Wave Motion, Speed and Classification
Progressive Waves
A progressive wave transfers energy away from a source of disturbance without transporting matter permanently — particles oscillate about their equilibrium positions.
Transverse Waves: Particle displacement is perpendicular to the wave's direction of travel (e.g., waves on a rope, light).
Longitudinal Waves: Particle displacement is parallel to the wave's direction of travel (e.g., sound in air, compression in a spring).
Solids vs Fluids: Both wave types exist in solids, but fluids support only longitudinal waves — sound in air is always longitudinal.
Pulse vs Periodic: A single disturbance creates a pulse; continuous rhythmic vibrations from a source (like a mass-spring system executing SHM) create periodic waves.
The Nature of Sound Waves
Sound is a mechanical wave that propagates as a longitudinal wave through a medium via successive compression and rarefaction cycles.
$$f = \frac{1}{T}$$
Relates the number of cycles per second to the time duration of one cycle.
$f$=Frequency(Hertz (Hz))
$T$=Period(Seconds (s))
$T \to 0$
→Frequency becomes infinitely high.
Longitudinal Motion: Particles oscillate parallel to the direction of energy transfer, creating alternating high- and low-pressure zones.
Pressure Variations: Compression regions have high pressure/density; rarefaction regions have low pressure/density.
Human Hearing Range: 20 Hz to 20,000 Hz. Below 20 Hz is infrasound; above 20,000 Hz is ultrasound.
The Wave Equation for Sound
The speed of sound ($v$) equals the product of its frequency and its wavelength, linking how fast energy propagates to the wave's spatial and temporal properties.
$$v = f\lambda$$
A crest moves one wavelength $\lambda$ in one period $T$, giving $v = \lambda / T = f\lambda$.
$v$=Wave speed(m/s)
$f$=Frequency(Hz)
$\lambda$=Wavelength(m)
Constant $v$ (same medium)
→$f$ and $\lambda$ are inversely proportional: $f \propto 1/\lambda$.
Wave crosses into a new medium
→Frequency stays constant (set by source); speed and wavelength both change.
Wavelength Definition: Distance between two consecutive identical points (e.g., compression to compression).
Inverse Proportionality Shortcut: For a fixed medium, doubling $f$ halves $\lambda$.
Phase Relationships
Two points on a wave are described by their phase difference, which determines whether they vibrate in unison or opposition.
$$\phi = \frac{2\pi x}{\lambda}$$
The phase lag of a point at distance $x$ from a reference point, measured in radians.
$\phi$=Phase difference(radians)
$x$=Separation between the two points(m)
$\lambda$=Wavelength(m)
$x = \lambda, 2\lambda, 3\lambda, \dots$
→Points are in phase ($\phi = 2\pi n$) — identical displacement and velocity at all times.
$x = \lambda/2, 3\lambda/2, \dots$
→Points are completely out of phase ($\phi = \pi + 2\pi n$) — opposite displacements at all times.
In Phase: Separated by whole multiples of $\lambda$; always have identical displacements.
Out of Phase: Separated by odd multiples of $\lambda/2$; when one is at max displacement, the other is at min.
Speed of Sound in Air — Newton vs Laplace
Newton assumed sound propagation in air is an isothermal process and derived the speed using the gas pressure as the elastic modulus.
$$v = \sqrt{\frac{P}{\rho}}$$
Newton's formula: treats compressions/rarefactions as isothermal (constant temperature). Predicts 280 m/s — about 16% too low.
$P$=Atmospheric pressure(Pa)
$\rho$=Density of air(kg/m³)
At STP ($P = 1.013 \times 10^5$ Pa, $\rho = 1.29$ kg/m³)
→$v = \sqrt{1.013 \times 10^5 / 1.29} \approx 280$ m/s (experimental value is 332 m/s).
Newton's Assumption: Temperature remains constant during compressions and rarefactions (isothermal process, Boyle's law: $PV = \text{const}$).
The Error: Newton's value of 280 m/s is significantly lower than the measured 332 m/s.
Why It Fails: Compressions happen too rapidly for heat to dissipate — the process is actually adiabatic, not isothermal.
