Displacement Wave and Pressure Wave

Q: What is the formula for pressure amplitude of a sound wave?

The pressure amplitude is B A k, which equals 2 pi B A over lambda, where B is the bulk modulus, A the displacement amplitude and lambda the wavelength. Using B = rho v squared it can also be written rho v omega A.

Q: How are pressure and density waves related?

The excess density equals the excess pressure divided by v squared, so the density wave is in phase with the pressure wave. Both are maximum at compressions and minimum at rarefactions, and both lead the displacement wave by a quarter cycle.

Q: What is the intensity of a sound wave in terms of pressure amplitude?

Intensity equals pressure amplitude squared divided by 2 rho v, or pressure amplitude squared times v divided by 2B. It depends on the square of the pressure amplitude and not on frequency. In terms of displacement amplitude, I = 2 pi squared f squared A squared rho v.

Q: How much power does a wave on a string carry?

The average power is one half mu omega squared A squared v, where mu is mass per unit length, A the amplitude and v the wave speed. Kinetic and potential energy each carry half. Doubling the amplitude or the frequency makes the power four times larger.

PhysicsWavesFor JEE aspirants

A sound wave can be written as a displacement wave, $s = A sin (ω t - k x)$ , or as a pressure wave, $Δ P = Δ P_{0} cos (ω t - k x)$ with $Δ P_{0} = B A k$ . The displacement wave and pressure wave describe the same sound, but they are $π /2$ out of phase: pressure is maximum where displacement is zero. This page derives the pressure wave, adds the density wave, compares real amplitudes, and finds the energy and intensity carried by waves on strings and in air, $I = \frac{Δ P _{0}^{2}}{2 ρ v}$ . Frequently tested in JEE Main and JEE Advanced.

On this page1Displacement description2Pressure wave3Phase relations4Density wave5Real magnitudes6Energy on a string7Intensity of sound

Key Formulas - Quick Reference

Displacement wave: $s = A sin (ω t - k x)$ ; $s$ is the shift of a layer along the wave
★ Must learnExcess pressure: $Δ P = - B \frac{\partial s}{\partial x} = B A k cos (ω t - k x)$
★ Must learnPressure amplitude: $Δ P_{0} = B A k = \frac{2 π B A}{λ} = ρ v ω A$ (using $B = ρ v^{2}$ )
★ Must learnPhase: pressure (and density) lead displacement by $\frac{π}{2}$ ; $Δ P$ is maximum where $s = 0$
Density wave: $Δ ρ = \frac{ρ}{B} Δ P = \frac{Δ P}{v ^{2}}$ , in phase with $Δ P$
Pressure and particle velocity (wave towards $+ x$ ): $Δ P = ρ v v_{p}$
String: kinetic and potential energy per length $= \frac{1}{2} μ ω^{2} A^{2} cos^{2} (k x - ω t)$ each
★ Must learnAverage power on a string: $P = \frac{1}{2} μ ω^{2} A^{2} v = 2 π^{2} f^{2} A^{2} μv$
★ Must learnIntensity: $I = \frac{1}{2} ρ ω^{2} A^{2} v = \frac{Δ P _{0}^{2}}{2 ρ v} = \frac{Δ P _{0}^{2} v}{2 B}$

1. Two Ways to Describe a Sound Wave

A longitudinal wave in a fluid can be described by the longitudinal displacements of the layers of the medium. Let a wave travel along $+ x$ . A layer whose undisturbed (mean) position is at $x$ is shifted, at time $t$ , by $s$ along $x$ , so its actual position is $x + s$ :

s (x, t) = A sin (ω t - k x)

x

is the distance of the layer's mean position from the origin;

s

is its displacement from that mean position. Fix

x = x_{0}

and the layer performs SHM of amplitude

A

and period

T

with initial phase

- k x_{0}

Figure 1: In a sound wave,

x

labels a layer by its mean position and

s (x, t)

is how far it has moved along the wave. The layer is actually at

x + s

;

s

is typically less than a micrometre.

