In short

A sound wave is a longitudinal mechanical wave — a travelling pattern of compressions (regions where the medium is momentarily denser and at higher pressure) and rarefactions (regions at lower density and lower pressure). Particles of the medium oscillate along the direction of propagation, not across it.

The speed of sound in any fluid is set by the medium's stiffness and inertia:

\boxed{\; v = \sqrt{\dfrac{B}{\rho}} \;}

where B is the bulk modulus (pressure needed to produce a unit relative change in volume, in Pa) and \rho is the mass density (kg/m³). In an ideal gas the relevant bulk modulus is adiabatic, B_{\text{ad}} = \gamma P, giving Laplace's formula

v = \sqrt{\dfrac{\gamma P}{\rho}} = \sqrt{\dfrac{\gamma R T}{M}},

with \gamma = C_P/C_V, T the absolute temperature (K), and M the molar mass (kg/mol). In dry air at 20 °C this gives v \approx 343 m/s, and v rises by roughly 0.6 m/s for every extra degree Celsius. Sound travels faster in warm air, faster in humid air (water vapour is lighter than air), and much faster in water (≈ 1480 m/s) and steel (≈ 5900 m/s) — denser, but much stiffer.

At constant temperature, pressure has no effect on the speed of sound in an ideal gas — a pressure increase raises P and \rho in exactly the same proportion, and the ratio P/\rho is fixed. Humans hear roughly 20 Hz to 20\,000 Hz; below that is infrasound (elephants, whales, earthquakes), above is ultrasound (bats, sonography, SONAR).

Stand in the courtyard of the Meenakshi temple at dawn. A bronze bell is struck. The sound rushes out in all directions, reaches your ear a fraction of a second later, and sets your eardrum vibrating — you hear a deep, ringing note. Nothing physical has travelled from the bell to you. No bronze, no air, no object at all covered that distance with you. What arrived is a pattern of pressure variation — a travelling disturbance in the air between the bell and your ear. That pattern is a sound wave.

If you have read the article on introduction to waves, you already know that every wave carries energy without carrying matter. Sound is the most familiar example — and, because the medium is normally air, which is invisible, it is also the wave that most strongly illustrates how odd the whole business really is. You cannot see the bell's ring travelling across the courtyard. But it is there, a faint compression of air molecules, racing outward at about 343 metres every second, and your eardrum registers each oscillation of pressure as a sensation of sound.

This article does four things. First, it builds the physical picture — how air molecules conspire to carry a sound wave, and why sound must be longitudinal. Second, it derives the speed-of-sound formula v = \sqrt{B/\rho} by applying Newton's second law to a thin slab of fluid. Third, it explains the classic mistake Newton made with the ideal-gas case and how Laplace corrected it, giving the formula v = \sqrt{\gamma P / \rho} that matches experiment. Fourth, it surveys the frequency bands — infrasound, audible, ultrasound — and the real factors that affect wave speed: temperature, humidity, medium.

You will need the wave equation for the derivation and a working notion of bulk modulus for the stiffness side of the formula. Both are linked where they first appear.

What a sound wave looks like

A sitar string wave is transverse — the string moves up and down while the wave runs along its length, so you can see the shape of the wave with your eyes. A sound wave is longitudinal: the air parcels move back and forth along the same direction the wave is travelling, and the displacements are tiny (a few micrometres for ordinary conversation). There is nothing to see. But there is something happening.

Picture a long, thin column of air — imagine the narrow bore of a bansuri laid on a table, or simply a long cardboard tube. At the left end, you tap a membrane sharply inward. That motion shoves the air molecules right next to the membrane a little to the right. Those molecules crowd into the molecules next to them, making a small region of slightly higher pressure and higher density — a compression. The newly-crowded molecules then push on the ones beyond them, and meanwhile, the original molecules (no longer being pushed) spring back leftward. A fraction of a second later, the compression has moved a little to the right, and behind it is a region that is slightly emptier than normal — a rarefaction — where the molecules briefly have more space than at equilibrium.

If you tap the membrane rhythmically — in, out, in, out — the air column fills with alternating compressions and rarefactions, spaced at regular intervals, all sliding down the tube at the speed of sound. That is exactly what your vocal cords do, what a shehnai's reed does at a Varanasi ghat wedding, what the skin of a tabla does when struck. A continuous sound wave is a continuous train of pressure oscillations, propagating through the medium.