Laplace corrected Newton by recognizing that sound compressions are adiabatic — heat doesn't escape fast enough — replacing $P$ with $\gamma P$.
$$v = \sqrt{\frac{\gamma P}{\rho}}$$
Laplace's corrected formula. The factor $\gamma$ accounts for the adiabatic nature of rapid compressions/rarefactions.
$\gamma$=Ratio of specific heats ($C_p / C_v$)(dimensionless)
$P$=Atmospheric pressure(Pa)
$\rho$=Density of gas(kg/m³)
Diatomic gas (air, O₂, N₂)
→$\gamma = 1.40$
Monatomic gas (He, Ne, Ar)
→$\gamma = 1.67$
Polyatomic gas (CO₂)
→$\gamma = 1.29$
The Correction Factor: Laplace multiplied Newton's elastic modulus by $\gamma$: $E = \gamma P$ instead of $E = P$.
Result for Air: $v = \sqrt{1.4} \times 280 \approx 333$ m/s, matching the experimental value of 332 m/s.
Dimensional Check: $[P/\rho] = \text{Pa}/(\text{kg/m}^3) = \text{m}^2/\text{s}^2$, so the square root gives m/s.
Factors Affecting Sound Speed in Gases
Changing the pressure of a gas at constant temperature does not change the speed of sound, because density changes proportionally with pressure.
Why No Effect: From $v = \sqrt{\gamma P / \rho}$, if $P$ doubles at constant $T$, then $\rho$ also doubles (ideal gas law), so $P/\rho$ stays the same.
At the same temperature and pressure, the speed of sound is inversely proportional to the square root of the gas density.
$$v \propto \frac{1}{\sqrt{\rho}}$$
Lighter Gas = Faster Sound: Sound in hydrogen ($\rho_H$) is 4× faster than in oxygen ($\rho_O = 16\rho_H$), since $\sqrt{16} = 4$.
Helium Voice Effect: Helium is ~7× less dense than air, so sound travels ~2.7× faster, shifting resonant frequencies of the vocal tract upward.
The speed of sound in a gas is proportional to the square root of its absolute temperature, and increases by approximately 0.61 m/s per °C rise.
$$\frac{v_t}{v_0} = \sqrt{\frac{T}{T_0}}$$
Relates the speed at temperature $t$ °C to the speed at 0 °C using absolute temperatures.
$v_t$=Speed at temperature $t$ °C(m/s)
$v_0$=Speed at 0 °C (332 m/s for air)(m/s)
$T$=Absolute temperature ($t + 273$ K)(K)
$T_0$=273 K (0 °C)(K)
Small temperature changes
→Use the linear approximation: $v_t \approx v_0 + 0.61t$
$v$ doubles
→Absolute temperature must quadruple ($v \propto \sqrt{T}$, so $2v \Rightarrow 4T$).
Linear Approximation: For moderate temperatures, $v_t \approx 332 + 0.61t$ m/s (derived via binomial expansion of $\sqrt{1 + t/273}$).
Proportionality Shortcut: $v \propto \sqrt{T}$ means to double the speed you need 4× the absolute temperature.
The speed of sound is determined by the elasticity and density of the medium, moving faster through stiffer materials.
$$v = \sqrt{\frac{E}{\rho}}$$
General formula: the elastic modulus $E$ represents resistance to compression, while $\rho$ represents inertia.
$E$=Elastic modulus ([bulk modulus] for fluids, Young's modulus for solids)(Pa)
$\rho$=Mass density of the medium(kg/m³)
Solid (e.g., steel)
→Very high elasticity dominates over higher density → fastest speed (~5000 m/s).
Gas (e.g., air)
→High compressibility (low $E$) → slowest speed (~340 m/s).
Phase Comparison: $v_{\text{solid}} > v_{\text{liquid}} > v_{\text{gas}}$ because the increase in elasticity from gas to solid far outweighs the increase in density.