Wherever neighbouring layers are displaced by different amounts, the air between them is squeezed or stretched, so its pressure changes. That gives a second, equivalent description: the pressure wave (also called the compression wave). Microphones and our eardrums respond to this pressure change.

2. Deriving the Pressure Wave

Consider a slice $A B$ of gas of cross-sectional area $S$ , with face $A$ at $x$ and face $B$ at $x + d x$ . As the wave passes, face $A$ moves by $s$ and face $B$ by $s + d s$ .

Figure 2: Faces

A

and

B

move by

s

and

s + d s

. The volume change is

d V = S d s

, so the volume strain is

\frac{d V}{V} = \frac{\partial s}{\partial x}

; a negative strain (squeeze) raises the pressure.

Original volume: $V = S d x$ . New thickness: $d x + d s$ , so the change in volume is $d V = S d s$ .
Volume strain: $\frac{d V}{V} = \frac{S d s}{S d x} = \frac{\partial s}{\partial x}$ .
Bulk modulus: $B = - \frac{Δ P}{d V / V}$ , so the excess pressure is
$Δ P = - B \frac{\partial s}{\partial x}$
From $s = A sin (ω t - k x)$ : $\frac{\partial s}{\partial x} = - A k cos (ω t - k x)$ , so
$Δ P = B A k cos (ω t - k x) = Δ P_{0} cos (ω t - k x)$

★ Must learnPressure amplitude:

Δ P_{0} = B A k = \frac{2 π}{λ} A B

Since

B = ρ v^{2}

and

v k = ω

, also

Δ P_{0} = ρ v^{2} A k = ρ v ω A

Δ P

is the excess pressure: the actual pressure at the point swings between

P_{atm} + Δ P_{0}

and

P_{atm} - Δ P_{0}

Exam Trick

Pressure goes with the slope. $Δ P = - B \partial s / \partial x$ : excess pressure is minus $B$ times the slope of the $s$ - $x$ graph. Where the $s$ - $x$ curve falls most steeply you have a compression; where it rises most steeply, a rarefaction; where it is flat (at the peaks of $∣ s ∣$ ), normal pressure. Same idea as $v_{p} = - v \partial y / \partial x$ in Concept 1.

3. Phase Relation Between Displacement and Pressure

$s = A sin (ω t - k x)$ and $Δ P = Δ P_{0} cos (ω t - k x) = Δ P_{0} sin (ω t - k x + \frac{π}{2})$ . So the pressure wave leads the displacement wave by $π /2$ , a shift of $λ /4$ in space.

Figure 3: At a compression

C

the displacement is zero but the excess pressure and density are maximum; where

∣ s ∣

is maximum,

Δ P = 0

. The pressure wave leads the displacement wave by

π /2

(a shift of

λ /4

At a point where	Displacement $s$	Excess pressure $Δ P$	Particle velocity $v_{p}$
Compression (centre)	$0$	$+ Δ P_{0}$ (maximum)	$+ A ω$ , along the wave
Rarefaction (centre)	$0$	$- Δ P_{0}$ (minimum)	$- A ω$ , against the wave
Layer at extreme ( $s = \pm A$ )	$\pm A$	$0$ (normal pressure)	$0$

Figure 4: Snapshot at

t = 0

(amplitudes scaled to the same height). The two curves are

λ /4

apart: zeros of one sit under the peaks and troughs of the other.

Pressure maxima (and minima) occur where the displacement is zero; displacement maxima occur where the pressure is at its normal level.
In a standing wave this becomes: a displacement node is a pressure antinode, and a displacement antinode is a pressure node. The closed end of a pipe is a displacement node, so it is a pressure antinode (Concept 5).
Particle velocity $v_{p} = \frac{\partial s}{\partial t} = A ω cos (ω t - k x)$ is in phase with the excess pressure for a wave towards $+ x$ .