Compressions and rarefactions in a travelling sound waveA horizontal tube is shown containing many small dots representing air molecules. Some regions of the tube have dots packed closely (compressions, higher pressure), alternating with regions where the dots are spread apart (rarefactions, lower pressure). An arrow above indicates the direction the wave travels. Below the tube, a smooth sinusoidal curve shows the local pressure variation: peaks coincide with compressions, troughs with rarefactions, and the horizontal axis matches the tube.compressionrarefactioncompressionrarefactioncompressiondirection of wave travelpp₀pressure variation along the tube at one instant
The tube shows a snapshot of a sound wave: air molecules are packed closely in the compressions and spaced out in the rarefactions. The curve below plots the pressure $p(x)$ at this instant — peaks line up with compressions, troughs with rarefactions. Both the dots and the curve slide rightward together at the speed of sound.

The "wave" can be described either by the displacement of the air parcels from their equilibrium positions — s(x, t) — or by the pressure variation from the ambient pressure p_0 — p(x, t). These two descriptions are equivalent but related in a subtle way you will see in a moment: the pressure wave is a quarter-wavelength out of phase with the displacement wave.

Watch a compression travel

Animated: a compression pulse moves down the tube at the speed of soundA small red dot starts near the left end of a horizontal tube and moves steadily to the right at 5 m/s in simulation time. A trail marks its path. Dashed guide lines mark 1 m, 5 m, 10 m, and 15 m along the tube.1 m5 m10 m15 m
A compression pulse released at $x = 1$ m travels to the right at 5 m/s (scaled for viewing; real sound is about 343 m/s). After one second it has reached 6 m; after three seconds, 16 m. The trail marks each position so you can read off the speed. Click replay to watch again.

Why sound is longitudinal

Whether a wave is transverse or longitudinal depends on which kind of restoring force the medium supplies. A string under tension can be bent sideways — the tension in the bent string supplies a sideways (transverse) restoring force, so string waves are transverse. A gas cannot be bent sideways — a gas offers no restoring force at all when you drag one layer laterally past another (gases have no shear stiffness). What a gas does offer is a restoring force when you compress it: squeeze a parcel of gas and its pressure rises, pushing back outward. That restoring force is along the direction in which you compressed it.

So in a gas (and in a liquid, to a very good approximation), the only wave that can propagate is one where the particles move along the direction of wave travel — a compression wave. That is a longitudinal wave, and it is what sound is. In solids, both transverse and longitudinal waves are possible, because solids resist both shear and compression — earthquakes produce both P-waves (longitudinal, primary, arrive first because they are faster) and S-waves (transverse, secondary, slower). But sound in the usual sense — sound you hear — is a wave in air, and air carries only compression.

Displacement, pressure, and density — three descriptions of the same wave

A sinusoidal sound wave of wavelength \lambda and angular frequency \omega can be written as a displacement wave:

s(x, t) = s_0 \sin(kx - \omega t),

where s(x, t) is how far the air parcel at equilibrium position x has moved along the direction of propagation at time t, and s_0 is the displacement amplitude (typically a few micrometres for ordinary speech, a few tens of micrometres for a loud sound). Here k = 2\pi/\lambda is the wave number and \omega = 2\pi f, exactly as in the wave equation.

The pressure variation p(x, t) — how much the local pressure deviates from the ambient p_0 — is related to how much the air is being squeezed. If neighbouring parcels are moving away from each other, the air thins out and pressure drops; if they are moving toward each other, pressure rises. Quantitatively, the local strain in the air is \partial s/\partial x (a positive \partial s/\partial x means the parcels further along are displaced more than those just behind, so the air is being stretched; negative means compressed). The bulk modulus B converts strain to pressure:

p(x, t) = -B\,\frac{\partial s}{\partial x}.

Why: stretching by a fraction \partial s/\partial x reduces the density and pressure; hence the minus sign. This is the fluid analogue of Hooke's law — pressure change equals minus the bulk modulus times volumetric strain.

Differentiate s(x, t) = s_0 \sin(kx - \omega t) with respect to x:

\frac{\partial s}{\partial x} = k s_0 \cos(kx - \omega t).

Why: the derivative of \sin is \cos, and the inner derivative \partial(kx - \omega t)/\partial x = k pulls down as a prefactor.

