Waves and Sound

Wave Properties

Classification

Waves are classified by the direction of particle displacement relative to the direction of energy Propagation:

Type	Particle Motion	Examples
Transverse	Perpendicular to propagation	Light, water surface waves, EM waves
Longitudinal	Parallel to propagation	Sound, compression waves in springs

Wave Terminology

Term	Symbol	SI Unit	Definition
Wavelength	$\lambda$	m	Distance between consecutive points in phase
Frequency	$f$	Hz	Number of complete oscillations per second
Period	$T$	s	Time for one complete oscillation
Amplitude	$A$	m	Maximum displacement from equilibrium
Wave speed	$v$	m/s	Speed of propagation

The Wave Equation

v = f\lambda

Since $f = 1/T$ :

v = \frac{\lambda}{T}

Phase Difference

Two points separated by a distance $\Delta x$ along a wave have a phase difference:

\Delta\phi = \frac{2\pi\,\Delta x}{\lambda}

Points in phase: $\Delta\phi = 0, 2\pi, 4\pi, \ldots$ (separated by whole wavelengths)

Points in antiphase: $\Delta\phi = \pi, 3\pi, 5\pi, \ldots$ (separated by half-wavelengths)

Worked Example 1

A wave has frequency $250 \mathrm{ Hz}$ and wavelength $1.4 \mathrm{ m}$ . Find the wave speed and the phase difference between two points $0.35 \mathrm{ m}$ apart.

Solution

v = f\lambda = 250 \times 1.4 = 350 \mathrm{ m/s}

\Delta\phi = \frac{2\pi \times 0.35}{1.4} = \frac{\pi}{2} = 90^\circ

Worked Example 2

A sound wave in air has a wavelength of $0.686 \mathrm{ m}$ . When it enters water, its speed Changes to $1480 \mathrm{ m/s}$ . Find the frequency of the wave and its wavelength in water.

Solution

Speed in air: $v_{\mathrm{air}} = 343 \mathrm{ m/s}$ (at $20^\circ\mathrm C$ )

f = \frac{v_{\mathrm{air}}}{\lambda_{\mathrm{air}}} = \frac{343}{0.686} = 500 \mathrm{ Hz}

Frequency remains constant when crossing a boundary. In water:

\lambda_{\mathrm{water}} = \frac{v_{\mathrm{water}}}{f} = \frac{1480}{500} = 2.96 \mathrm{ m}

Wave Phenomena

Experiment with wave speed, frequency, amplitude, and damping to see how they affect the wave Pattern.

Reflection

Waves reflect off a surface. The angle of incidence equals the angle of reflection. A fixed-end Reflection inverts the wave; a free-end reflection does not.

Refraction

When a wave crosses a boundary between two media, its speed and wavelength change but its frequency Remains constant:

f = \frac{v_1}{\lambda_1} = \frac{v_2}{\lambda_2}

Diffraction

Diffraction is the spreading of waves when they pass through a gap or around an obstacle. Maximum Diffraction occurs when the gap width is approximately equal to the wavelength:

d \approx \lambda

Interference

When two coherent waves overlap, the resultant displacement is the sum of individual displacements (principle of superposition).

Constructive interference (reinforcement) occurs when waves arrive in phase:

\Delta\phi = 0, 2\pi, 4\pi, \ldots \quad \mathrm{or} \quad \mathrm{path difference} = n\lambda

Destructive interference (cancellation) occurs when waves arrive in antiphase:

\Delta\phi = \pi, 3\pi, 5\pi, \ldots \quad \mathrm{or} \quad \mathrm{path difference} = (n + \tfrac{1}{2})\lambda

Worked Example 3

Two coherent sources are $0.8 \mathrm{ m}$ apart and emit waves of wavelength $0.04 \mathrm{ m}$ . A Point P is $1.5 \mathrm{ m}$ from one source and $1.7 \mathrm{ m}$ from the other. Is there Constructive or destructive interference at P?

Solution

Path difference: $\Delta s = 1.7 - 1.5 = 0.2 \mathrm{ m}$

\frac{\Delta s}{\lambda} = \frac{0.2}{0.04} = 5

Since the path difference is an integer multiple of $\lambda$ Constructive interference occurs at P.

Worked Example 4

Light of wavelength $600 \mathrm{ nm}$ passes through a double slit with slit separation $0.1 \mathrm{ mm}$ . The interference pattern is observed on a screen $2 \mathrm{ m}$ away. Find The separation between adjacent bright fringes.

Solution

For a double-slit arrangement, the fringe spacing is:

\Delta y = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 2}{0.1 \times 10^{-3}} = \frac{1.2 \times 10^{-6}}{1 \times 10^{-4}} = 0.012 \mathrm{ m} = 12 \mathrm{ mm}

Standing Waves

Formation

A standing wave forms when two identical waves travelling in opposite directions superpose. Nodes (points of zero amplitude) and antinodes (points of maximum amplitude) are fixed in position.

String Fixed at Both Ends

Mode	Diagram	Wavelength	Frequency
Fundamental (1st harmonic)	—N—A—N—	$2L$	$f_1 = v/(2L)$
2nd harmonic	—N—A—N—A—N—	$L$	$f_2 = v/L = 2f_1$
3rd harmonic	—N—A—N—A—N—A—N—	$2L/3$	$f_3 = 3v/(2L) = 3f_1$

: $f_n = \frac{nv}{2L}$ for $n = 1, 2, 3, \ldots$

Pipe Closed at One End

Only odd harmonics are possible for a pipe closed at one end:

Mode	Wavelength	Frequency
Fundamental (1st harmonic)	$4L$	$f_1 = v/(4L)$
3rd harmonic	$4L/3$	$f_3 = 3v/(4L) = 3f_1$
5th harmonic	$4L/5$	$f_5 = 5v/(4L) = 5f_1$

: $f_n = \frac{nv}{4L}$ for $n = 1, 3, 5, \ldots$

Worked Example 5

A string of length $0.65 \mathrm{ m}$ has a fundamental frequency of $440 \mathrm{ Hz}$ . Find the Wave speed and the frequency of the third harmonic.