Figure 5: Reading an

s

x

graph (wave towards

+ x

). At

B

(

s = 0

, slope most negative) the layers crowd: compression,

Δ P

and density maximum, layers moving forward. At

D

(

s = 0

, slope most positive): rarefaction. At

A

C

E

(

∣ s ∣

maximum) the pressure is normal.

JEE Advanced

Pressure and particle velocity. Comparing $Δ P = B A k cos (ω t - k x)$ with $v_{p} = A ω cos (ω t - k x)$ :

Δ P = \frac{B k}{ω} v_{p} = \frac{B}{v} v_{p} = ρ v v_{p}

Layers in a compression move forward, layers in a rarefaction move backward. For a wave travelling towards $- x$ the sign flips: $Δ P = - ρ v v_{p}$ . The ratio $ρ v$ is called the acoustic impedance of the medium.

Quick Recall: tap to check

Where in a sound wave is the excess pressure zero?

Where the displacement of the layers is maximum (

s = \pm A

What is the phase difference between the pressure wave and the displacement wave?

π /2

; the pressure wave leads.

At a compression, which way are the layers moving?

Forward, along the direction of the wave, at maximum speed

A ω

4. The Density Wave

Where the pressure rises, the air is squeezed and its density rises too. From the definition of bulk modulus, $B = - \frac{Δ P}{d V / V}$ .

For a fixed mass $m$ : $V = \frac{m}{ρ}$ , so $d V = - \frac{m}{ρ ^{2}} d ρ = - \frac{V}{ρ} d ρ$ , that is $\frac{d V}{V} = - \frac{d ρ}{ρ}$ .
Substitute: $B = \frac{ρ Δ P}{Δ ρ}$ , so
$Δ ρ = \frac{ρ}{B} Δ P = \frac{Δ P}{v ^{2}}$
If $Δ P = Δ P_{0} cos (ω t - k x)$ , then $Δ ρ = Δ ρ_{0} cos (ω t - k x)$ with $Δ ρ_{0} = \frac{ρ}{B} Δ P_{0} = \frac{Δ P _{0}}{v ^{2}} = ρ A k$ .

Key idea

Density and pressure waves are in phase; both are

9 0^{\circ}

out of phase with the displacement wave.

5. How Big Are These Quantities?

For a $1 kHz$ tone in air ( $ρ \approx 1.21 kg m^{- 3}$ , $v \approx 343 m s^{- 1}$ ), use $A = \frac{Δ P _{0}}{ρ v ω}$ :

Figure 6: For a

1 kHz

tone in air. Even the loudest bearable sound changes the pressure by only about

0.03%

of atmospheric pressure, and the faintest audible one moves air by less than an atom's width.

Why we measure sound by pressure. Displacement amplitudes of ordinary sounds are far below a micrometre and impossible to see, but pressure changes of $1 0^{- 5}$ to $10 Pa$ are easy to measure. The ear, microphones and sound-level meters all respond to $Δ P$ . For the same $A$ , $Δ P_{0} = ρ v ω A$ grows with frequency, so high notes need smaller displacements for the same pressure swing.

6. Energy Carried by a Wave on a String

A wave carries energy along the medium. The cleanest case is a string of mass per unit length $μ$ and tension $T$ , carrying $y = A sin (k x - ω t)$ .

6.1 Kinetic energy per unit length

Particle velocity $v_{p} = \frac{\partial y}{\partial t} = - A ω cos (k x - ω t)$ . An element $d x$ of mass $μ d x$ has kinetic energy $d K = \frac{1}{2} (μ d x) v_{p}^{2}$ , so

\frac{d K}{d x} = \frac{1}{2} μ ω^{2} A^{2} cos^{2} (k x - ω t)

It is greatest where the element crosses its mean position and zero at the extremes.

6.2 Potential energy per unit length

The potential energy of the string is elastic: an element of horizontal length $d x$ is stretched to $d s = d x^{2} + d y^{2}$ . The work done by the tension in stretching it is $d U = T (d s - d x)$ .

Binomial expansion: $d s = d x 1 + (\partial y / \partial x)^{2} \approx d x [1 + \frac{1}{2} (\frac{\partial y}{\partial x})^{2}]$ .
So $d U = \frac{1}{2} T (\frac{\partial y}{\partial x})^{2} d x$ .
With $\frac{\partial y}{\partial x} = k A cos (k x - ω t)$ and $T = μ v^{2}$ , $T k^{2} = μ ω^{2}$ :
$\frac{d U}{d x} = \frac{1}{2} μ ω^{2} A^{2} cos^{2} (k x - ω t)$

This equals the kinetic energy per unit length at every point and every instant. The element is stretched most where the slope is steepest, which is at the mean position, not at the extremes: the opposite of a single particle in SHM.

Figure 7: Why a string element has the most energy at

y = 0

. At a crest it lies flat (

d s = d x

) and is momentarily at rest: no KE, no PE. Crossing

y = 0

it is tilted most, so

d s = d x^{2} + d y^{2}

is longest (maximum stretch, maximum PE) and it moves fastest (maximum KE).

Figure 8: On a travelling wave, kinetic and potential energy per unit length are equal at every point, both

\propto cos^{2} (k x - ω t)

: maximum at

y = 0

, zero at crests and troughs. This is unlike SHM of a single particle.

6.3 Total energy, power and intensity

Mechanical energy per unit length: $\frac{d E}{d x} = μ ω^{2} A^{2} cos^{2} (k x - ω t)$ . The average of $cos^{2}$ over whole wavelengths is $\frac{1}{2}$ , so the average energy per unit length is $\frac{1}{2} μ ω^{2} A^{2}$ . This energy moves along at speed $v$ , so the average power transmitted is

P_{avg} = \frac{1}{2} μ ω^{2} A^{2} v = 2 π^{2} f^{2} A^{2} μv

Kinetic and potential energy each carry half of it:

⟨ \frac{d K}{d t} ⟩ = ⟨ \frac{d U}{d t} ⟩ = \frac{1}{4} μ ω^{2} A^{2} v

For a medium of density $ρ$ , replace $μ$ by $ρS$ ( $S$ = cross-section). Then the energy per unit volume and the power per unit area are

u = \frac{1}{2} ρ ω^{2} A^{2}, I = \frac{P}{S} = \frac{1}{2} ρ ω^{2} A^{2} v = uv

Key idea

Power and intensity grow as $A^{2} f^{2}$ : double the amplitude, four times the power; double the frequency at the same amplitude, four times the power.

7. Intensity of a Sound Wave in Terms of Pressure

Sound in air follows the same formula, $I = \frac{1}{2} ρ ω^{2} A^{2} v$ . Put $A = \frac{Δ P _{0}}{ρ v ω}$ :

I = \frac{1}{2} ρ ω^{2} v \frac{Δ P _{0}^{2}}{ρ ^{2} v ^{2} ω ^{2}} = \frac{Δ P _{0}^{2}}{2 ρ v} = \frac{Δ P _{0}^{2} v}{2 B}

(using

B = ρ v^{2}

). For a given medium, intensity depends only on the pressure amplitude, not on the frequency.

In terms of displacement

$I = 2 π^{2} f^{2} A^{2} ρ v$ . At fixed $A$ , $I \propto f^{2}$ . Use when $A$ is given.

In terms of pressure

$I = \frac{Δ P _{0}^{2}}{2 ρ v}$ . At fixed $Δ P_{0}$ , $I$ does not depend on $f$ . Use when $Δ P_{0}$ is given.

Exam Trick

Decibels and pressure. Since $I \propto Δ P_{0}^{2}$ , a rise of $20 dB$ (intensity $\times 100$ ) means $Δ P_{0} \times 10$ ; a rise of $6 dB$ (intensity $\times 4$ ) means $Δ P_{0} \times 2$ . In general $\frac{Δ P _{0, 2}}{Δ P _{0, 1}} = 1 0^{(β_{2} - β_{1}) /20}$ .

Figure 9: One chain links every sound-wave quantity:

A \to Δ P_{0} \to I \to β

and back. Keep

B = ρ v^{2}

and

k = \frac{ω}{v}

at hand.

Quick Recall: tap to check

The intensity of a sound doubles. By what factor does

Δ P_{0}

change?

2

, since

I \propto Δ P_{0}^{2}

The frequency doubles at the same displacement amplitude. What happens to

I

It becomes 4 times, since

I \propto A^{2} f^{2}

The frequency doubles at the same pressure amplitude. What happens to

I

Nothing:

I = \frac{Δ P _{0}^{2}}{2 ρ v}

does not depend on

f

Figure 10: Revision map:

s

Δ P

and

Δ ρ

waves, their phases, and the energy and intensity they carry.

8. Solved Examples

Solved Example 1

A sound wave of wavelength

40 cm

travels in air. The difference between the maximum and minimum pressures at a point is

2.0 \times 1 0^{- 3} N m^{- 2}

. Find the amplitude of vibration of the particles of the medium. Bulk modulus of air

= 1.4 \times 1 0^{5} N m^{- 2}

Solution:

Pressure swings from $P + Δ P_{0}$ to $P - Δ P_{0}$ , so $2Δ P_{0} = 2.0 \times 1 0^{- 3}$ and $Δ P_{0} = 1.0 \times 1 0^{- 3} N m^{- 2}$ .

$Δ P_{0} = B A k = \frac{2 π B A}{λ}$ , so

A = \frac{Δ P _{0} λ}{2 π B} = \frac{( 1.0 \times 1 0 ^{- 3} ) ( 0.40 )}{2 π ( 1.4 \times 1 0 ^{5} )} \approx 4.5 \times 1 0^{- 10} m

Answer: $A \approx 4.5 \times 1 0^{- 10} m = 4.5 \overset{˚}{A}$ , about the size of a few atoms.

Solved Example 2

The displacement wave in air is

s = (2 \times 1 0^{- 6} m) sin (1000 π t - 3 π x)

, with

x

in m and

t

in s. The density of air is

1.2 kg m^{- 3}

. Write the pressure wave.

Solution:

$v = \frac{ω}{k} = \frac{1000 π}{3 π} \approx 333 m s^{- 1}$ ; $B = ρ v^{2} = 1.2 \times (333.3)^{2} \approx 1.33 \times 1 0^{5} Pa$ .

$Δ P_{0} = B A k = (1.33 \times 1 0^{5}) (2 \times 1 0^{- 6}) (3 π) \approx 2.51 Pa$ .

Answer: $Δ P = (2.51 Pa) cos (1000 π t - 3 π x)$ , leading the displacement wave by $π /2$ .

Solved Example 3

The pressure wave in a gas is

Δ P = (0.01 Pa) sin (1000 t - 3 x)

(SI units). The density of the gas is

1.3 kg m^{- 3}

. Find the displacement amplitude and write the displacement wave.

Solution:

$v = \frac{1000}{3} \approx 333.3 m s^{- 1}$ ; $B = ρ v^{2} = 1.3 \times (333.3)^{2} \approx 1.44 \times 1 0^{5} Pa$ .

$A = \frac{Δ P _{0}}{B k} = \frac{0.01}{( 1.44 \times 1 0 ^{5} ) ( 3 )} \approx 2.3 \times 1 0^{- 8} m$ .

Phase: $Δ P = - B \frac{\partial s}{\partial x}$ , so $\frac{\partial s}{\partial x} = - \frac{Δ P _{0}}{B} sin (1000 t - 3 x)$ . Integrating in $x$ : $s = - \frac{Δ P _{0}}{B k} cos (1000 t - 3 x)$ .

