFERENCE

17.4.5 The most asked idea of this chapter

12345
1, 2, 3 are in the same phase; 4 and 5 are in the same phase
Δφ = 2πλ Δx
Δφ = 2πT Δt
  1. Δx = λ for Δφ = 2π; Δx = λ2 for Δφ = π.
  2. Points in the same phase are separated by whole wavelengths.
  3. Particles always having the same speed (opposite phase) are separated by λ2.
ΔφΔx
π3λ6
π2λ4
πλ2
2πλ
  1. Between two equations: write both as sines first.
  2. cos θ = sin(θ + 90°): this adds π2 to the phase.
  3. Then Δφ = difference of the phase constants, and path difference = λ2π × Δφ.
  4. Total phase at a point: put the given x and t into (ωt − kx) + φ0.
TENSION

17.4.6 String wave given by an equation

  1. Find v = ωk from the equation, then use v = √Tμ.
T = μω2k2
  1. Always change μ to kgm; gcm is 0.1 kgm.
JEE · NEET

Corner

Differential wave equation
∂2y∂t2 = v2 ∂2y∂x2
  1. Every travelling wave satisfies it; v is the wave speed.
Slope and particle velocity
vp = −v ∂y∂x
  1. Particle velocity = −(wave speed) × (slope of the wave at that point).
Practise PYQs on this topicProgressive Wave Equation · 109 questions with solutions
05
CONCEPT

Reflection, Superposition & Interference

In one line
Waves pass through each other: at a point, the displacements simply add. Reflection only changes the phase.
REFLECTION

17.5.1 Rigid end and free end

Rigid endincidentphase πFree endincidentno change
Same medium: v, n and λ do not change on reflection
Rigid endFree end
crest returns astroughcrest
phase changeπzero
path changeλ2zero
  1. A wave reflected from a wall stays transverse and is inverted at a rigid end.
  2. Pressure waves show no phase change at a rigid end.
  3. In refraction, n stays fixed and both v and λ change.
ECHO

17.5.2 Reflected sound heard separately

t = 2dv
  1. Hearing persists for 0.1 s, so the reflector must be at d > 17 m.
  2. Person between two cliffs, echoes at t1 and t2:
d = v(t1 + t2)2
  1. Engine approaching a hill at speed u, whistle at distance d, echo after t: (v + u) t = 2d.
SUPERPOSITION

17.5.3 Displacements add

y = y1 + y2 + y3 + …
  1. Different phase: interference.
  2. Opposite direction: stationary wave.
  3. Slightly different frequency: beats.
RESULTANT

17.5.4 Amplitude of two waves of the same frequency

A2 = a12 + a22 + 2a1a2 cos φ
  1. Equal amplitudes: A = 2a cos φ2.
  2. Resultant equal to each wave, A = a, when φ = 120°.
  3. A = √2 a when φ = 90°.
  4. A sin ωt + B cos ωt has amplitude √(A2 + B2).
INTERFERENCE

17.5.5 Maximum and minimum

Constructive: φ = 0A = a1​ + a2​Destructive: φ = πA = a1​ − a2​
Blue and pink waves add to give the purple wave
I = I1 + I2 + 2√(I1I2) cos φ
ConstructiveDestructive
phase2mπ(2m + 1)π
pathmλ(2m + 1)λ2
amplitudea1 + a2|a1 − a2|
intensity(√I1 + √I2)2(√I1 − √I2)2
ImaxImin = (a1 + a2a1 − a2)2
  1. Amplitude ratio is the root of the intensity ratio.
  2. Imax + Imin = 2(I1 + I2).
  3. Energy is not destroyed; it is only redistributed.
  4. Sources must be coherent to give steady interference.
JEE · NEET

Corner

Wave going into a second string
  1. v1 and v2 are the speeds in the two strings; the frequency stays the same.
Ar = v2 − v1v2 + v1 Ai
At = 2v2v1 + v2 Ai
  1. Ar is negative when the second string is denser: the reflected pulse is inverted.
Quincke’s tube
  1. Sound splits into two paths and meets again at the detector.
  2. Pulling the sliding tube by x changes the path by 2x.
  3. For two successive maxima, the path changes by λ.
λ = 2x
Practise PYQs on this topicReflection, Superposition & Interference · 22 questions with solutions
06
CONCEPT

Stationary Waves

In one line
Two identical waves moving in opposite directions make a pattern that stays in place: fixed nodes and antinodes, with no energy flow.
THE EQUATION

