Transverse and Longitudinal Waves

1.0What is a Wave?

Waves are disturbances that transfer energy from one point to another without the transfer of matter. They are fundamental in mechanical physics and an important topic in JEE Physics, particularly in Oscillations and Waves.

• Waves transport energy, not matter.
• They are classified based on particle motion relative to wave propagation:
• • Transverse Waves
  • Longitudinal Waves

2.0What are Transverse Waves?

Transverse Waves are waves in which the particles of the medium oscillate perpendicular to the direction of wave propagation.

Key Features:

• Particle motion: perpendicular to wave direction
• Examples: light waves, water waves, waves on a string
• Can propagate in solids and on surfaces of liquids
• Characterized by crest, trough, amplitude, wavelength, and frequency

Formula for wave speed:
v = f \lambda
Where:

• ( v ) = wave speed
• ( f ) = frequency
• ( $\lambda$ ) = wavelength

3.0What are Longitudinal Waves?

Longitudinal Waves are waves in which the particles of the medium oscillate parallel to the direction of wave propagation.

Key Features:

• Particle motion: parallel to wave direction
• Examples: sound waves, ultrasound waves, seismic P-waves
• Require a medium (solid, liquid, or gas) to propagate
• Characterized by compressions, rarefactions, wavelength, frequency, and amplitude

Wave speed formula (for longitudinal waves in a medium):

$v=\sqrt{\frac{B}{\rho }}$
Where:

• ( B ) = bulk modulus of the medium
• ( $\rho$ ) = density of the medium

4.0Differences Between Transverse and Longitudinal Waves

5.0Based on the motion of particles of medium:

Waves are of two types on the basis of motion of particles of the medium:
(a) Longitudinal waves
(b) Transverse waves

Transverse waves	Longitudinal waves
Particles of the medium vibrate in a direction perpendicular to the direction of propagation of wave.	Particles of a medium vibrate in the direction of wave motion.
It travels in the form of crests (C) and troughs (T).	It travels in the form of compression (C) and rarefaction (R).
Transverse waves can be transmitted through solids, they can be set up on the surface of liquids. But they cannot be transmitted into liquids and gases.	These waves can be transmitted through solids, liquids and gases because for these waves propagation, volume elasticity is necessary.

Speed of Transverse wave on Strings

This velocity depends on two properties of the string:
(a) Elasticity (measured by tension F in the string)
(b) Inertia (measured by mass per unit length)

Take a small element Δℓ of the string in which a wave is moving with speed v. Suppose this element forms an arc of radius R. A force of tension pulls this element tangentially by which horizontal components cancel and vertical components add to form a restoring force i.e.

$F_r=F\sin \theta +F\sin \theta$

$F_r=2F\sin \theta \approx 2F\cdot \frac{\Delta \ell /2}{R}=\frac{F,\Delta \ell }{R}$

If μ is mass per unit length of string and Δm is the mass of element then downward acceleration is given by:

$a=\frac{F_r}{\Delta m}$

$=\frac{F,\Delta \ell /R}{\Delta m}$

$=\frac{F}{\mu R}$

$\text{where}$

$\mu =\frac{\Delta m}{\Delta \ell }$

But the element is moving in a circle of radius R.
So centripetal acceleration:

$a=\frac{v^2}{R}$

∴ $\frac{v^2}{R}=\frac{F}{\mu R}$

$v=\sqrt{\frac{F}{\mu }}$

Note:

If A is area of cross-section and ρ is density:

$\mu =\rho A$

So, velocity:

$v=\sqrt{\frac{F}{\rho A}}$

Speed of Longitudinal Sound Wave

In a solid medium ( $v=\sqrt{\frac{k+\frac{4}{3}\eta }{\rho }}$ ) where  k = Bulk modulus; ($\eta =$) Modulus of rigidity; ( $\rho =$ ) Density.

When the solid is in the form of long bar ( $v=\sqrt{\frac{Y}{\rho }}$ ) where ( Y = ) Young's modulus of material of rod.