Answer: $s = - (2.3 \times 1 0^{- 8} m) cos (1000 t - 3 x)$ .

Solved Example 4

The threshold of pain corresponds to a pressure amplitude of about

28 Pa

. For a

1 kHz

sound in air (

ρ = 1.21 kg m^{- 3}

v = 343 m s^{- 1}

), find the displacement amplitude, the intensity and the sound level.

Solution:

$A = \frac{Δ P _{0}}{ρ v ω} = \frac{28}{( 1.21 ) ( 343 ) ( 2 π \times 1000 )} \approx 1.1 \times 1 0^{- 5} m$ .

$I = \frac{Δ P _{0}^{2}}{2 ρ v} = \frac{2 8 ^{2}}{2 ( 1.21 ) ( 343 )} \approx 0.94 W m^{- 2}$ ; $β = 10 lo g_{10} \frac{0.94}{1 0 ^{- 12}} \approx 120 dB$ .

Answer: $A \approx 11 μ m$ , $I \approx 0.94 W m^{- 2}$ , $β \approx 120 dB$ . At the threshold of hearing ( $Δ P_{0} \approx 2.8 \times 1 0^{- 5} Pa$ ) the same formula gives $A \approx 1.1 \times 1 0^{- 11} m$ .

Solved Example 5

The phase difference between the pressure wave and the displacement wave of a sound is
(A)

0

(B)

π /4

(C)

π /2

(D)

π

Solution:

$Δ P = - B \partial s / \partial x$ turns $sin$ into $cos$ : a quarter-cycle shift.

Answer: (C).

Solved Example 6

At a point in a sound wave where the displacement of the air layer is maximum, the excess pressure is
(A) maximum
(B) minimum
(C) zero
(D)

Δ P_{0} /2

Solution:

At $s = \pm A$ the $s$ - $x$ graph is flat, so $\partial s / \partial x = 0$ and $Δ P = 0$ : the pressure is at its normal value.

Answer: (C).

Solved Example 7

Two sound waves travel in the same medium. How does the intensity change if (a) the amplitude is kept the same and the frequency is doubled, (b) the pressure amplitude is kept the same and the frequency is doubled?

Solution:

(a) $I = 2 π^{2} f^{2} A^{2} ρ v \propto f^{2}$ at fixed $A$ : the intensity becomes 4 times.

(b) $I = \frac{Δ P _{0}^{2}}{2 ρ v}$ does not contain $f$ : the intensity is unchanged (the displacement amplitude halves).

Solved Example 8

A sound wave in air (

v = 340 m s^{- 1}

ρ = 1.2 kg m^{- 3}

) has a pressure amplitude of

10 Pa

. Find the density amplitude and the fractional change in density.

Solution:

$Δ ρ_{0} = \frac{Δ P _{0}}{v ^{2}} = \frac{10}{34 0 ^{2}} \approx 8.7 \times 1 0^{- 5} kg m^{- 3}$ .

$\frac{Δ ρ _{0}}{ρ} = \frac{8.7 \times 1 0 ^{- 5}}{1.2} \approx 7.2 \times 1 0^{- 5}$ .

Answer: $8.7 \times 1 0^{- 5} kg m^{- 3}$ , a change of only about $0.007%$ .

Solved Example 9

500 Hz

sound wave in air (

ρ = 1.3 kg m^{- 3}

v = 330 m s^{- 1}

) has displacement amplitude

1.0 μ m

. Find its intensity and sound level.

Solution:

$I = 2 π^{2} f^{2} A^{2} ρ v = 2 π^{2} (500)^{2} (1 0^{- 6})^{2} (1.3) (330) \approx 2.1 \times 1 0^{- 3} W m^{- 2}$ .

$β = 10 lo g_{10} (2.1 \times 1 0^{9}) \approx 93 dB$ .