Typical Sound Speeds
•
Air (STP): 332 m/s
•
Water (20 °C): 1483 m/s
•
Iron: 5130 m/s
•
Hydrogen (STP): 1286 m/s
•
Helium (STP): 972 m/s
Principle of Superposition
When two or more waves overlap in a medium, the resultant displacement at any point is the algebraic sum of the individual displacements — this is the superposition principle.
$$Y = y_1 + y_2 + \cdots + y_n$$
Constructive Interference: When waves are in phase, displacements add — the resultant amplitude is larger.
Destructive Interference: When waves are completely out of phase ($y_1 = -y_2$), the resultant displacement is zero.
Waves Pass Through Each Other: After overlapping, each wave continues unchanged in shape and direction.
Three Key Phenomena: Superposition leads to (1) interference, (2) beats, and (3) stationary waves.
Beats
When two sound waves of slightly different frequencies are heard together, the resulting sound alternates between loud and soft — these periodic intensity fluctuations are called beats.
$$f_{\text{beat}} = |f_1 - f_2|$$
The number of loudness maxima per second equals the absolute difference of the two source frequencies.
$f_{\text{beat}}$=Beat frequency (beats per second)(Hz)
$f_1, f_2$=Frequencies of the two sources(Hz)
$|f_1 - f_2| > 10$ Hz
→Beats become too rapid for the human ear to distinguish; a rough sound is heard instead.
$f_1 = f_2$
→No beats — a steady tone is heard (this is the goal when tuning instruments).
Mechanism: At instants when both waves are in phase, constructive interference produces a loud sound; when out of phase, destructive interference produces silence.
Tuning Application: Musicians use beats to tune instruments — adjust the string until beats vanish ($f_{\text{beat}} = 0$).
Loading a Tuning Fork: Adding wax lowers the fork's frequency. If beat frequency increases after loading, the fork originally had a frequency lower than the reference.
Reflection of Waves
When a wave reaches the boundary between two media, part of it is reflected back with the same frequency and wavelength, but the phase may change depending on boundary type.
Fixed End (Rarer to Denser): Reflected wave undergoes a 180° phase change — an incident crest returns as a trough.
Free End (Denser to Rarer): Reflected wave has no phase change — an incident crest returns as a crest.
Echo: A practical consequence of sound reflection from a distant surface.
Reverberation: When the reflecting surface is close (round-trip time < 0.1 s), the reflected sound blends with the original rather than being heard as a separate echo.
An echo occurs when sound reflects off a surface; the total distance traveled is twice the distance to the obstacle.
$$d = \frac{v \times t}{2}$$
Calculates the distance to a reflector based on the round-trip time of a sound pulse.
$d$=Distance to obstacle(m)
$v$=Speed of sound(m/s)
$t$=Total round-trip time(s)
$t < 0.1$ s
→Human ear perceives reverberation instead of a distinct echo.
Sonar/Ultrasound: Application of echoes for underwater depth measurement and medical imaging.
Application Strategy
Worked Examples
Problem Statement
A bat emits an ultrasound chirp at 40,000 Hz. If the speed of sound is 340 m/s, find the wavelength of the chirp.
Wave EquationUltrasound
Strategic Toolkit
$$v = f\lambda$$
Step 1
Identify Knowns
Extract frequency and speed from the problem statement.
$f = 40{,}000$ Hz, $v = 340$ m/s
Step 2
Rearrange Formula
Solve the wave equation for wavelength ($\lambda$).
$\lambda = \frac{v}{f}$
Step 3
Substitution
Plug the values into the equation.
$\lambda = \frac{340}{40{,}000}$
Step 4
Final Calculation
Perform division and state units.
$\lambda = 0.0085$ m $= 8.5$ mm
Problem Statement
A hiker claps their hands in front of a vertical cliff and hears the echo 1.5 seconds later. If the air temperature is 20 °C (speed = 343 m/s), how far is the cliff?
Echo ReflectionRound-Trip Distance
Strategic Toolkit
$$d = \frac{v \times t}{2}$$
Step 1
Extract Variables
Note the total round-trip time and the speed of sound.
$t = 1.5$ s, $v = 343$ m/s
Step 2
Apply Echo Formula
Use the reflection formula to find the one-way distance.