17.6.1 Adding incident and reflected waves

  1. Reflection at a free end: y1 = a sin(ωt − kx) and y2 = a sin(ωt + kx).
y = 2a cos kx sin ωt
  1. Reflection at a rigid end:
y = 2a sin kx cos ωt
  1. Amplitude depends on x, not on t.
  2. The product form with x and t in separate factors tells you it is stationary.
  3. Each component wave has amplitude half of the 2a printed in the equation.
v = ωk
  1. This v is the speed of the component waves.
NODES AND ANTINODES

17.6.2 Where the amplitude is zero or largest

NNNNAAANN = λ2
Nodes (N) stay at rest; antinodes (A) have amplitude 2a
SeparationValue
node to next nodeλ2
antinode to next antinodeλ2
node to next antinodeλ4
EquationNodes atAntinodes at
2a sin kxx = 0, λ2, λ, …x = λ4, 3λ4, …
2a cos kxx = λ4, 3λ4, …x = 0, λ2, λ, …
  1. Two waves of frequency n meeting head-on: NN = v2n.
  2. At nodes, the pressure and density change is largest.
  3. At antinodes, the pressure and density change is smallest.
  4. Reflecting wall: a displacement node at the wall; the first antinode is at λ4 from it.
PROPERTIES

17.6.3 What to remember

  1. Can be transverse or longitudinal.
  2. No net transfer of energy across a node.
  3. All particles in one loop vibrate in phase.
  4. Particles in neighbouring loops differ by phase π.
  5. All particles (except nodes) reach their mean position together, twice in each period.
  6. Different particles have different amplitudes: zero at nodes, 2a at antinodes.
  7. Unequal amplitudes give non-zero minima at nodes.
ProgressiveStationary
transfers energyno energy transfer
same amplitude for allamplitude varies with position
phase changes point to pointsame phase within a loop
no point always at restnodes always at rest
Practise PYQs on this topicStationary Waves & Vibrating Strings · 118 questions with solutions
07
CONCEPT

Vibrating Strings & Sonometer

In one line
A string fixed at both ends has nodes at both ends, so only whole numbers of half-wavelengths fit: l = p λ2.
TERMS

17.7.1 The language of musical questions

  1. Harmonics: integer multiples of the fundamental frequency.
  2. Overtones: the frequencies above the fundamental. The first overtone is the next one actually produced.
  3. Octave: frequency double. x octaves up = 2x times.
  4. Unison: equal frequencies.
  5. Resonance: a body is forced at its own natural frequency and vibrates strongly.
MODES

17.7.2 Standing waves on a string

fundamentalfirst harmonicn1​first overtonesecond harmonic2n1​second overtonethird harmonic3n1​
Nodes at both ends; p loops in the pth harmonic
np = p2l√Tμ
λp = 2lp
pNameNodesAntinodes
1fundamental21
21st overtone32
32nd overtone43
p(p − 1)th overtonep + 1p
  1. All harmonics are present: 1 : 2 : 3 : …
  2. pth overtone = (p + 1)th harmonic.
  3. Antinodes = nodes − 1.
  4. Longest wavelength on a string = 2l.
  5. Consecutive resonances differ by the fundamental: n1 = np+1 − np.
  6. A fork of frequency n resonates only with its multiples: n, 2n, 3n, …
  7. Plucked at l2p from one end, the string vibrates in p loops.
LAWS OF STRINGS

17.7.3 What the fundamental depends on

n = 12lr√Tπρ
LawKeep fixedResult
lengthT, μn ∝ 1l
tensionl, μn ∝ √T
massl, Tn ∝ 1√μ
radiusl, T, ρn ∝ 1r
densityl, T, rn ∝ 1√ρ
Δnn = 12 ΔTT
  1. Use the half rule only for a small change.
  2. The second rule: nl = constant, so a shorter string has higher frequency.
SONOMETER

17.7.4 Resonance of a wire with a fork

hollow resonance boxMlbridgebridge
The hollow box increases the loudness; the bridges set the vibrating length
  1. Length between bridges is adjusted till the wire vibrates with the fork.
  2. Same fork, same length: p2 × T is constant.
p12M1 = p22M2
  1. Second overtone means p = 3: maximum amplitude at l6, l2 and 5l6.
LOAD IN A LIQUID

17.7.5 Immersed bob

n′n = √1 − σρ
  1. Bob of specific gravity s in water: σρ = 1s.
  2. At the poles and at the equator (same mass, g differs): for the same fork, l ∝ √g.
SEGMENTS

17.7.6 Bridges dividing one wire

  1. Tension and μ stay the same, so nl = constant.
  2. Frequencies 1 : 3 : 4 mean lengths ratio 12 : 4 : 3.
1n = 1n1 + 1n2 + 1n3
COMPOSITE WIRE

17.7.7 Two wires joined with a node at the joint

joint = nodel1​, m1​ : p1​ loopsl2​, m2​ : p2​ loops
Same tension and same frequency; loops p1 and p2 differ
p1l1√Tμ1 = p2l2√Tμ2
  1. Smallest whole numbers p1 and p2 give the lowest common frequency.
  2. Same material, radii r and 2r: μ2 = 4μ1.
Practise PYQs on this topicStationary Waves & Vibrating Strings · 118 questions with solutions
08
CONCEPT

Organ Pipes & Resonance Tube

In one line
Air column modes: a closed end is a node, an open end is an antinode. Closed pipes have odd harmonics only; open pipes have all.
CLOSED PIPE

17.8.1 Node at the closed end, antinode at the open end

fundamentall = λ4n1​1st overtonel = 3λ43n1​2nd overtonel = 5λ45n1​
Curves show the displacement envelope
nN = (2N − 1)v4l
  1. Only odd harmonics: 1 : 3 : 5 : …
  2. pth overtone = (2p + 1)th harmonic.
  3. Longest wavelength = 4l.
  4. pth overtone has (p + 1) nodes and (p + 1) antinodes.
  5. Allowed lengths for a fork: l, 3l, 5l, …
  6. Time for sound to go from open end to closed end: t = lv, so n = 14t.
OPEN PIPE

17.8.2 Antinodes at both ends

fundamentall = λ2n1​1st overtonel = 2λ22n1​2nd overtonel = 3λ23n1​
All harmonics are present, so the sound is richer
nN = Nv2l
  1. All harmonics: 1 : 2 : 3 : …
  2. pth overtone = (p + 1)th harmonic.
  3. Longest wavelength = 2l.
  4. pth overtone has (p + 1) nodes and (p + 2) antinodes.
  5. Successive harmonics differ by v2l.
COMPARE

17.8.3 Closed and open pipes

ClosedOpen
fundamentalv4lv2l
harmonicsodd onlyall
pth overtone(2p + 1)n1(p + 1)n1
nodes in pth overtonep + 1p + 1
antinodesp + 1p + 2
  1. Same length: fundamental of open = 2 × that of closed.
  2. Unison: equate the frequencies and cancel v.
2p + 14lc = q + 12lo
  1. Here p and q are the overtone numbers of the closed and open pipes.
  2. Opening a closed end or closing an open end: use the two fundamentals v4l and v2l.
  3. Hearing limit: largest (2p + 1)n1 ≤ 20 kHz gives the audible overtones of a closed pipe.
  4. Two harmonics given: if they are 150 and 250 with difference 100, the fundamental is half the difference: 50 Hz, and the pipe is closed.
WATER AND CUTTING

17.8.4 What happens to the length

  1. Open pipe dipped so that a fraction f is in water: it becomes closed with length (1 − f)l.
n′ = v4(1 − f)l
  1. Half in water: n′ = n, the fundamental is unchanged.
  2. Closed pipe with water filling fraction f: the length becomes (1 − f)l.
  3. Closed pipe cut into two: the closed part has 2n, and the open part has 4n.
END CORRECTION

17.8.5 The antinode is just outside the open end

antinodeelclosedeeopen
The antinode forms a distance e beyond each open end
e = 0.6r
  1. Closed pipe: effective length l + e. Open pipe: effective length l + 2e.
  2. A wider pipe has more end correction and lower frequency.
nclosed = v4(l + 0.6r)
nopen = v2(l + 1.2r)
RESONANCE TUBE

17.8.6 Measuring the speed of sound

l1​l1​ + e = λ4l2​l2​ + e = 3λ4
The water level sets the length of the closed column
v = 2n(l2 − l1)
e = l2 − 3l12
  1. Next resonance length: l2 = 3l1 + 2e.
  2. So l2 is slightly more than 3l1.
  3. Without end correction: l1 = λ4 and l2 = 3λ4.
  4. Water poured in a tube of height H: find the longest resonating air column; the water height is H minus that.
JOINS AND GASES