Answer: $2.1 \times 1 0^{- 3} W m^{- 2}$ , about $93 dB$ .

Solved Example 10

For a sound of pressure amplitude

28 Pa

in air (

ρ = 1.21 kg m^{- 3}

v = 343 m s^{- 1}

), find the maximum speed of the air layers. Compare it with the speed of sound.

Solution:

$Δ P_{0} = ρ v (v_{p})_{m a x}$ , so $(v_{p})_{m a x} = \frac{28}{( 1.21 ) ( 343 )} \approx 0.067 m s^{- 1}$ .

Answer: about $6.7 cm s^{- 1}$ , roughly $2 \times 1 0^{- 4}$ of the wave speed, even for the loudest bearable sound.

Solved Example 11

Two sounds have levels

60 dB

and

40 dB

. Find the ratio of their pressure amplitudes.

Solution:

$β_{1} - β_{2} = 20 dB$ , so $\frac{I _{1}}{I _{2}} = 100$ . Since $I \propto Δ P_{0}^{2}$ , $\frac{Δ P _{0, 1}}{Δ P _{0, 2}} = 100 = 10$ .

Answer: $10 : 1$ .

Solved Example 12

A string of linear mass density

0.05 kg m^{- 1}

under tension

80 N

carries a sinusoidal wave of amplitude

5 mm

and frequency

60 Hz

. Find the wave speed, the average power transmitted and the average energy in one wavelength.

Solution:

$v = T / μ = 80/0.05 = 40 m s^{- 1}$ ; $ω = 2 π (60) \approx 377 rad s^{- 1}$ .

$P = \frac{1}{2} μ ω^{2} A^{2} v = \frac{1}{2} (0.05) (377)^{2} (5 \times 1 0^{- 3})^{2} (40) \approx 3.55 W$ .

Energy in one wavelength $= \frac{1}{2} μ ω^{2} A^{2} λ = \frac{P}{f} = \frac{3.55}{60} \approx 0.059 J$ ( $λ = 40/60 \approx 0.67 m$ ).

Answer: $40 m s^{- 1}$ , $3.55 W$ , $0.059 J$ .

Solved Example 13

Figure 5 shows the

s

x

graph of a longitudinal wave travelling towards

+ x

at one instant. At which point is the density of the medium maximum?
(A)

A

(B)

B

(C)

C

(D)

D

Solution:

Excess density follows $- \frac{\partial s}{\partial x}$ , so it is greatest where the $s$ - $x$ graph falls most steeply: at $B$ , where $s = 0$ on a falling slope. The layer just before $B$ is pushed forward and the one just after is pulled back, so layers crowd at $B$ . At $A$ and $C$ the slope is zero (normal density); at $D$ the slope is most positive (rarefaction).

Answer: (B).

Practice Questions

A sound wave has displacement amplitude $1 0^{- 8} m$ and wavelength $0.5 m$ in air ( $B = 1.4 \times 1 0^{5} Pa$ ). Find the pressure amplitude.Answer: $Δ P_{0} = 2 π B A / λ \approx 0.018 Pa$ .
The pressure amplitude of a sound is doubled at the same frequency. What happens to (a) the displacement amplitude (b) the intensity?Answer: (a) doubles (b) becomes 4 times.
Write the pressure wave for $s = A cos (ω t + k x)$ in a medium of bulk modulus $B$ .Answer: $Δ P = - B \partial s / \partial x = B A k sin (ω t + k x)$ .
At a rarefaction, what are the signs of $s$ , $Δ P$ and $v_{p}$ for a wave travelling towards $+ x$ ?Answer: $s = 0$ , $Δ P = - Δ P_{0}$ , $v_{p} = - A ω$ (layers move backward).
A $60 dB$ sound in air ( $ρ = 1.21$ , $v = 343$ , SI) has what pressure amplitude?Answer: $I = 1 0^{- 6} W m^{- 2}$ ; $Δ P_{0} = 2 ρ v I \approx 0.029 Pa$ .
A string wave's amplitude and frequency are both doubled at the same tension. By what factor does the power change?Answer: $P \propto A^{2} f^{2}$ : 16 times.