$d = \frac{343 \times 1.5}{2}$
Step 3
Result
Calculate the product and divide by 2.
$d = 257.25$ m
Problem Statement
Find the temperature at which the speed of sound in air is twice its value at 10 °C.
Speed-Temperature RelationshipAbsolute Temperature
Strategic Toolkit
$$\frac{v_T}{v_{T_0}} = \sqrt{\frac{T}{T_0}}$$
Step 1
Convert to Kelvin
Convert the reference temperature to absolute scale.
$T_0 = 10 + 273 = 283$ K
Step 2
Set Up the Ratio
We want $v_T = 2v_0$, so the ratio is 2.
$\frac{v_T}{v_0} = \sqrt{\frac{T}{283}} = 2$
Step 3
Solve for T
Square both sides and solve.
$\frac{T}{283} = 4 \Rightarrow T = 1132$ K
Step 4
Convert Back
Express the answer in Celsius.
$T = 1132 - 273 = 859$ °C
Problem Statement
Tuning fork A (320 Hz) produces 4 beats/s with fork B. Loading B with wax increases the beat frequency to 6 beats/s. Find the original and loaded frequencies of B.
Beat FrequencyLoading Effect
Strategic Toolkit
$$f_{\text{beat}} = |f_1 - f_2|$$
Step 1
List Possibilities
4 beats/s means B is either 4 Hz above or below A.
$f_B = 320 + 4 = 324$ Hz or $f_B = 320 - 4 = 316$ Hz
Step 2
Apply Loading Logic
Wax lowers B's frequency. If original was 324 Hz, loading brings it closer to 320 → fewer beats. If original was 316 Hz, loading moves it further from 320 → more beats. Beats increased, so original must be 316 Hz.
$f_B = 316$ Hz (original)
Step 3
Find Loaded Frequency
The new beat frequency is 6, and B is below A.
$f_{B,\text{loaded}} = 320 - 6 = 314$ Hz
Problem Statement
Calculate the speed of sound in air at STP using Laplace's correction. Given: $P = 1.013 \times 10^5$ Pa, $\rho = 1.29$ kg/m³, $\gamma = 1.4$.
Laplace CorrectionAdiabatic Process
Strategic Toolkit
$$v = \sqrt{\frac{\gamma P}{\rho}}$$
Step 1
Newton's Value First
Calculate what Newton's isothermal formula gives.
$v_{\text{Newton}} = \sqrt{\frac{P}{\rho}} = \sqrt{\frac{1.013 \times 10^5}{1.29}} \approx 280$ m/s
Step 2
Apply Laplace Correction
Multiply under the root by $\gamma = 1.4$.
$v = \sqrt{\gamma} \times v_{\text{Newton}} = \sqrt{1.4} \times 280$
Step 3
Final Result
Evaluate the expression.
$v = 1.183 \times 280 \approx 331$ m/s (experimental: 332 m/s)
Problem Statement
If the density of oxygen is 16 times that of hydrogen, and both are diatomic gases at the same temperature and pressure, find the ratio of the speed of sound in hydrogen to that in oxygen.
Density EffectProportional Reasoning
Strategic Toolkit
$$v \propto \frac{1}{\sqrt{\rho}}$$
Step 1
Write the Ratio
Since $\gamma$ and $P$ are the same for both diatomic gases at the same conditions.
$\frac{v_H}{v_O} = \sqrt{\frac{\rho_O}{\rho_H}}$
Step 2
Substitute
Given $\rho_O = 16\rho_H$.
$\frac{v_H}{v_O} = \sqrt{\frac{16\rho_H}{\rho_H}} = \sqrt{16} = 4$
Step 3
Result
Sound travels 4 times faster in hydrogen than in oxygen.
$v_H = 4 \times v_O$
Memorize formulas
The Universal Wave Equation
Speed is the product of frequency and wavelength.
Formula
$v = f\lambda$
Newton's Formula (Isothermal — Incorrect)
Assumes constant temperature during compression. Gives 280 m/s, 16% too low.