Velocity of sound waves in a fluid medium (liquid or gas) is given by
( $v=\sqrt{\frac{B}{\rho }}$ )

where, ( $\rho =$ ) density of the medium and ( B = ) Bulk modulus of the medium given by,
( $B=-V\frac{dP}{dV}$ )

Newton's formula : Newton assumed propagation of sound through a gaseous medium to be an isothermal process.
( PV = ) constant
($\frac{dP}{dV}=-\frac{P}{V}$)

Wave on String and Sound Waves

and hence ( B = P ) using equation ( $B=-V\frac{dP}{dV}$ )

and thus velocity of sound in a gas,
( $v=\sqrt{\frac{P}{\rho }}=\sqrt{\frac{RT}{M}}$ ) where ( M = ) molar mass

the density of air at ( $0^{\circ }$ ) and pressure 76 cm of Hg column is ( $\rho =1.293kg/m^3$ ). This temperature and pressure is called standard temperature and pressure or STP. Speed of sound in air is 280 m/s. This value is less than measured speed of sound in air 332 m/s then Laplace suggested the correction.

Laplace’s correction : Later Laplace established that propagation of sound in a gas is not an isothermal but an adiabatic process and hence ( $P{V}^{\gamma }=$ ) constant.

( $\frac{dP}{dV}=-\gamma \frac{P}{V}$ )

where, ( $B=-V\frac{dP}{dV}=\gamma P$ )

and hence speed of sound in a gas,
( $v=\sqrt{\frac{\gamma P}{\rho }}=\sqrt{\frac{\gamma RT}{M}}$ )

6.0Factors affecting speed of sound in atmosphere

(a) Effect of temperature: as temperature (T) increases velocity (v) increases.

$v\propto \sqrt{T}$

For small change in temperature above room temperature v increases linearly by 0.6 m/s for every 1∘C1∘C rise in temp.

$v=\sqrt{\frac{\gamma R}{M}}\times T^{1/2}$

$\frac{\Delta v}{v}=\frac{1}{2}\frac{\Delta T}{T}$

$\Delta v=(0.6)\Delta T$

(b) Effect of pressure: The speed of sound in a gas is given by

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

So at constant temperature, if P changes then ρ also changes in such a way that P/ρP/ρ remains constant. Hence pressure does not have any effect on velocity of sound as long as temperature is constant.

(c) Effect of humidity: With increase in humidity density decreases. This is because the molar mass of water vapour is less than the molar mass of air.

Some points regarding velocity of sounds:

(1) As solids are most elastic while gases least i.e. ($E_S>E_L>E_G$). So, the velocity of sound is maximum in solids and minimum in gases

$V_{steel}>V_{water}>V_{air}$

$5000\text{ m/s}>1500\text{ m/s}>330\text{ m/s}$

As for sound ( $v_{water}>v_{Air}$ ) while for light ( $v_w<v_A$ ).
Water is rarer than air for sound and denser for light.

The concept of rarer and denser media for a wave is through the velocity of propagation (and not density). Lesser the velocity, denser is said to be the medium and vice-versa.

(2) Sound of any frequency or wavelength travels through a given medium with the same velocity.
For a given medium velocity remains constant. All other factors like phase, loudness pitch, quality etc. have practically no effect on sound velocity.

(3) Relation between velocity of sound and root mean square velocity.

So,

$v_{\text{sound}}=\sqrt{\frac{\gamma RT}{M}}$

$v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$

$\frac{v_{\text{rms}}}{v_{\text{sound}}}=\sqrt{\frac{3}{\gamma }}\text{or}{v}_{\text{sound}}=[\gamma /3{]}^{1/2}{v}_{\text{rms}}$

(4) There is no atmosphere on moon, therefore propagation of sound is not possible there. To do conversation on moon, the astronaut uses an instrument which can transmit and detect electromagnetic waves.

7.0Wave Parameters and Formulas

Key Parameters of Waves:

1. Amplitude (A): Maximum displacement of particles from mean position
2. Wavelength (λ): Distance between two consecutive crests or compressions
3. Frequency (f): Number of oscillations per second
4. Wave speed (v): Distance travelled per unit time

Common Formulas:

• Wave speed: ( $v=f\lambda$ )
• Frequency: ( $f=\frac{1}{T}$ ) (T = time period)
• For longitudinal waves in a medium: ( $v=\sqrt{\frac{B}{\rho }}$ )

8.0Properties of Transverse and Longitudinal Waves

Transverse Waves

• Wavelength (λ) = Distance between two consecutive crests or troughs
• Amplitude (A) = Maximum displacement from mean position
• Particles move up and down

Longitudinal Waves

• Wavelength (λ) = Distance between two consecutive compressions or rarefactions
• Amplitude (A) = Maximum displacement along the wave direction
• Particles move back and forth

9.0Examples of Transverse and Longitudinal Waves

Transverse Waves Examples:

• Light waves
• Waves on a stretched string
• Water surface waves

Longitudinal Waves Examples:

• Sound waves in air
• Ultrasound waves in medicine
• Seismic P-waves during earthquakes

Keyword: transverse and longitudinal waves examples

10.0Applications of Waves in Real Life 

• Sound Engineering: Longitudinal waves for speakers and musical instruments
• Communication: Radio and light waves (transverse) for transmission
• Medical Imaging: Ultrasound (longitudinal waves)
• Seismology: P-waves (longitudinal) and S-waves (transverse) for earthquake detection

11.0Related Questions

1. Fig. shows a string of linear mass density (1\ gm/cm) on which a wave pulse is travelling. Find the time taken by the pulse in travelling through a distance of (50\ cm) on the string. Take ( g = 10\ m/s^2 ).

Solution:
Tension in the string
F = mg = 10 N

Mass per unit length = (1\ g/cm = 0.1\ kg/m)

Velocity of wave, $v=\sqrt{\frac{F}{\mu }}=\sqrt{\frac{10}{0.1}}=10m/s$

Time taken by pulse in travelling a distance of (50\ cm = \frac{0.5}{10} = .05\ seconds)

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2. The diameter of an iron wire is (1.2\ mm). If the speed of transverse wave in the wire be (50\ m/s) then what is the tension in the wire. The density of iron is ($7.7\times 10^{3}kg/m^3$).

Solution:
Diameter of wire = ($1.2mm=1.2\times 10^{-3}m$)  ;  Radius of wire = ($0.6\times 10^{-3}m$)

Speed of wave = (50\ m/s)  ;  Density of wire ($\rho =7.7\times 10^{3}Kg/m^3$)

Speed of transverse wave is given by

$v=\sqrt{\frac{F}{\mu }}$

But

$\mu =\rho A$

$\therefore F=v^2\cdot \rho \cdot A=(50{)}^2\times 7.7\times 10^3\times 3.14\times (0.6\times 10^{-3}{)}^2=21.76N$

3. Find the speed of sound in (H_2) at temperature (T), if the speed of sound in ($O_2$) is (450\ m/s) at this temperature.

Solution:

$v=\sqrt{\frac{\gamma RT}{M}}$

Since temperature, (T) is constant,

$\frac{v_{H_2}}{v_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}=\sqrt{\frac{32}{2}}=4v_{H_2}=4\times 450=1800m/s$

4. At which temperature the speed of sound in hydrogen will be same as that of speed of sound in oxygen ($100^{\circ }C$)
(A) –148°C  (B) –212.5°C  (C) –317.5°C  (D) –249.7°C
Ans. (D)

Solution:
Speed of sound in gases is $v=\sqrt{\frac{\gamma RT}{M}}$ As (v) and $(\gamma )$ are same for both gases in this example

$T\propto M$

$\frac{T_{H_2}}{T_{O_2}}=\frac{M_{H_2}}{M_{O_2}}$

$\frac{T_{H_2}}{(273+100)}=\frac{2}{32}\Rightarrow T_{H_2}=23.2K=-249.7^{\circ }C$

5. A gas mixture has 24% of Argon, 32% of oxygen, and 44% of ($C{O}_2$) by mass. Find the velocity of sound in the gas mixture at ($27^{\circ }C$). Given (R = 8.4) in S.I. units.
Molecular weight of ($Ar=40,O_2=32,C{O}_2=44$) and ( ${\gamma }_{Ar}=5/3,{\gamma }_{O_2}=7/5,{\gamma }_{C{O}_2}=4/3$ )

Solution:

$\begin{array}{rl}{M}_{mix}& =\frac{M_{total}}{m_{Ar}/M_{Ar}+m_{O_2}/M_{O_2}+m_{C{O}_2}/M_{C{O}_2}}=\frac{100}{24/40+32/32+44/44}=\frac{100}{2.6}\\ f_{mix}& =\frac{n_1{f}_1+n_2{f}_2+n_3{f}_3}{n_1+n_2+n_3}=\frac{0.6\times 3+1\times 5+1\times 6}{2.6}=\frac{12.8}{2.6}\\ {\gamma }_{mix}& =\frac{f_{mix}+2}{f_{mix}}=\frac{12.8/2.6+2}{12.8/2.6}=\frac{18}{12.8}\\ v& =\sqrt{\frac{{\gamma }_{mix}RT}{M_{mix}}}=\sqrt{\frac{18\times 8.4\times 300}{12.8\times 100\times 10^{-3}}\times 2.6}\\ & \approx 303.5\text{ m/s}\end{array}$