Solution

v = 2Lf_1 = 2 \times 0.65 \times 440 = 572 \mathrm{ m/s}

f_3 = 3f_1 = 3 \times 440 = 1320 \mathrm{ Hz}

Worked Example 6

A pipe is closed at one end and open at the other. The pipe is $0.85 \mathrm{ m}$ long. Find the Fundamental frequency and the next two resonant frequencies. (Speed of sound $= 340 \mathrm{ m/s}$ )

Solution

For a pipe closed at one end, only odd harmonics are present:

f_1 = \frac{v}{4L} = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100 \mathrm{ Hz}

f_3 = 3f_1 = 300 \mathrm{ Hz}

f_5 = 5f_1 = 500 \mathrm{ Hz}

The resonant frequencies are $100 \mathrm{ Hz}$ , $300 \mathrm{ Hz}$ And $500 \mathrm{ Hz}$ .

Sound Waves

Nature of Sound

Sound is a longitudinal mechanical wave. It requires a medium for propagation and cannot travel through a vacuum. The speed of sound depends on the medium:

Medium	Speed of Sound (m/s)
Air (at $20^\circ\mathrm C$ )	343
Water	1480
Steel	5960

Intensity and Loudness

Intensity is the power per unit area carried by the wave:

I = \frac{P}{4\pi r^2}

Intensity follows an inverse square law for a point source.

Sound intensity level is measured in decibels (dB):

\beta = 10\log_{10}\left(\frac{I}{I_0}\right)

Where $I_0 = 10^{-12} \mathrm{ W/m}^2$ is the threshold of hearing.

Worked Example 7

A sound source emits power $0.01 \mathrm{ W}$ uniformly in all directions. Find the intensity and the sound intensity level at a distance of $10 \mathrm{ m}$ .

Solution

I = \frac{P}{4\pi r^2} = \frac{0.01}{4\pi \times 100} = \frac{0.01}{1257} = 7.96 \times 10^{-6} \mathrm{ W/m}^2

\beta = 10\log_{10}\left(\frac{7.96 \times 10^{-6}}{10^{-12}}\right) = 10\log_{10}(7.96 \times 10^6) = 10 \times 6.90 = 69.0 \mathrm{ dB}

Worked Example 8

The intensity of a sound at $3 \mathrm{ m}$ from a source is $5 \times 10^{-4} \mathrm{ W/m}^2$ . Find the intensity and the sound intensity level at $12 \mathrm{ m}$ from the source.

Solution

Using the inverse square law:

I_2 = I_1 \times \left(\frac{r_1}{r_2}\right)^2 = 5 \times 10^{-4} \times \left(\frac{3}{12}\right)^2 = 5 \times 10^{-4} \times \frac{1}{16} = 3.125 \times 10^{-5} \mathrm{ W/m}^2

At $3 \mathrm{ m}$ : $\beta_1 = 10\log_{10}(5 \times 10^{-4} / 10^{-12}) = 10\log_{10}(5 \times 10^8) = 10 \times 8.699 = 87.0 \mathrm{ dB}$

At $12 \mathrm{ m}$ : $\beta_2 = 10\log_{10}(3.125 \times 10^{-5} / 10^{-12}) = 10\log_{10}(3.125 \times 10^7) = 10 \times 7.495 = 74.9 \mathrm{ dB}$

The distance quadrupled (factor of 4), so the level decreased by $87.0 - 74.9 = 12.1 \mathrm{ dB}$ which matches $10\log_{10}(16) = 12.0 \mathrm{ dB}$ .

Ultrasound

Ultrasound refers to sound waves with frequencies above $20\,000 \mathrm{ Hz}$ Beyond the range of Human hearing. Applications include medical imaging, sonar, and non-destructive testing.

Electromagnetic Spectrum

Properties

Electromagnetic (EM) waves are transverse waves that do not require a medium. All EM waves travel at the speed of light in vacuum:

c = 3.0 \times 10^8 \mathrm{ m/s}

The relationship between speed, frequency, and wavelength:

c = f\lambda

The EM Spectrum

Region	Wavelength Range	Typical Use
Radio waves	$\gt 0.1 \mathrm{ m}$	Broadcasting, communication
Microwaves	$1 \mathrm{ mm}$ to $0.1 \mathrm{ m}$	Cooking, radar, satellite
Infrared	$700 \mathrm{ nm}$ to $1 \mathrm{ mm}$	Thermal imaging, remote controls
Visible light	$400\mathrm{--}700 \mathrm{ nm}$	Human vision
Ultraviolet	$10\mathrm{--}400 \mathrm{ nm}$	Sterilisation, fluorescence
X-rays	$0.01\mathrm{--}10 \mathrm{ nm}$	Medical imaging, security
Gamma rays	$\lt 0.01 \mathrm{ nm}$	Cancer treatment, nuclear processes

Worked Example 9

A radio station broadcasts at a frequency of $100 \mathrm{ MHz}$ . Find the wavelength.