So the pressure wave is

p(x, t) = -B k s_0 \cos(kx - \omega t) = p_0^{\text{amp}} \cos(kx - \omega t + \pi),

with pressure amplitude

\boxed{\; p_0^{\text{amp}} = B k s_0 = B \frac{2\pi}{\lambda}\, s_0. \;}

Two useful takeaways. First, the pressure wave is a quarter-period out of phase with the displacement wave: wherever the air is most displaced (a peak of s), the difference between the displacements of neighbouring parcels is zero, and the pressure is at its equilibrium value. Wherever the air is at its equilibrium position (s = 0), the slope \partial s/\partial x is largest, and the pressure deviation is largest — either a pressure peak (compression) or trough (rarefaction). A standing wave in a closed pipe puts this to dramatic use, as you will see in the next chapter on vibrations of strings and pipes.

Second, the pressure amplitude grows with the bulk modulus and with the spatial frequency k. For a fixed displacement amplitude, higher-frequency sounds carry larger pressure swings — one reason a dog whistle at 30 kHz, with displacements invisible to the eye, still delivers enough pressure variation to hurt a dog's ear.

Deriving the speed of sound — Newton's second law on a slab

Now derive the speed formula. Take a fluid (air, water, whatever) and consider a thin slab of it between x and x + \Delta x, with cross-sectional area A. In equilibrium the slab has volume A\Delta x, mass \rho A \Delta x, and the pressure on both faces is the ambient p_0.

Free-body diagram of a thin slab of fluid in a sound waveA rectangular slab of fluid of cross-sectional area A and thickness delta-x. On the left face, pressure p of x pushes rightward; on the right face, pressure p of x plus delta x pushes leftward. The net horizontal force is minus the derivative of p with respect to x times A times delta x. Mass of the slab is rho times A times delta x.slabarea A, thickness Δxmass ρ·A·Δxp(x)·Ap(x+Δx)·Axx + Δxposition →pressure on each face → net force on the slab
A thin slab of fluid between $x$ and $x + \Delta x$, cross-sectional area $A$. The pressure on the left face is $p(x)$, pushing the slab to the right with force $p(x)\,A$; the pressure on the right face is $p(x+\Delta x)$, pushing it to the left with force $p(x+\Delta x)\,A$. The net force is what accelerates the slab.

Assumptions: The wave is one-dimensional (plane wave along x). The displacements are small compared to the wavelength, so the slab's mass barely changes. The fluid is inviscid (no internal friction). The wavelength is much longer than the mean free path of molecules (so a continuum description applies). Temperature-pressure coupling is addressed later with Laplace's correction.

Step 1. Write Newton's second law for the slab.

When a sound wave is passing, the slab is displaced by s(x, t) and the pressure on its faces is no longer uniform. Let the pressure at position x be p(x, t) = p_0 + p'(x, t), where p' is the small deviation from the ambient p_0. The net horizontal force on the slab is

F_{\text{net}} = p(x)\,A - p(x + \Delta x)\,A = -\big[p(x + \Delta x) - p(x)\big]\,A \approx -\frac{\partial p}{\partial x}\, A\, \Delta x,

where the last step uses the first-order Taylor expansion of p about x. Why: the pressure on the left face pushes the slab to the right (+), and the pressure on the right face pushes it to the left (-). Their difference is the gradient of pressure times the slab's thickness, in the limit \Delta x \to 0.

The slab's mass is \rho A \Delta x and its acceleration is \partial^2 s/\partial t^2 (because every point in the slab has moved by s(x,t)). So:

\rho A \Delta x\, \frac{\partial^2 s}{\partial t^2} = -\frac{\partial p}{\partial x}\, A\, \Delta x.

Cancel A \Delta x:

\rho \, \frac{\partial^2 s}{\partial t^2} = -\frac{\partial p}{\partial x}. \tag{1}

Why: force per unit volume equals density times acceleration. This is Newton's second law on a per-unit-volume basis, sometimes called the Euler equation for a fluid.

Step 2. Relate pressure change to displacement change.

You already did this above: the constitutive relation between pressure and strain is p' = -B\,\partial s/\partial x. So \partial p/\partial x = -B\,\partial^2 s/\partial x^2 (since p_0 is constant and drops out).

Step 3. Substitute into (1).