Formula
$v = \sqrt{\frac{P}{\rho}}$
Laplace Correction (Adiabatic — Correct)
Accounts for adiabatic compression with the heat capacity ratio $\gamma$.
Formula
$v = \sqrt{\frac{\gamma P}{\rho}}$
Speed-Temperature Ratio
Speed of sound is proportional to the square root of absolute temperature.
Formula
$\frac{v_t}{v_0} = \sqrt{\frac{T}{T_0}}$
Speed-Temperature Linear Approximation
Sound gains about 0.61 m/s for every degree Celsius above 0 °C.
Formula
$v_t \approx 332 + 0.61t$
Echo Half-Rule
For echoes, the clock measures the round trip — divide by 2 for one-way distance.
Formula
$d = \frac{v \times t}{2}$
Beat Frequency
Number of beats per second equals the difference of the two frequencies.
Formula
$f_{\text{beat}} = |f_1 - f_2|$
Phase Difference
Phase lag in radians for two points separated by distance $x$.
Formula
$\phi = \frac{2\pi x}{\lambda}$
Review
1
Using Laplace's corrected formula $v = \sqrt{\gamma P / \rho}$, calculate the speed of sound in air at STP given $\gamma = 1.4$, $P = 1.01 \times 10^5$ Pa, and $\rho = 1.29$ kg/m³:
Ans
≈ 332 m/s
Review
Tap to reveal
Substitute the gas data into Laplace's formula
Use the corrected relation for sound speed in a gas and plug in the given values of $\gamma$, $P$, and $\rho$.
$$v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{1.4 \times 1.01 \times 10^5}{1.29}}$$
Evaluate the numerical expression
Compute the ratio inside the square root first, then take the square root to get the speed.
$$v = \sqrt{1.096 \times 10^5} \approx 331\ \text{m/s} \approx 332\ \text{m/s}$$
2
Two containers hold gas at the same temperature: container X has air ($M = 29$ g/mol) and container Y has carbon dioxide ($M = 44$ g/mol). Both are diatomic-like with $\gamma = 1.4$. The ratio of the speed of sound $v_X / v_Y$ is approximately:
Ans
1.23
Review
Tap to reveal
Use the temperature-based sound-speed relation
At fixed temperature and the same $\gamma$, sound speed varies inversely with the square root of molar mass.
$$v \propto \frac{1}{\sqrt{M}}$$
Form the ratio and simplify
Take the ratio of the two speeds and substitute the molar masses of carbon dioxide and air.
$$\frac{v_X}{v_Y} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{\frac{44}{29}} \approx 1.23$$
3
The speed of sound at 0 °C is 332 m/s. At what temperature would the speed be approximately 344 m/s?
Ans
≈ 20 °C
Review
Tap to reveal
Apply the linear temperature approximation
Use the standard near-room-temperature relation between sound speed and Celsius temperature.
$$v \approx 332 + 0.61t$$
Solve for the temperature
Set the speed equal to 344 m/s and isolate $t$.
$$344 = 332 + 0.61t \Rightarrow t = \frac{12}{0.61} \approx 19.7^\circ\text{C} \approx 20^\circ\text{C}$$
4
A sound wave of frequency 500 Hz travels from air ($v = 340$ m/s) into water ($v = 1500$ m/s). The wavelength in water is:
Ans
3.0 m
Review
Tap to reveal
Keep the frequency unchanged across the boundary
When a wave enters a new medium, its frequency stays fixed because the source has not changed.
$$f = 500\ \text{Hz}$$
Use $\lambda = v/f$ in water
Substitute the speed of sound in water and the unchanged frequency to get the wavelength.
$$\lambda = \frac{1500}{500} = 3.0\ \text{m}$$
5
Fork A (440 Hz) produces 5 beats/s with fork B. When fork B is filed (slightly sharpened), the beat frequency decreases to 3 beats/s. The original frequency of B is:
Ans
445 Hz
Review
Tap to reveal
Use the beat-frequency formula
The initial 5 beats/s means the other tuning fork differs from 440 Hz by 5 Hz.