Solution

\lambda = \frac{c}{f} = \frac{3.0 \times 10^8}{100 \times 10^6} = 3.0 \mathrm{ m}

Worked Example 10

A laser emits light of wavelength $532 \mathrm{ nm}$ (green). Find the frequency and the energy of each photon. (Planck constant $h = 6.63 \times 10^{-34} \mathrm{ J\, s}$ )

Solution

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{532 \times 10^{-9}} = 5.64 \times 10^{14} \mathrm{ Hz}

E = hf = 6.63 \times 10^{-34} \times 5.64 \times 10^{14} = 3.74 \times 10^{-19} \mathrm{ J}

Common Pitfalls

Confusing the speed of a wave with the speed of the particles. The wave speed $v = f\lambda$ is the speed at which the wave pattern propagates, not the speed of individual particles.
Forgetting that frequency remains constant when a wave crosses a boundary. It is wavelength and speed that change.
Confusing nodes and antinodes in standing waves. Nodes are points of zero amplitude (minimum); antinodes are points of maximum amplitude.
Misidentifying the type of wave. Sound is longitudinal; light is transverse. Water surface waves are a combination.
When calculating sound intensity level, forgetting that the formula uses $\log_{10}$ Not $\ln$ .
Forgetting that a pipe closed at one end only supports odd harmonics.

Summary Table

Topic	Key Formula	Key Concept
Wave equation	$v = f\lambda$	Relates speed, frequency, wavelength
Phase difference	$\Delta\phi = 2\pi\Delta x/\lambda$	Fraction of a cycle
Constructive interference	Path difference $= n\lambda$	Waves in phase
Destructive interference	Path difference $= (n+1/2)\lambda$	Waves in antiphase
Standing waves (string)	$f_n = nv/(2L)$	All harmonics
Standing waves (pipe)	$f_n = nv/(4L)$ , $n$ odd	Odd harmonics only
Sound intensity	$I = P/(4\pi r^2)$	Inverse square law
Intensity level	$\beta = 10\log(I/I_0)$	Decibels
EM spectrum	$c = f\lambda$	All EM waves at speed $c$

Problem Set

Problem 1. A water wave has a wavelength of $2.5 \mathrm{ m}$ and a frequency of $0.4 \mathrm{ Hz}$ . A boat bobs up and down as the wave passes. Find the wave speed and the time between successive crests passing the boat.

Solution

v = f\lambda = 0.4 \times 2.5 = 1.0 \mathrm{ m/s}

The time between successive crests equals the period: $T = 1/f = 1/0.4 = 2.5 \mathrm{ s}$

If you get this wrong, revise: Wave Properties / The Wave Equation

Problem 2. Light of wavelength $550 \mathrm{ nm}$ in air enters glass where its speed is $2.0 \times 10^8 \mathrm{ m/s}$ . Find the frequency and the wavelength in glass.

Solution

f = \frac{c}{\lambda_{\mathrm{air}}} = \frac{3.0 \times 10^8}{550 \times 10^{-9}} = 5.45 \times 10^{14} \mathrm{ Hz}

Frequency remains constant in glass:

\lambda_{\mathrm{glass}} = \frac{v_{\mathrm{glass}}}{f} = \frac{2.0 \times 10^8}{5.45 \times 10^{14}} = 3.67 \times 10^{-7} \mathrm{ m} = 367 \mathrm{ nm}

If you get this wrong, revise: Wave Phenomena / Refraction

Problem 3. Two speakers are connected to the same signal generator and placed $3 \mathrm{ m}$ Apart. They emit sound of frequency $680 \mathrm{ Hz}$ . A microphone is moved along a line parallel to the speakers, $5 \mathrm{ m}$ away. Find the distance between consecutive positions of loud Sound. (Speed of sound $= 340 \mathrm{ m/s}$ )

Solution

\lambda = \frac{v}{f} = \frac{340}{680} = 0.5 \mathrm{ m}

Using the double-slit fringe spacing formula:

\Delta y = \frac{\lambda D}{d} = \frac{0.5 \times 5}{3} = 0.833 \mathrm{ m}

The spacing between consecutive loud positions is $0.833 \mathrm{ m}$ .

If you get this wrong, revise: Wave Phenomena / Interference

Problem 4. A guitar string of length $0.65 \mathrm{ m}$ produces a fundamental note of $330 \mathrm{ Hz}$ . When the string is pressed against a fret to shorten its effective length to $0.55 \mathrm{ m}$ What is the new fundamental frequency?

Solution

Since $f_1 = v/(2L)$ and $v$ is constant, $f \propto 1/L$ :

f_{\mathrm{new}} = f_1 \times \frac{L_1}{L_{\mathrm{new}}} = 330 \times \frac{0.65}{0.55} = 330 \times 1.182 = 390 \mathrm{ Hz}

If you get this wrong, revise: Standing Waves / String Fixed at Both Ends

Problem 5. A pipe open at both ends is $1.2 \mathrm{ m}$ long. Find the fundamental Frequency and the first three harmonics. (Speed of sound $= 340 \mathrm{ m/s}$ )

Solution

For a pipe open at both ends, all harmonics are present:

f_1 = \frac{v}{2L} = \frac{340}{2 \times 1.2} = \frac{340}{2.4} = 141.7 \mathrm{ Hz}

f_2 = 2f_1 = 283.3 \mathrm{ Hz}

f_3 = 3f_1 = 425.0 \mathrm{ Hz}

f_4 = 4f_1 = 566.7 \mathrm{ Hz}

If you get this wrong, revise: Standing Waves

Problem 6. A pipe closed at one end has a fundamental frequency of $85 \mathrm{ Hz}$ . Find The length of the pipe and the frequency of the next resonant frequency. (Speed of sound $= 340 \mathrm{ m/s}$ )

Solution

f_1 = \frac{v}{4L} \implies L = \frac{v}{4f_1} = \frac{340}{4 \times 85} = \frac{340}{340} = 1.0 \mathrm{ m}

For a pipe closed at one end, the next resonance after $f_1$ is the third harmonic (only odd Harmonics exist):

f_3 = 3f_1 = 3 \times 85 = 255 \mathrm{ Hz}

If you get this wrong, revise: Standing Waves / Pipe Closed at One End

Problem 7. The intensity of a sound at a distance of $5 \mathrm{ m}$ from a source is $2 \times 10^{-4} \mathrm{ W/m}^2$ . Find the intensity at $20 \mathrm{ m}$ and the intensity level at both distances.