\rho \, \frac{\partial^2 s}{\partial t^2} = -\left(-B\,\frac{\partial^2 s}{\partial x^2}\right) = B\,\frac{\partial^2 s}{\partial x^2}
\boxed{\; \frac{\partial^2 s}{\partial t^2} = \frac{B}{\rho}\,\frac{\partial^2 s}{\partial x^2}. \;}

Why: this is the wave equation \partial^2 s/\partial t^2 = v^2\,\partial^2 s/\partial x^2 with wave speed v^2 = B/\rho. Every longitudinal disturbance in the fluid propagates at that speed, independent of amplitude or frequency.

Step 4. Read off the wave speed.

\boxed{\; v = \sqrt{\dfrac{B}{\rho}} \;} \tag{2}

This is the master formula for the speed of sound in any fluid. The same universal pattern — v = \sqrt{\text{restoring stiffness}/\text{inertia}} — that gives v = \sqrt{T/\mu} for a string gives v = \sqrt{B/\rho} for a fluid. Stiffer medium pushes the wave faster; heavier medium slows it down.

Checking the pattern in three familiar fluids

Medium B (Pa) \rho (kg/m³) v = \sqrt{B/\rho} (m/s)
Air (20 °C, 1 atm) \sim 1.4\times 10^5 1.20 \approx 343
Water (20 °C) 2.2\times 10^9 998 \approx 1483
Sea water 2.3\times 10^9 1025 \approx 1500
Steel (longitudinal) 1.6\times 10^{11} 7850 \approx 4510

Water is a thousand times denser than air, but its bulk modulus is ten thousand times larger, so sound travels four times faster in water than in air. That is why whales can call each other across hundreds of kilometres — ocean acoustics is a field built on this one formula. Steel is denser again, but far stiffer still; the pure longitudinal wave speed in steel is roughly five to six kilometres per second, so if a rail at Howrah junction is tapped with a hammer, the click reaches an observer three kilometres away through the rail in about half a second, while the airborne sound takes almost nine seconds.

Newton's formula and Laplace's correction

When Newton first worked out the theory of sound in the Principia, he wrote down equation (2) but assumed the compressions and rarefactions were isothermal — that is, that the temperature of a parcel of air does not change as it is compressed, because heat has time to flow in and out. For an ideal gas undergoing an isothermal change, p V = constant, so p \propto 1/V and the isothermal bulk modulus is B_{\text{iso}} = P (the ambient pressure itself).

Newton then plugged B = P = 1.013\times 10^5 Pa and \rho = 1.29 kg/m³ (roughly, for air at 0 °C) into v = \sqrt{B/\rho} and got about 280 m/s. But careful experiment showed the speed of sound in air is really about 331 m/s at 0 °C. Newton's answer was low by about 15%. He never reconciled this discrepancy.

More than a century later, Laplace noticed what Newton had missed: a sound wave oscillates far too quickly for heat to flow. A 1000 Hz sound wave has a period of a millisecond; in that time, the compressed air in one compression cannot equilibrate its temperature with the rarefied air in a nearby rarefaction a few tens of centimetres away, because thermal conduction in air is slow. So each parcel of air is, in effect, thermally isolated during the sound wave — each compression is adiabatic, not isothermal. The air heats up a little in a compression and cools a little in a rarefaction, and the wave equation must use the adiabatic bulk modulus.

For an adiabatic change of an ideal gas, p V^{\gamma} = constant, where \gamma = C_P/C_V is the ratio of specific heats. Differentiate:

p\, \gamma V^{\gamma - 1}\, dV + V^{\gamma}\, dp = 0
\Rightarrow\quad dp = -\gamma\, \frac{p}{V}\, dV.

Why: take the differential of pV^\gamma = constant using the product rule; the left side must vanish, giving a relation between dp and dV.

Rearranging gives the adiabatic bulk modulus:

B_{\text{ad}} = -V \frac{dp}{dV} = \gamma\, p.

Why: the definition of bulk modulus is B = -V\,dp/dV — the pressure needed to produce a fractional volume change. The minus sign makes B positive because compressing the gas (negative dV) raises the pressure.

So the correct speed of sound in an ideal gas is

\boxed{\; v = \sqrt{\dfrac{\gamma P}{\rho}} \;} \tag{3}

which is called Laplace's formula. For dry air, \gamma = 1.40 (a diatomic gas: N_2 and O_2), and at 0 °C, P = 1.013\times 10^5 Pa, \rho = 1.29 kg/m³. Plugging in:

v = \sqrt{\frac{1.40 \times 1.013\times 10^5}{1.29}} = \sqrt{1.099 \times 10^5} \approx 331 \text{ m/s}.