17.8.7 Other ways of asking

  1. Open pipes n1 and n2 joined end to end:
1n = 1n1 + 1n2
  1. Closed pipe n1 joined to open pipe n2 makes a closed pipe of length l1 + l2:
1n = 1n1 + 2n2
  1. Different gases at equal compressibility: v ∝ 1√ρ; put this into the pipe formula.
  2. String resonating with a pipe: equate the two formulas, and use v = √Tμ for the string.
Practise PYQs on this topicOrgan Pipes & Resonance Tube · 121 questions with solutions
09
CONCEPT

Beats

In one line
Two notes of slightly different frequency make the loudness rise and fall; the number of rises per second is the difference of the frequencies.
BEATS

17.9.1 Waxing and waning

T = 1n1​ − n2​loudsoft
The amplitude varies slowly; one maximum to the next is one beat
x = n1 − n2
Tbeat = 1n1 − n2
  1. Hearing persists for 0.1 s, so distinct beats need a difference below 10 per second.
  2. Beats need nearly equal frequencies; amplitudes should be nearly equal.
  3. Time between maximum and next minimum = half the beat period.
  4. Waves of frequency n − 1, n and n + 1 together give 2 beats per second.
A = 2a cos π(n1 − n2)t
FROM EQUATIONS

17.9.2 Read each frequency

  1. y = a sin ωt: n = ω2π. So x = ω1 − ω22π.
  2. If y = a sin 2π(…): the coefficient of t inside is n.
  3. Intensity ratio between waxing and waning:
ImaxImin = (a1 + a2a1 − a2)2
FROM WAVELENGTHS

17.9.3 Same speed of sound

x = v(1λ1 − 1λ2)
  1. Given the beats, solve this for v.
  2. Wavelength difference for two forks: Δλ = v(1n1 − 1n2).
SERIES OF FORKS

17.9.4 Each fork x beats from the next

nlast = nfirst + (N − 1)x
  1. Last fork is the octave of the first: nlast = 2nfirst.
nfirst = (N − 1)x
TUNING FORK

17.9.5 What changes its frequency

  1. Wax on a prong (loading): frequency falls.
  2. Filing a prong: frequency rises.
  3. Rise in temperature: frequency falls slightly.
  4. Shorter prongs: higher frequency.
  5. A fork does not vibrate well if one prong is broken.
FINDING THE UNKNOWN

17.9.6 The ± decision

  1. Fork A is known, B is unknown, and they give x beats. So nB = nA ± x.
  2. Load or file, and watch the new beats x′.
Change on BBeatsnB was
wax (n falls)risenA − x
wax (n falls)fallnA + x
file (n rises)risenA + x
file (n rises)fallnA − x
  1. If A is treated instead of B, the sign is reversed.
  2. Beats unchanged after wax on B: nB was nA + x, and it is now nA − x.
STRINGS AND WIRES

17.9.7 Beats from a change in length or tension

  1. Fork with two lengths l1 and l2 of the wire, x beats each (l1 > l2): the shorter wire is above the fork and the longer below.
n = xl1 + l2l1 − l2
  1. Length 25 cm in unison and 24 cm with x beats: nwire = n + x.
  2. Two forks in unison with l1 and l2, with x beats: use n1l1 = n2l2.
  3. Tension increased: n rises. If beats fall, the string was below the fork.
  4. Two wires in tension T1 and T2 with x beats: n1 − n2 = x and n ∝ √T.
ORGAN PIPES

17.9.8 Beats from two pipes

x = v2(1l1 − 1l2)
  1. This is for two open pipes at the fundamental.
  2. Open and closed of the same length: x = v4l.
  3. Compare harmonics carefully: overtones first change into harmonics.
  4. Changing a length changes that pipe only. Find vl from the first case, then put it into the second.
Practise PYQs on this topicBeats · 78 questions with solutions
10
CONCEPT

Doppler Effect

In one line
Relative motion along the line of sound changes the frequency heard; the source’s own frequency stays the same.
THE IDEA

17.10.1 Moving source squeezes the waves

vₛbehind: λ′ larger, n′ lowerin front: λ′ smaller, n′ higher
Moving source: wavelength is shorter in front and longer behind
  1. Moving source changes the wavelength; the speed of sound stays v.
  2. Moving observer changes the rate at which wave crests meet him; λ stays the same.
GENERAL FORMULA

17.10.2 One formula, signs from the direction

n′ = n · v ± vov ∓ vs
  1. Upper signs when moving towards each other.
  2. Lower signs when moving away.
  3. Approaching gives a higher pitch, and receding a lower pitch.
  4. v is the speed of sound relative to the ground.
MOVING SOURCE