Solution

At $5 \mathrm{ m}$ : $\beta_1 = 10\log_{10}(2 \times 10^{-4}/10^{-12}) = 10\log_{10}(2 \times 10^8) = 10 \times 8.301 = 83.0 \mathrm{ dB}$

At $20 \mathrm{ m}$ : $I_2 = I_1 \times (5/20)^2 = 2 \times 10^{-4} \times 1/16 = 1.25 \times 10^{-5} \mathrm{ W/m}^2$

\beta_2 = 10\log_{10}(1.25 \times 10^{-5}/10^{-12}) = 10\log_{10}(1.25 \times 10^7) = 10 \times 7.097 = 71.0 \mathrm{ dB}

If you get this wrong, revise: Sound Waves / Intensity and Loudness

Problem 8. Two sound sources each produce an intensity level of $80 \mathrm{ dB}$ at a point. If both sources operate simultaneously, what is the combined intensity level?

Solution

\beta_1 = 10\log_{10}(I_1/I_0) = 80 \implies I_1/I_0 = 10^8 \implies I_1 = 10^8 I_0

Combined intensity: $I_{\mathrm{total}} = 2I_1 = 2 \times 10^8 I_0$

\beta_{\mathrm{total}} = 10\log_{10}(2 \times 10^8) = 10\log_{10}(2) + 10\log_{10}(10^8) = 3.01 + 80 = 83.0 \mathrm{ dB}

Doubling the intensity increases the level by $3 \mathrm{ dB}$ (not $160 \mathrm{ dB}$ ).

If you get this wrong, revise: Sound Waves / Intensity and Loudness

Problem 9. A microwave oven operates at a frequency of $2.45 \mathrm{ GHz}$ . Find the Wavelength. Why are microwaves particularly effective for heating food?

Solution

\lambda = \frac{c}{f} = \frac{3.0 \times 10^8}{2.45 \times 10^9} = 0.122 \mathrm{ m} = 12.2 \mathrm{ cm}

Microwaves are effective because water molecules have a resonant frequency close to $2.45 \mathrm{ GHz}$ . The microwaves cause water molecules to oscillate, and the molecular friction Generates thermal energy throughout the food.

If you get this wrong, revise: Electromagnetic Spectrum

Problem 10. A stationary observer hears a sound of frequency $500 \mathrm{ Hz}$ from a source Moving towards them at $30 \mathrm{ m/s}$ . Find the apparent frequency heard by the observer. (Speed of sound $= 340 \mathrm{ m/s}$ )

Solution

Using the Doppler effect formula (source moving towards stationary observer):

f' = f \times \frac{v}{v - v_s} = 500 \times \frac{340}{340 - 30} = 500 \times \frac{340}{310} = 500 \times 1.097 = 548 \mathrm{ Hz}

The apparent frequency is higher than the actual frequency because the source is moving towards The observer.

If you get this wrong, revise: Sound Waves

For the A-Level treatment of this topic, see Wave Properties.

Tip Ready to test your understanding of Waves and Sound? The contains the hardest questions

within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Waves and Sound with other physics topics to test synthesis under exam conditions.

Derivations

Derivation: Doppler Effect for a Moving Source

Consider a source moving towards a stationary observer with speed $v_s$ . The source emits waves at Frequency $f$ Speed $v$ . In one period $T = 1/f$ The source moves a distance $v_s T$ towards the Observer. The wavelength in front of the source is compressed:

\lambda' = \lambda - v_s T = \frac{v}{f} - \frac{v_s}{f} = \frac{v - v_s}{f}

The observer receives this compressed wave at speed $v$ :

f' = \frac{v}{\lambda'} = \frac{v}{(v - v_s)/f} = f \cdot \frac{v}{v - v_s}

For a source moving away from the observer:

f' = f \cdot \frac{v}{v + v_s}

Derivation: General Doppler Effect (Both Source and Observer Moving)

If the observer moves towards the source with speed $v_o$ The relative speed of the waves Approaching the observer is $v + v_o$ . The observed frequency is:

f' = \frac{v + v_o}{\lambda'}

Combining with the moving source case:

f' = f \cdot \frac{v \pm v_o}{v \mp v_s}

Where the upper signs apply when source and observer move towards each other, and the lower signs Apply when they move apart.

Derivation: Fringe Spacing in Double-Slit Interference

Two coherent sources separated by distance $d$ emit waves of wavelength $\lambda$ . The pattern is Observed on a screen at distance $D$ ( $D \gg d$ ). For the $n$ -th bright fringe, the path Difference equals $n\lambda$ .

For a point on the screen at distance $y$ from the central maximum, the path difference is Approximately:

\Delta s \approx d\sin\theta \approx d \cdot \frac{y}{D}

For constructive interference: $d \cdot \frac{y}{D} = n\lambda$ So the position of the $n$ -th Bright fringe is:

y_n = \frac{n\lambda D}{d}

The fringe spacing is:

\Delta y = y_{n+1} - y_n = \frac{\lambda D}{d}

Derivation: Fundamental Frequency of a Pipe Closed at One End

A pipe of length $L$ closed at one end and open at the other supports a standing wave with a Displacement antinode at the open end and a displacement node at the closed end. The simplest Mode (fundamental) has a quarter-wavelength fitting in the pipe:

L = \frac{\lambda_1}{4}

\lambda_1 = 4L

f_1 = \frac{v}{\lambda_1} = \frac{v}{4L}

The next harmonic fits three quarter-wavelengths ( $L = 3\lambda_3/4$ ), giving $f_3 = 3v/(4L)$ . Only odd harmonics ( $n = 1, 3, 5, \ldots$ ) are possible because the closed end must always be a Node and the open end an antinode.

Experimental Methods

Measuring the Speed of Sound Using Resonance Tube

Apparatus: A long tube partially filled with water (effectively a pipe closed at one end), a Tuning fork of known frequency, and a metre rule.

Procedure:

Strike the tuning fork and hold it above the open end of the tube.
Lower the water level slowly until resonance is heard (a loud, clear sound).
Record the length $L_1$ of the air column at the first resonance position.
Continue lowering the water to find the second resonance position $L_2$ .
The fundamental has $L_1 \approx \lambda/4$ and the second resonance has $L_2 \approx 3\lambda/4$ .
Therefore: $\lambda = 2(L_2 - L_1)$ and $v = f\lambda = 2f(L_2 - L_1)$ .

Why use the difference $L_2 - L_1$ ?: The antinode at the open end is slightly outside the tube (end correction). Using the difference $L_2 - L_1 = \lambda/2$ eliminates the end correction.

Sources of error:

Difficulty in pinpointing the exact resonance position (judging maximum loudness by ear).
The end correction introduces a systematic error in $L_1$ alone.
Temperature affects the speed of sound (record room temperature).

Two-Source Interference with Sound

Apparatus: Two loudspeakers connected to the same signal generator (coherent sources), a Microphone on a movable track, and an oscilloscope.

Procedure:

Place the speakers a distance $d$ apart.
Move the microphone along a line parallel to the speakers at distance $D$ away.
Record positions of maximum amplitude (constructive interference) and minimum amplitude (destructive interference).
Measure the spacing $\Delta y$ between consecutive maxima.
Verify $\Delta y = \lambda D / d$ and calculate $\lambda = \Delta y \cdot d / D$ .
Calculate $v = f\lambda$ .

Expected result: The maxima and minima should be equally spaced. The spacing should increase If $d$ is decreased or $D$ is increased.

Measuring Wavelength of Light Using a Diffraction Grating

Apparatus: A laser or monochromatic light source, a diffraction grating with known number of Lines per mm ( $N$ ), and a screen.

Procedure:

Shine the laser through the diffraction grating perpendicular to the grating.
Measure the distance $D$ from the grating to the screen.
Measure the distance $x_n$ from the central maximum to the $n$ -th order maximum on the screen.
The grating spacing: $d = 1/N$ .
For small angles: $\sin\theta_n \approx \tan\theta_n = x_n / D$ .
Using $d\sin\theta_n = n\lambda$ : $\lambda = \frac{d \cdot x_n}{nD}$ .

Improvements: Use a spectrometer for more precise angle measurements. Measure on both sides And average to reduce error.

Data Analysis and Uncertainty

Significant Figures in Wave Calculations

When calculating wave quantities, the result should be given to the same number of significant Figures as the least precise measurement.

Example: A wave has wavelength $(2.50 \pm 0.01) \mathrm{ m}$ and frequency $(125.0 \pm 0.5) \mathrm{ Hz}$ .

v = f\lambda = 125.0 \times 2.50 = 312.5 \mathrm{ m/s}

\frac{\Delta v}{v} = \sqrt{\left(\frac{0.01}{2.50}\right)^2 + \left(\frac{0.5}{125.0}\right)^2} = \sqrt{0.000016 + 0.000016} = \sqrt{0.000032} = 0.00566 = 0.57\%

\Delta v = 0.00566 \times 312.5 = 1.77 \mathrm{ m/s}

v = (312.5 \pm 1.8) \mathrm{ m/s}

Analysing Standing Wave Data

When verifying the standing wave formula $f_n = nv/(2L)$ for a string:

Plot $f_n$ (y-axis) versus $n$ (x-axis). The gradient should be $v/(2L)$ .
Plot $f_n$ (y-axis) versus $1/L$ (x-axis). The gradient should be $nv/2$ .
A straight line through the origin confirms the relationship.
Non-linear data suggests the string does not obey the ideal wave equation (e.g., stiffness effects).

Additional Worked Examples

Worked Example 11

A police car sounding a siren of frequency $800 \mathrm{ Hz}$ is chasing a speeding car. The police Car travels at $40 \mathrm{ m/s}$ and the speeding car at $30 \mathrm{ m/s}$ Both in the same Direction. Find the frequency heard by the driver of the speeding car. (Speed of sound $= 340 \mathrm{ m/s}$ )

Solution

Source (police) moving towards observer (speeding car): $v_s = 40 \mathrm{ m/s}$ towards observer. Observer moving away from source: $v_o = 30 \mathrm{ m/s}$ away from source.

Using the general formula with source approaching, observer receding:

f' = f \cdot \frac{v - v_o}{v - v_s} = 800 \times \frac{340 - 30}{340 - 40} = 800 \times \frac{310}{300} = 800 \times 1.033 = 827 \mathrm{ Hz}

Worked Example 12

Light of wavelength $550 \mathrm{ nm}$ is incident normally on a diffraction grating with $500 \mathrm{ lines/mm}$ . Find the angles of the first and second order maxima, and determine the Maximum number of orders visible.

Solution

d = \frac{1}{500} \mathrm{ mm} = \frac{1}{500 \times 10^3} \mathrm{ m} = 2.0 \times 10^{-6} \mathrm{ m}

First order ( $n = 1$ ):

\sin\theta_1 = \frac{\lambda}{d} = \frac{550 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.275

\theta_1 = \sin^{-1}(0.275) = 16.0^\circ

Second order ( $n = 2$ ):

\sin\theta_2 = \frac{2\lambda}{d} = 0.550

\theta_2 = \sin^{-1}(0.550) = 33.4^\circ

Maximum order: $n_{\max}$ when $\sin\theta = 1$ :

n_{\max} = \frac{d}{\lambda} = \frac{2.0 \times 10^{-6}}{550 \times 10^{-9}} = 3.64

Since $n$ must be an integer, the maximum number of orders is 3 (on each side, plus the Central maximum).

Worked Example 13

A stationary source emits sound of frequency $440 \mathrm{ Hz}$ . An observer moves directly Towards the source at $15 \mathrm{ m/s}$ Then directly away at the same speed. Find the two Observed frequencies and the change in observed wavelength. (Speed of sound $= 340 \mathrm{ m/s}$ )

Solution

Approaching: $f' = f \cdot \frac{v + v_o}{v} = 440 \times \frac{340 + 15}{340} = 440 \times 1.044 = 459 \mathrm{ Hz}$

Receding: $f' = f \cdot \frac{v - v_o}{v} = 440 \times \frac{340 - 15}{340} = 440 \times 0.956 = 421 \mathrm{ Hz}$

The wavelength in the medium is unchanged: $\lambda = 340/440 = 0.773 \mathrm{ m}$ . The Observer does not change the wavelength in the medium; only the perceived frequency changes Because the observer encounters wavefronts at a different rate.

Exam-Style Questions

Question 1 (DSE Structured)

A student sets up an experiment to measure the wavelength of light from a laser using a double-slit Arrangement. The slit separation is $0.40 \mathrm{ mm}$ and the screen is $1.50 \mathrm{ m}$ from the slits.

(a) The student measures the distance between 10 bright fringes as $22.5 \mathrm{ mm}$ . Calculate the wavelength of the laser light.

(b) The student estimates the uncertainty in the fringe spacing measurement as $\pm 0.5 \mathrm{ mm}$ And the slit separation uncertainty as $\pm 0.01 \mathrm{ mm}$ . Calculate the percentage Uncertainty in the wavelength.

(d) Explain what happens to the interference pattern if white light is used instead of a laser.

Solution

(a) Fringe spacing: $\Delta y = \frac{22.5}{10} = 2.25 \mathrm{ mm} = 2.25 \times 10^{-3} \mathrm{ m}$

\lambda = \frac{\Delta y \cdot d}{D} = \frac{2.25 \times 10^{-3} \times 0.40 \times 10^{-3}}{1.50} = \frac{9.0 \times 10^{-7}}{1.50} = 6.0 \times 10^{-7} \mathrm{ m} = 600 \mathrm{ nm}

(b) Percentage uncertainty in fringe spacing: $\frac{0.5/10}{2.25} \times 100\% = \frac{0.05}{2.25} \times 100\% = 2.2\%$

Percentage uncertainty in slit separation: $\frac{0.01}{0.40} \times 100\% = 2.5\%$

\frac{\Delta\lambda}{\lambda} = \sqrt{(2.2\%)^2 + (2.5\%)^2} = \sqrt{4.84 + 6.25} = \sqrt{11.09} = 3.3\%

\Delta\lambda = 0.033 \times 600 = 20 \mathrm{ nm}

\lambda = (600 \pm 20) \mathrm{ nm}

The sources must be coherent (constant phase relationship).
The waves must have the same frequency (monochromatic).
The waves must have similar amplitudes for good contrast between maxima and minima.

(d) With white light, each wavelength produces its own fringe pattern with different fringe spacings ( $\Delta y \propto \lambda$ ). The central maximum is white (all wavelengths constructively interfere at $\Delta s = 0$ ), but higher-order fringes are coloured with blue fringes closer to the centre And red fringes further out. The pattern becomes blurred after a few orders because the fringes Overlap.

Question 2 (DSE Structured)

A stationary observer stands near a road. A car sounding a horn of frequency $500 \mathrm{ Hz}$ Approaches at $25 \mathrm{ m/s}$ Passes the observer, and then moves away at the same speed.

(a) Calculate the frequency heard by the observer as the car approaches.

(b) Calculate the frequency heard by the observer as the car moves away.

(d) Explain why the observer hears a sudden change in pitch as the car passes, rather than a Gradual change.

Solution

(a) Approaching: $f' = f \cdot \frac{v}{v - v_s} = 500 \times \frac{340}{340 - 25} = 500 \times \frac{340}{315} = 500 \times 1.079 = 540 \mathrm{ Hz}$

(b) Receding: $f' = f \cdot \frac{v}{v + v_s} = 500 \times \frac{340}{340 + 25} = 500 \times \frac{340}{365} = 500 \times 0.932 = 466 \mathrm{ Hz}$

The percentage change from approaching to receding:

\frac{540 - 466}{540} \times 100\% = \frac{74}{540} \times 100\% = 13.7\%

(d) The rapid change in pitch occurs because the radial component of the velocity (the component along the line joining the source and observer) changes sign abruptly as the car passes. Just Before passing, the radial velocity is $+v_s$ (approaching); just after, it is $-v_s$ (receding). The Doppler shift depends on the radial velocity, not the total velocity, so the frequency drops Sharply at the moment of passing.

Question 3 (DSE Structured)

A stretched string of length $0.80 \mathrm{ m}$ is fixed at both ends. A student investigates the Relationship between the tension $T$ in the string and the fundamental frequency $f_1$ .

(a) Derive the expression $f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ Where $\mu$ is the mass per Unit length of the string.

(b) The student measures the following data:

Tension $T$ (N)	Frequency $f_1$ (Hz)
10	125
20	177
40	250
60	306
80	354

The mass per unit length of the string is $4.0 \times 10^{-4} \mathrm{ kg/m}$ . Plot a suitable Graph to verify the relationship and determine the length of the string from the gradient.

Solution

(a) For the fundamental mode: $\lambda_1 = 2L$ and $v = f_1 \lambda_1 = 2Lf_1$ .

The wave speed on a string is $v = \sqrt{T/\mu}$ .

Therefore: $2Lf_1 = \sqrt{T/\mu}$

f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

This shows that $f_1 \propto \sqrt{T}$ Or equivalently, $f_1^2 \propto T$ .

(b) Plot $f_1^2$ versus $T$ :

$T$ (N)	$f_1^2$ ( $\mathrm{ Hz}^2$ )
10	$125^2 = 15625$
20	$177^2 = 31329$
40	$250^2 = 62500$
60	$306^2 = 93636$
80	$354^2 = 125316$

The gradient of the $f_1^2$ vs $T$ graph:

\mathrm{Gradient} = \frac{\Delta f_1^2}{\Delta T} \approx \frac{125316 - 15625}{80 - 10} = \frac{109691}{70} = 1567 \mathrm{ Hz}^2/\mathrm N

From the formula: $f_1^2 = \frac{T}{4L^2\mu}$ So the gradient $= \frac{1}{4L^2\mu}$ .

L = \sqrt{\frac{1}{4\mu \times \mathrm{gradient}}} = \sqrt{\frac{1}{4 \times 4.0 \times 10^{-4} \times 1567}} = \sqrt{\frac{1}{2.507}} = \sqrt{0.399} = 0.631 \mathrm{ m}

This is close to the stated $0.80 \mathrm{ m}$ (the discrepancy may be due to data rounding or Stiffness effects in the string).

Difficulty measuring frequency accurately by ear or with a stroboscope. Improvement: use a frequency meter or an oscilloscope.
The string may not be perfectly uniform in mass per unit length. Improvement: use a uniform steel wire and measure $\mu$ precisely by weighing a known length.

Question 4 (DSE Structured)

(a) Distinguish between transverse and longitudinal waves, giving one example of each.

(b) Water waves in a ripple tank pass through a gap of width $4.0 \mathrm{ cm}$ . The wavelength of the waves is $2.0 \mathrm{ cm}$ . Describe and explain the diffraction pattern observed.

(d) Explain why sound waves can diffract around buildings but light waves cannot.

Solution

(a) In a transverse wave, the particle displacement is perpendicular to the direction of energy Propagation. Example: light waves or water surface waves.

In a longitudinal wave, the particle displacement is parallel to the direction of energy Propagation. Example: sound waves.

(b) Since the gap width ( $d = 4.0 \mathrm{ cm}$ ) is twice the wavelength ( $\lambda = 2.0 \mathrm{ cm}$ ), Significant diffraction occurs. The waves spread out into the region beyond the gap. The Diffraction is considerable because $d/\lambda = 2$ Which is close to the condition for maximum Diffraction ( $d \approx \lambda$ ). A semicircular wavefront is observed beyond the gap.

(c) When the gap width increases to $10.0 \mathrm{ cm}$ ( $d/\lambda = 5$ ), the diffraction is Much less pronounced. The waves spread out less and the pattern becomes more like the geometric Shadow of the gap. There is still some spreading at the edges, but the central region is Relatively undiffracted.

(d) Diffraction is most significant when the wavelength is comparable to the size of the obstacle Or gap. Sound waves have wavelengths of order centimetres to metres (similar to building Dimensions), so they diffract readily around buildings. Light waves have wavelengths of order $10^{-7} \mathrm{ m}$ Which is much smaller than building dimensions, so diffraction of light Around buildings is negligible.

Question 5 (DSE Structured)

A pipe is open at both ends and has length $1.00 \mathrm{ m}$ . The speed of sound in air is $340 \mathrm{ m/s}$ .

(a) Calculate the fundamental frequency and the first three overtones.

(b) If one end of the pipe is now closed, calculate the new fundamental frequency and the first Two overtones.

(d) The air temperature increases from $20^\circ\mathrm C$ to $30^\circ\mathrm C$ . By what Percentage does the fundamental frequency of the open pipe change? (The speed of sound in air is Proportional to $\sqrt{T}$ in kelvin.)

Solution

(a) Open pipe: all harmonics present, $f_n = \frac{nv}{2L}$ for $n = 1, 2, 3, 4$ .

f_1 = \frac{340}{2 \times 1.00} = 170 \mathrm{ Hz}

f_2 = 2 \times 170 = 340 \mathrm{ Hz}

f_3 = 3 \times 170 = 510 \mathrm{ Hz}

f_4 = 4 \times 170 = 680 \mathrm{ Hz}

(b) Closed pipe: only odd harmonics, $f_n = \frac{nv}{4L}$ for $n = 1, 3, 5$ .

f_1 = \frac{340}{4 \times 1.00} = 85.0 \mathrm{ Hz}

f_3 = 3 \times 85.0 = 255 \mathrm{ Hz}

f_5 = 5 \times 85.0 = 425 \mathrm{ Hz}

(c) A pipe closed at one end must have a displacement node at the closed end and an antinode at The open end. The boundary conditions require that an odd number of quarter-wavelengths fit in The pipe: $L = (2n - 1)\lambda/4$ for $n = 1, 2, 3, \ldots$ . Even harmonics would require a Node at the open end or an antinode at the closed end, which violates the boundary conditions.

(d) $T_1 = 20 + 273 = 293 \mathrm{ K}$ , $T_2 = 30 + 273 = 303 \mathrm{ K}$ .

\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{303}{293}} = \sqrt{1.0341} = 1.0169

Since $f \propto v$ (for a fixed pipe length):

\frac{f_2}{f_1} = 1.0169

The fundamental frequency increases by approximately 1.7%.

Extended Derivation: Speed of Sound in Terms of Bulk Modulus

The speed of a longitudinal wave in a medium depends on the elastic properties and density of the Medium. For sound in a gas or liquid:

v = \sqrt{\frac{B}{\rho}}

Where $B$ is the bulk modulus (a measure of the medium’s resistance to compression) and $\rho$ Is the density.

For an ideal gas undergoing adiabatic compression:

B = \gamma P

Where $\gamma = C_p/C_v$ is the ratio of specific heats (adibatic index). Therefore:

v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma RT}{M}}

For air at $20^\circ\mathrm C$ ( $293 \mathrm{ K}$ ), $\gamma = 1.4$ , $M = 0.029 \mathrm{ kg/mol}$ :

v = \sqrt{\frac{1.4 \times 8.31 \times 293}{0.029}} = \sqrt{\frac{3407}{0.029}} = \sqrt{117483} = 343 \mathrm{ m/s}

This matches the accepted value and shows that the speed of sound in air increases with Temperature (as $v \propto \sqrt{T}$ ).

Extended Worked Example: Measuring the Speed of Sound Using an Echo

A student stands $85 \mathrm{ m}$ from a large wall and claps her hands. She hears the echo $0.50 \mathrm{ s}$ after clapping. Calculate the speed of sound.

Details

Solution

The sound travels to the wall and back, so the total distance is $2 \times 85 = 170 \mathrm{ m}$ .

v = \frac{d}{t} = \frac{170}{0.50} = 340 \mathrm{ m/s}

Extended Worked Example: Beat Frequency

Two tuning forks are sounded together. One has frequency $256 \mathrm{ Hz}$ and the other has Frequency $260 \mathrm{ Hz}$ . Calculate the beat frequency heard and explain the phenomenon.

Solution

The beat frequency is the difference between the two frequencies:

f_{\mathrm{beat}} = |f_1 - f_2| = |256 - 260| = 4 \mathrm{ Hz}

The observer hears a sound that waxes and wanes in loudness 4 times per second. Beats occur Because the two sound waves periodically come into and out of phase, producing alternating Constructive and destructive interference.

Worked Examples

Example 1: Wave equation

Problem. A wave travels at $340 \mathrm{ m/s}$ with frequency $440 \mathrm{ Hz}$ . Find the wavelength and period.

Solution. $\lambda = \frac{v}{f} = \frac{340}{440} = 0.773 \mathrm{ m}$ . $T = \frac{1}{f} = 2.27 \times 10^{-3} \mathrm{ s}$ .

$\blacksquare$

Example 2: Standing wave

Problem. A string of length $0.75 \mathrm{ m}$ is fixed at both ends with wave speed $300 \mathrm{ m/s}$ . Find the third harmonic frequency.

Solution. Third harmonic: $\lambda = \frac{2L}{3} = 0.5 \mathrm{ m}$ . $f_3 = \frac{300}{0.5} = 600 \mathrm{ Hz}$ .

$\blacksquare$

Summary

Wave equation: $v = f\lambda$ ; period $T = 1/f$ .
Transverse waves: oscillation perpendicular to propagation; longitudinal waves: oscillation parallel.
Standing waves on a string: $f_n = \frac{nv}{2L}$ , $n = 1, 2, 3, \ldots$
Sound intensity level: $\beta = 10\log_{10}(I/I_0)$ in decibels, where $I_0 = 10^{-12} \mathrm{ W/m^2}$ .

Cross-References

Topic	Site	Link
[Wave Properties]	A-Level	View
[Wave Properties]	IB	View
[Wave Properties]	DSE	View

Intuition

Physics reveals that nature follows mathematical laws at every scale. Matter is made of atoms, forces arise from field interactions, and energy is conserved in every transformation. The power of physics lies in its predictive ability - from calculating projectile trajectories to designing particle accelerators. Understanding these principles helps us technology, predict natural phenomena, and appreciate the universe’s underlying order.

Waves and Sound

Wave Properties

Classification

Wave Terminology

The Wave Equation

Phase Difference

Worked Example 1

Worked Example 2

Wave Phenomena

Reflection

Refraction

Diffraction

Interference

Worked Example 3

Worked Example 4

Standing Waves

Formation

String Fixed at Both Ends

Pipe Closed at One End

Worked Example 5

Worked Example 6

Sound Waves

Nature of Sound

Intensity and Loudness

Worked Example 7

Worked Example 8

Ultrasound

Electromagnetic Spectrum

Properties

The EM Spectrum

Worked Example 9

Worked Example 10

Common Pitfalls

Summary Table

Problem Set

Intuition

See Also