This matches the measured value to better than 1%. The factor \sqrt{\gamma} = \sqrt{1.4} \approx 1.183 is precisely the correction Newton was missing, and it is a genuinely new piece of physics: sound waves heat and cool the air as they pass, and the gas responds to those tiny temperature swings by being effectively stiffer than it would be at a fixed temperature. That is the physics Laplace caught.

Recasting in temperature — the useful form

Combine Laplace's formula with the ideal gas law PV = nRT, which gives P/\rho = RT/M where M is the molar mass (kg/mol). Substituting:

\boxed{\; v = \sqrt{\dfrac{\gamma R T}{M}} \;} \tag{4}

This is the form used in almost every practical calculation. For dry air with M = 0.02897 kg/mol and \gamma = 1.40, equation (4) becomes:

v \approx 20.05\,\sqrt{T}\ \text{m/s},

with T in kelvin. At T = 293 K (20 °C) this gives v \approx 343 m/s, matching the standard reference value.

A useful rule-of-thumb, obtained by Taylor-expanding around 0 °C, is

v(\theta) \approx 331 + 0.6\,\theta\ \text{m/s},

where \theta is the temperature in Celsius. In a Rajasthan desert at noon (say 45 °C) sound travels at roughly 331 + 0.6\times 45 \approx 358 m/s; on a Shimla winter night at -5 °C, about 328 m/s. A difference of a few percent, but detectable with a stopwatch over a kilometre.

What speed depends on — and what it doesn't

The temperature-only form in equation (4) makes the dependencies crisp.

Temperature: sound travels faster in warm air

As the Celsius rule above shows, a hotter gas carries sound faster because the same kinetic energy is spread among atoms, each moving faster on average — the pressure-restoration chain responds more snappily. Quantitatively, v \propto \sqrt{T}, so doubling the absolute temperature multiplies the speed by \sqrt{2} \approx 1.41. Between a Leh winter (250 K) and a Pune May afternoon (310 K), the speed of sound goes up by about 11%.

Pressure: no effect (at constant temperature)

A naive reading of equation (3) would suggest that higher pressure means faster sound. But in an ideal gas at fixed temperature, increasing the pressure also increases the density in exactly the same proportion (since P/\rho = RT/M depends only on T and M). So P/\rho is fixed, and the speed does not change. That is why the speed of sound is the same at sea level in Mumbai and atop the 8000-metre Kanchenjunga base camp (temperatures aside) — denser air down low, thinner air up high, same speed if the temperature is the same.

This is a place where physics intuition can trip you. "Higher pressure → stiffer → faster" is the instinct, but in an ideal gas the density tracks the pressure and the effects cancel.

Humidity: moist air is slightly faster

Water vapour (M = 0.018 kg/mol) is lighter than the dry-air average (M = 0.029 kg/mol). At fixed temperature and pressure, replacing some dry air with water vapour lowers the average molar mass M, and v \propto 1/\sqrt{M} goes up. The effect is small — fully saturated air at 30 °C is only about 0.4% faster than dry air at the same temperature — but it is not nothing, and it is why sound carries slightly further on a humid monsoon evening in Kerala than on a bone-dry afternoon in Jaisalmer at the same temperature.

Medium: liquids faster than gases, solids faster still

From equation (2), v = \sqrt{B/\rho}. Liquids have bulk moduli thousands of times larger than gases, and although they are also much denser, the stiffness wins; sound in water is about four times faster than in air. In metals, the relevant modulus is Young's modulus (for a thin rod) or the combination K + \frac{4}{3}G (for an extended solid); both are enormous compared to fluids, so sound races through steel at several kilometres per second.

Frequency: (almost) no effect

For the idealised fluid in the derivation above, the wave speed is independent of frequency — audible 200 Hz and ultrasonic 100 kHz sound travel at the same 343 m/s in air. A medium with this property is called non-dispersive. Real air is very slightly dispersive because of tiny viscous and relaxation effects, but at JEE level you can treat the speed as frequency-independent.

Explore how temperature changes the speed

Interactive: how the speed of sound in air depends on temperatureA plot of speed of sound versus Celsius temperature, running from minus twenty degrees to plus fifty degrees. The curve shown is the exact formula, approximately linear over this range. A draggable red point on the horizontal axis sets the temperature; dashed guides extend up from the point to the curve and across to the y-axis, and a readout panel shows the speed.temperature θ (°C)speed v (m/s)315325335345355365−2002040v = 20.05 √(T)drag the red point along the axis
Drag the red point along the horizontal axis to set the air temperature between $-20$ °C and $+50$ °C. The red curve plots the exact formula $v = 20.05\sqrt{T}$ m/s (with $T$ in kelvin), and the readout gives the speed of sound at that temperature. At 0 °C, $v \approx 331$ m/s; at 20 °C, $\approx 343$ m/s; at 40 °C, $\approx 355$ m/s.