Common Mistakes to Avoid

Watch out

Placing pressure maxima where displacement is maximum. They are $λ /4$ apart: pressure is maximum where $s = 0$ .
Using $Δ P_{0} = B A$ . The correct amplitude is $B A k = 2 π B A / λ$ ; check the units (Pa needs a factor per metre).
Reading the peak-to-peak pressure difference as the amplitude. Maximum minus minimum equals $2Δ P_{0}$ .
Taking the density wave out of phase with pressure. $Δ ρ = Δ P / v^{2}$ : they are in phase.
Saying intensity always grows with frequency. At fixed pressure amplitude, $I = Δ P_{0}^{2} / (2 ρ v)$ has no $f$ .
Assuming potential energy on a string is largest at the crest (as in SHM). It is largest where the string is steepest, at $y = 0$ .
Forgetting that power is proportional to the square of amplitude and frequency: $P = 2 π^{2} f^{2} A^{2} μv$ .
Mixing up $B$ and $P$ : the bulk modulus of air for sound is $γ P \approx 1.4 \times 1 0^{5} Pa$ , not $P$ .

Frequently Asked Questions

What is the difference between a displacement wave and a pressure wave?

Both describe the same sound. The displacement wave gives how far each air layer moves from its mean position, $s = A sin (ω t - k x)$ . The pressure wave gives the excess pressure, $Δ P = B A k cos (ω t - k x)$ . They have the same frequency and wavelength but differ in phase by $π /2$ .

Why is pressure maximum where displacement is zero in a sound wave?

Excess pressure depends on how much a layer is squeezed, which is the change of displacement with position, not the displacement itself. At the centre of a compression, the layers on both sides are displaced towards it, so the layer there does not move but is squeezed the most.

What is the formula for pressure amplitude of a sound wave?

The pressure amplitude is $Δ P_{0} = B A k = 2 π B A / λ$ , where $B$ is the bulk modulus, $A$ the displacement amplitude and $λ$ the wavelength. Using $B = ρ v^{2}$ it can also be written $Δ P_{0} = ρ v ω A$ .

How are pressure and density waves related?

The excess density is $Δ ρ = Δ P / v^{2}$ , so the density wave is in phase with the pressure wave. Both are maximum at compressions and minimum at rarefactions, and both lead the displacement wave by a quarter cycle.

What is the intensity of a sound wave in terms of pressure amplitude?

Intensity is $I = Δ P_{0}^{2} / (2 ρ v)$ , or $Δ P_{0}^{2} v / (2 B)$ . It depends on the square of the pressure amplitude and on the medium, but not on the frequency. In terms of displacement amplitude, $I = 2 π^{2} f^{2} A^{2} ρ v$ .

How much power does a wave on a string carry?

The average power is $P = \frac{1}{2} μ ω^{2} A^{2} v$ , where $μ$ is mass per unit length, $A$ the amplitude and $v$ the wave speed. Kinetic and potential energy each carry half. Doubling the amplitude or the frequency makes the power four times larger.

How is the pressure wave tested in JEE Main?

JEE Main questions give a displacement or pressure equation and ask for the other, the pressure amplitude $B A k$ , the phase difference of $π /2$ , the point of maximum pressure, or intensity from $Δ P_{0}^{2} / (2 ρ v)$ . Energy and power on a string, $\frac{1}{2} μ ω^{2} A^{2} v$ , is also common.

What does JEE Advanced ask about displacement and pressure waves?

JEE Advanced links these ideas to standing waves in pipes, where a displacement node is a pressure antinode, and uses relations such as $Δ P = ρ v v_{p}$ , the density wave, and intensity comparisons in decibels. Sign care, $Δ P = - B \partial s / \partial x$ , decides many answers.

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