17.10.3 Observer at rest

towards: n′ = n · vv − vs
away: n′ = n · vv + vs
λ′ = v − vsn
  1. Half frequency needs source speed v, moving away.
  2. Source with speed v10 towards observer: n′ = 10n9.
MOVING OBSERVER

17.10.4 Source at rest

towards: n′ = n · v + vov
away: n′ = n · v − vov
  1. Percent rise in frequency = vov × 100: vo = v5 gives 20%.
  2. Triple frequency: vo = 2v towards the source.
  3. Same speed approach then recede: n1 − n2 = 2n vov.
BOTH MOVING

17.10.5 Common cases

Casen′ formula
towards each othern · v + vov − vs
away from each othern · v − vov + vs
source chases observern · v − vov − vs
observer chases sourcen · v + vov + vs
  1. Same speed in the same direction: n′ = n.
  2. A passenger in the train hears the true frequency of its whistle.
  3. Medium alone moving: no Doppler effect.
  4. Moving at right angles to the line joining them: n′ = n.
CROSSING

17.10.6 Source passes a stationary observer

nbeforenafter = v + vsv − vs
  1. For vs ≪ v: drop in pitch = 2n vsv.
  2. Observer passing a stationary source: ratio = v + vov − vo.
REFLECTION

17.10.7 Car and wall, sound off an obstacle

wallS, Oun′n″
The wall hears n′ and returns it without change; the driver then hears n″
n″ = n · v + uv − u
  1. Car moving towards the wall at speed u, with source and driver together.
  2. Small u: n″ − n = 2unv.
  3. Wavelength reflected from an obstacle moving towards the source: λ′ = λ · v − uv + u.
  4. Source moving away towards a cliff: echo frequency rises to n · vv − vs.
DOPPLER BEATS

17.10.8 Two notes, one listener

  1. Two stationary sources of frequency n, listener moving from one to the other with speed u:
x = 2nuv
  1. The same result in terms of wavelength: x = 2uλ.
  2. Direct sound and echo: a source moving towards a wall at vs with the observer behind it:
x = n v(1v − vs − 1v + vs)
  1. A vehicle accelerating from a stationary siren: first get the required vo from the frequency, then use v2 = 2as.
JEE · NEET

Corner

Source moving off the line of sight
vₛACBOθ
Only the component along the line to O counts
n′ = n · vv − vs cos θ
  1. As the source goes from A to B, θ grows and n′ keeps falling.
  2. At C, θ = 90° and n′ = n.
Source on a rotating string
  1. Speed of the source: vs = rω.
nmax = n · vv − rω
nmin = n · vv + rω
SONAR
  1. Ultrasonic waves are reflected by a submarine and received back.
  2. The target acts first as observer, then as source.
n″ = n · v + vtv − vt
Wind
  1. Wind speed w in the direction of sound adds to v in both places.
n′ = n · v + w − vov + w − vs
Practise PYQs on this topicDoppler Effect · 83 questions with solutions
EXAM-DAY CARD

Just see this before the exam

Wave Motion
v = nλ
wave relation
v = √Tμ
string
v = √γRTM
sound
n = p2l√Tμ
string modes
n = v4l, v2l
closed, open
n′ = n · v ± vov ∓ vs
Doppler
FORMULA SHEET

Whole chapter on one page

Basics & speed

v = nλ, I ∝ a2n2

I ∝ 1r2, L = 10 log II0

string: v = √Tμ

rope: v = √gy, t = 2√Lg

Sound

v = √γPρ = √γRTM

v ∝ √T, no effect of P

vt = v0 + 0.61 t

vvrms = √γ3

Wave equation

y = a sin(ωt − kx)

v = ωk, k = 2πλ

vp,max = aω, vp,maxv = 2πaλ

Δφ = 2πλ Δx

Superposition

A2 = a12 + a22 + 2a1a2 cos φ

ImaxImin = (a1 + a2a1 − a2)2

y = 2a sin kx cos ωt

NN = λ2, NA = λ4

String & pipes

string: n = p2l√Tμ

closed: (2N − 1)v4l

open: Nv2l

v = 2n(l2 − l1), e = 0.6r

Beats & Doppler

x = n1 − n2

nfirst = (N − 1)x

n′ = n · v ± vov ∓ vs

wall: n″ = n · v + uv − u

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Overtone number used as harmonic number

Closed pipe with even harmonics

Half rule for a large change in T

Sign of vo and vs by memory

CORRECT ✓

Closed: pth overtone = (2p + 1)th harmonic

Closed pipe has odd harmonics only

Use n ∝ √T as a ratio for large changes

Moving towards raises n′, away lowers it