Worked examples

Example 1: Speed of sound on a 40 °C afternoon in Jaipur

A Jaipur summer afternoon hits 40 °C. What is the speed of sound in the air?

Setup for Example 1: sound speed at 40 °CA thermometer mark at 40 degrees Celsius on the left. An arrow and a label v equals 355 m per second on the right, labelled speed of sound.40 °Cθspeed of sound in airv = ?
Jaipur on a hot afternoon: air at 40 °C. The speed of sound is set by the absolute temperature through $v = \sqrt{\gamma RT/M}$.

Step 1. Convert temperature to kelvin.

T = 40 + 273.15 = 313.15 \text{ K}

Why: Laplace's formula (4) requires absolute temperature, not Celsius. The physics is about total molecular kinetic energy, which is zero only at T = 0 K.

Step 2. Apply equation (4).

v = \sqrt{\frac{\gamma R T}{M}}

Plug in \gamma = 1.40 (diatomic air), R = 8.314 J/(mol·K), T = 313.15 K, and M = 0.02897 kg/mol:

v = \sqrt{\frac{1.40 \times 8.314 \times 313.15}{0.02897}}

Why: these are standard values for dry air. The ratio R/M is the specific gas constant for air, about 287 J/(kg·K).

Step 3. Evaluate.

v = \sqrt{\frac{3645.5}{0.02897}} = \sqrt{125\,836} \approx 354.7 \text{ m/s}

Step 4. Check with the rule of thumb.

v \approx 331 + 0.6 \times 40 = 331 + 24 = 355 \text{ m/s}

Why: the approximation agrees with the exact value to about 0.1%, which confirms the calculation and gives you an on-the-fly mental check for any future problem.

Result: The speed of sound in Jaipur at 40 °C is about 355 m/s.

What this shows: on a hot day, sound travels a few percent faster than the "standard" 343 m/s quoted for 20 °C. Between a winter morning at 10 °C and a summer afternoon at 40 °C, the speed of sound changes by about 18 m/s — enough that a ranger measuring the distance of a thunderclap by counting seconds needs to account for the season.

Example 2: A shot is fired in Lonar Crater — echo from the rim

A visitor at Lonar Crater in Maharashtra fires a starter pistol and hears the echo from the far rim 4.20 seconds later. The air temperature is 28 °C. How far away is the rim?

Setup for Example 2: echo in Lonar CraterA visitor stands on the left at a pistol. An arrow labelled going goes rightward to a rock wall on the right. Another arrow labelled returning goes leftward from the wall back to the visitor. Distance d between visitor and wall. Round-trip time is 4.20 s.visitorcrater rimgoing (t/2)returning (t/2)distance d
The pistol's report travels a distance $d$ to the rim, reflects, and travels $d$ back. Total path length is $2d$; total time is 4.20 s.

Step 1. Find the speed of sound at 28 °C.

Using the rule of thumb v(\theta) \approx 331 + 0.6\theta:

v \approx 331 + 0.6 \times 28 = 331 + 16.8 \approx 347.8 \text{ m/s}.

Why: a 1% accurate value is plenty for this problem, and the rule of thumb saves unit conversions.

Step 2. Set up the round-trip relation.

The sound goes out a distance d and returns a distance d, so the total distance is 2d and the total time is 4.20 s. Using distance = speed × time:

2d = v\,t_{\text{total}}.

Step 3. Solve for d.

d = \frac{v\, t_{\text{total}}}{2} = \frac{347.8 \times 4.20}{2} = \frac{1460.8}{2} \approx 730\text{ m}.

Why: the factor of 2 is the whole trick. Students who forget the round-trip double the answer and place the rim 1460 m away instead of 730 m.

Result: The far rim of Lonar Crater is about 730 metres from the visitor.

What this shows: an echo measurement with a stopwatch and a basic knowledge of the speed of sound is a genuine distance-measuring tool. Sonar works on exactly this principle in water (with v \approx 1480 m/s); SONAR-equipped fishing boats in the Bay of Bengal use it to map the sea floor and locate shoals of pomfret.

Example 3: Pressure amplitude of a loud conversation

A normal conversation has a displacement amplitude s_0 \approx 1.1 \times 10^{-8} m and a dominant frequency around 500 Hz. Estimate the pressure amplitude p_0^{\text{amp}} at 20 °C.

Step 1. Gather the needed quantities.

At 20 °C, v = 343 m/s. The angular frequency is \omega = 2\pi f = 2\pi(500) \approx 3142 rad/s. The wave number is k = \omega/v = 3142/343 \approx 9.16 m^{-1}. The bulk modulus of air is

B = \gamma P = 1.40 \times 1.013 \times 10^5 \approx 1.418 \times 10^5 \text{ Pa}.

Step 2. Apply p_0^{\text{amp}} = B k s_0.

p_0^{\text{amp}} = (1.418 \times 10^5)(9.16)(1.1 \times 10^{-8}) \approx 1.43 \times 10^{-2} \text{ Pa}.

Why: the product brings together the stiffness of the medium (B), how rapidly the wave varies spatially (k), and how far the air parcels swing (s_0). Each factor contributes one power to the pressure amplitude.

Step 3. Compare to atmospheric pressure.

The ratio is

\frac{p_0^{\text{amp}}}{P_0} = \frac{1.43 \times 10^{-2}}{1.013 \times 10^5} \approx 1.4 \times 10^{-7},

or about one part in seven million. Why: sound is a tiny perturbation of the ambient pressure. The human ear is an exquisitely sensitive pressure detector — it can register fractional pressure changes as small as 10^{-10} at the threshold of hearing.

Result: The pressure amplitude is about 0.014 Pa, or roughly 10^{-7} of atmospheric pressure.

What this shows: even audible sounds are extraordinarily small pressure disturbances. A jet engine at 10 metres might produce a pressure amplitude of about 20 Pa — a thousand times larger than conversation — but still only one part in five thousand of atmospheric pressure. The ear's enormous dynamic range — a factor of 10^{12} in intensity from whisper to jet engine — is why intensity is measured on a logarithmic scale (decibels), which you will meet in intensity and loudness of sound.

Common confusions

Infrasound, audible, and ultrasound

The human ear responds to pressure variations between roughly 20 Hz and 20 000 Hz. Below this band is infrasound — vibrations too low for the ear to register as tone, though very loud infrasound can be felt in the chest (a large organ pipe, a blue whale's call, the seismic rumble before an earthquake). Above is ultrasound — frequencies above about 20 kHz, inaudible to humans but important in medicine, industry, and nature.

Band Frequency Real-world source
Infrasound < 20 Hz Elephants communicating at 5–10 Hz over kilometres; whale calls; earthquake waves; wind turbines
Audible 20 Hz – 20 kHz Everything you can hear; speech centred near 500 Hz, bat chirps at the upper edge
Ultrasound 20 kHz – 1 GHz Medical sonography (2–15 MHz), industrial cleaning, SONAR (20–500 kHz), bat echolocation (20–200 kHz), dog whistles (around 30 kHz)

The upper audible limit drops with age: children can hear up to about 20 kHz, but by age 40 the limit has typically fallen to 15 kHz, and older adults rarely hear above 12 kHz. This is not because sound stops propagating — the high-frequency waves reach the ear exactly as they always did — but because the tiny hair cells at the cochlea's base, which respond to the highest frequencies, are the first to age.

Indian medicine and research make heavy use of ultrasound. AIIMS and Tata Memorial centres use sonography (typically 3–7 MHz) to image soft tissue during pregnancy. The Indian Navy uses SONAR (Sound Navigation and Ranging) to map the sea floor and detect submarines, relying on the same speed-distance relation as Example 2 but in sea water. The physics is identical in all cases — equation (2) just with a different B and \rho.

If you came here to understand what a sound wave is, how fast it travels, and why Newton's formula needed Laplace's correction, you have what you need. What follows is the full pressure-wave derivation, the energy carried by a sound wave, and a few subtle points about real gases.

Deriving the wave equation for the pressure directly

You can write the wave equation in terms of p' rather than s. Start from the density-continuity equation for a thin slab: if the slab's endpoints are at x + s(x,t) and x + \Delta x + s(x + \Delta x, t), the slab's length is

\Delta x\big(1 + \partial s/\partial x\big),

so conservation of mass gives the fractional density change:

\frac{\rho - \rho_0}{\rho_0} = -\frac{\partial s}{\partial x}.

Why: the mass of the slab is fixed; if its length grows by a fraction \partial s/\partial x, its density must shrink by the same fraction.

Combining with p' = -B\,\partial s/\partial x gives p'/B = (\rho - \rho_0)/\rho_0, so pressure deviations and density deviations are proportional.

Differentiating the Euler equation \rho_0\,\partial^2 s/\partial t^2 = -\partial p'/\partial x once more with respect to x, and using \partial^2 s/\partial x \partial t = \partial^2 s/\partial t \partial x:

\rho_0\, \frac{\partial^2}{\partial t^2}\left(\frac{\partial s}{\partial x}\right) = -\frac{\partial^2 p'}{\partial x^2}.

Substitute \partial s/\partial x = -p'/B:

-\frac{\rho_0}{B}\,\frac{\partial^2 p'}{\partial t^2} = -\frac{\partial^2 p'}{\partial x^2}
\Rightarrow\quad \frac{\partial^2 p'}{\partial t^2} = \frac{B}{\rho_0}\,\frac{\partial^2 p'}{\partial x^2}.

Same wave equation, same speed, expressed in pressure. A sinusoidal solution is p'(x,t) = p_0^{\text{amp}}\cos(kx - \omega t + \pi), with the same \omega/k = v.

Energy and intensity carried by a sound wave

A plane sound wave carries energy. The intensity I is the energy per unit area per unit time passing through a surface perpendicular to the wave's direction. For a sinusoidal wave, the result is

\boxed{\; I = \tfrac{1}{2}\rho v \omega^2 s_0^2 = \frac{p_0^{\text{amp}\,2}}{2\rho v} \;}

in watts per square metre. The first form shows intensity scales as the square of the displacement amplitude and the square of the angular frequency — the same A^2 f^2 scaling you met for waves on a string. The second form, in terms of pressure amplitude, is the one typically used in acoustics. A full derivation takes a step through the kinetic-energy density of the oscillating parcels; it appears in the article on intensity and loudness of sound.

Why \gamma = 1.40 for air

For an ideal gas, \gamma = C_P/C_V = 1 + R/C_V. The molar heat capacity at constant volume depends on how many degrees of freedom each molecule has. For a diatomic molecule like N_2 or O_2 at room temperature, there are 3 translational and 2 rotational degrees of freedom (the rotation about the molecular axis contributes negligibly for a linear molecule), giving C_V = \tfrac{5}{2}R and \gamma = 1 + 2/5 = 1.40. Monatomic gases (helium, argon) have 3 translational degrees only, C_V = \tfrac{3}{2}R, \gamma = 1 + 2/3 = 1.67. So sound in helium at the same T is faster than in air by a factor \sqrt{\gamma_{He}/\gamma_{air} \times M_{air}/M_{He}} \approx \sqrt{(1.67/1.40)(29/4)} \approx 2.9 — which is why your voice sounds squeaky after inhaling helium: your vocal-tract resonances shift upward by that factor.

Real gases: where the ideal approximation fails

Equation (4) assumes the gas is ideal — molecules do not interact except during elastic collisions, and the heat capacity is constant. Real gases violate both assumptions slightly. At very high pressure (say, 10^5 Pa in a diving cylinder) or very low temperature (say, liquid-nitrogen range), corrections become significant. For ordinary atmospheric conditions in India, from sea level to Himalayan passes, the ideal-gas form is accurate to better than 0.2%.

A more serious correction is vibrational relaxation: in polyatomic gases at audible frequencies, the vibrational modes of the molecules do not have time to equilibrate with the translational modes during one oscillation period. The effective \gamma ends up slightly frequency-dependent, making air slightly dispersive — different frequencies travel at different speeds. The effect is tiny (parts per thousand) and only matters for precision acoustics, but it is a real departure from equation (4).

A historical footnote

C.V. Raman's 1922 monograph On the Molecular Diffraction of Light laid the acoustic foundations that he later used in his 1928 Raman-effect discovery. The Acoustical Society of India is named in part for his contributions. Raman also studied the acoustics of Indian musical instruments — the tabla, the mridangam, the veena — and explained why their struck membranes, unlike a Western kettledrum, can produce harmonic overtones. These investigations use exactly the speed-of-sound formula above, applied to the specific boundary conditions of a circular membrane.

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