$$|f_B - 440| = 5 \Rightarrow f_B = 435\ \text{Hz or } 445\ \text{Hz}$$
Use the effect of filing
Filing a fork raises its frequency, so the new beat frequency must be smaller only if the original fork was above 440 Hz and moved closer to it.
$$f_B = 445\ \text{Hz}$$
6
Two sound waves of equal amplitude and frequency but with a phase difference of $\pi$ radians meet at a point. The resultant displacement is:
Ans
Zero (complete destructive interference)
Review
Tap to reveal
Recognize the out-of-phase condition
A phase difference of $\pi$ radians means one wave is exactly inverted relative to the other.
$$\Delta\phi = \pi$$
Add the two displacements
Equal amplitudes cancel term by term under superposition, so the resultant displacement vanishes.
$$Y = A\sin(\omega t) + A\sin(\omega t + \pi) = 0$$
7
A transverse wave pulse (crest) travelling along a string hits a rigid wall. The reflected pulse is:
Ans
A trough (inverted, 180° phase change)
Review
Tap to reveal
Identify the boundary as a fixed end
A rigid wall does not move, so the string behaves like a wave reflecting from a fixed boundary.
$$\text{fixed end} \Rightarrow \text{phase inversion}$$
Apply the reflection rule
A crest returns inverted after reflection, so the pulse becomes a trough.
$$\text{crest} \rightarrow \text{trough}$$
8
Sound travels at 332 m/s in air (diatomic, $\gamma = 1.4$). What is the approximate speed of sound in helium ($\gamma = 1.67$, density = 1/7 that of air) at the same temperature and pressure?
Ans
≈ 970 m/s
Review
Tap to reveal
Use the ratio form for two gases
At the same temperature and pressure, compare helium and air through the dependence on both $\gamma$ and density.
$$\frac{v_{He}}{v_{air}} = \sqrt{\frac{\gamma_{He}}{\gamma_{air}} \cdot \frac{\rho_{air}}{\rho_{He}}}$$
Substitute the given values
Helium has $\gamma = 1.67$ and density one-seventh that of air, so the ratio becomes about 2.89.
$$\frac{v_{He}}{v_{air}} = \sqrt{\frac{1.67}{1.4} \times 7} \approx 2.89$$
Multiply by the air speed
Scale the air speed by the ratio to get the approximate speed in helium.
$$v_{He} \approx 2.89 \times 332 \approx 960\ \text{m/s} \approx 970\ \text{m/s}$$
9
The speed of sound in air at 20 °C is 343 m/s. What is the approximate speed of sound at −40 °C?
Ans
≈ 306 m/s
Review
Tap to reveal
Use the square-root temperature dependence
For a gas, the speed of sound is proportional to the square root of absolute temperature.
$$v_2 = v_1 \sqrt{\frac{T_2}{T_1}}$$
Convert Celsius to Kelvin and substitute
Replace the temperatures by 293 K and 233 K and evaluate the ratio.
$$v_2 = 343 \sqrt{\frac{233}{293}} \approx 343 \times 0.892 \approx 306\ \text{m/s}$$
10
Two points P and Q on a sound wave are separated by $\frac{\lambda}{3}$. When P has zero displacement and is moving in the positive direction, Q has:
Ans
Positive displacement equal to $\frac{A\sqrt{3}}{2}$
Review
Tap to reveal
Convert separation into phase difference
A distance of one-third of a wavelength corresponds to a phase shift of $2\pi/3$.
$$\Delta\phi = 2\pi\left(\frac{1}{3}\right) = \frac{2\pi}{3}$$
Read P's phase as the reference point
Since P is at zero displacement and moving in the positive direction, take its phase as 0 and shift forward by $2\pi/3$ for Q.
$$y_Q = A\sin\left(\frac{2\pi}{3}\right) = \frac{A\sqrt{3}}{2}$$
Question 1 of 10
0:00
1
Using Laplace's corrected formula $v = \sqrt{\gamma P / \rho}$, calculate the speed of sound in air at STP given $\gamma = 1.4$, $P = 1.01 \times 10^5$ Pa, and $\rho = 1.29$ kg/m³: