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Waves and Oscillations (Stationary Waves and Vibrations of a Stretched String)

Speed of Longitudinal Wave, Bulk Modulus, Boyle’s Law and Newton–Laplace Formula for Velocity of Sound in Gas

The speed of a longitudinal wave depends on the elastic property and density of the medium through which the wave travels. In general, the speed of a longitudinal wave is determined by the square root of the ratio of the appropriate modulus of elasticity to the density of the medium.

General Expression for the Speed of a Longitudinal Wave

The speed of a longitudinal wave travelling through an elastic medium is given by:

\[ \boxed{v=\sqrt{\frac{E}{\rho}}} \]

Here, E represents the appropriate modulus of elasticity of the medium and ρ represents its density.

The particular modulus of elasticity depends on the nature of the medium. Young’s modulus is used for longitudinal waves in solids, whereas bulk modulus is used for longitudinal waves in liquids and gases.

Speed of Longitudinal Waves in Solids

In a solid, a longitudinal wave produces alternate compressions and extensions along the direction of propagation. The relevant elastic property is Young’s modulus Y.

Therefore, the speed of a longitudinal wave in a solid is:

\[ \boxed{v=\sqrt{\frac{Y}{\rho}}} \]

Here, Y is Young’s modulus of the material and ρ is its density.

Speed of Longitudinal Waves in Liquids and Gases

Liquids and gases cannot sustain a longitudinal deformation described by Young’s modulus. The relevant elastic property for these fluids is the bulk modulus of elasticity B.

Therefore, the speed of a longitudinal wave in a liquid or gas is:

\[ \boxed{v=\sqrt{\frac{B}{\rho}}} \]

Here, B is the bulk modulus of the medium and ρ is its density.

Bulk Modulus of Elasticity

Bulk modulus of elasticity is a measure of the resistance offered by a substance to a change in its volume. It is defined as the ratio of the change in pressure to the corresponding volumetric strain.

The bulk modulus is expressed as:

\[ \boxed{B=-\frac{\Delta P}{\Delta V/V}} \]

The negative sign indicates that an increase in pressure generally produces a decrease in volume. The magnitude of the bulk modulus is therefore considered positive for ordinary compressions.

The pressure applied to a fluid is force per unit area:

\[ P=\frac{F}{A} \]

The SI unit of bulk modulus is the pascal (Pa), which is equivalent to newton per square metre.

Boyle’s Law and Isothermal Bulk Modulus

For a fixed mass of an ideal gas at constant temperature, Boyle’s law states that the product of pressure and volume remains constant.

\[ PV=\text{constant} \]

Differentiating the above equation:

\[ d(PV)=0 \]

Using the product rule of differentiation:

\[ P\,dV+V\,dP=0 \]

Rearranging:

\[ V\,dP=-P\,dV \]

Dividing both sides by V:

\[ dP=-P\frac{dV}{V} \]

Therefore:

\[ -\frac{dP}{dV/V}=P \]

Comparing this result with the definition of bulk modulus:

\[ B=-\frac{dP}{dV/V} \]

We obtain:

\[ \boxed{B=P} \]

Thus, for an ideal gas undergoing an isothermal change, the bulk modulus is equal to the pressure of the gas.

Newton’s Formula for the Velocity of Sound in a Gas

For a gas, the velocity of a longitudinal wave is given by:

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

Newton assumed that the propagation of sound through a gas is an isothermal process. Under isothermal conditions, the bulk modulus of the gas is:

\[ B=P \]

Substituting this value of bulk modulus:

\[ v=\sqrt{\frac{P}{\rho}} \]

Therefore, Newton’s formula for the velocity of sound in a gas is:

\[ \boxed{v=\sqrt{\frac{P}{\rho}}} \]

Limitation of Newton’s Formula

Newton’s formula did not agree with the experimentally observed velocity of sound in gases. The calculated value was considerably lower than the experimental value.

The reason for this discrepancy was Newton’s assumption that the propagation of sound in a gas is an isothermal process. In reality, the compressions and rarefactions produced during the propagation of sound occur very rapidly, so there is insufficient time for significant heat exchange with the surroundings.

Laplace’s Correction to Newton’s Formula

Laplace corrected Newton’s formula by considering the propagation of sound in a gas to be an adiabatic process rather than an isothermal process.

For an adiabatic process, the bulk modulus of a gas is:

\[ \boxed{B=\gamma P} \]

Here, γ is the ratio of the specific heat capacity of the gas at constant pressure to its specific heat capacity at constant volume.

\[ \gamma=\frac{C_P}{C_V} \]

The velocity of a longitudinal wave in a gas is:

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

Substituting the adiabatic bulk modulus:

\[ v=\sqrt{\frac{\gamma P}{\rho}} \]

Newton–Laplace Formula for the Velocity of Sound in a Gas

Therefore, the corrected expression for the velocity of sound in a gas is:

\[ \boxed{v=\sqrt{\frac{\gamma P}{\rho}}} \]

This equation is known as the Newton–Laplace formula for the velocity of sound in a gas.

The formula shows that the velocity of sound depends on the pressure, density and ratio of specific heats of the gas.

Newton–Laplace Formula in Terms of Temperature and Molecular Mass

The ideal gas equation for a given mass of gas is:

\[ PV=nRT \]

If the mass of the gas is m and its molar mass is M, then the number of moles is:

\[ n=\frac{m}{M} \]

Therefore:

\[ PV=\frac{m}{M}RT \]

Dividing both sides by V:

\[ P=\frac{m}{V}\frac{RT}{M} \]

Since the density of the gas is:

\[ \rho=\frac{m}{V} \]

Therefore:

\[ P=\rho\frac{RT}{M} \]

Dividing by density:

\[ \frac{P}{\rho}=\frac{RT}{M} \]

The Newton–Laplace formula is:

\[ v=\sqrt{\frac{\gamma P}{\rho}} \]

Substituting \(\frac{P}{\rho}=\frac{RT}{M}\):

\[ v=\sqrt{\frac{\gamma RT}{M}} \]

Hence, the Newton–Laplace formula can also be written as:

\[ \boxed{v=\sqrt{\frac{\gamma RT}{M}}} \]

Here, R is the universal gas constant, T is the absolute temperature of the gas, M is the molar mass of the gas and γ is the ratio of specific heats.

Thus, the Newton–Laplace formula establishes that the velocity of sound in an ideal gas depends on the ratio of specific heats, absolute temperature and molar mass of the gas.

Factors Affecting the Speed of Longitudinal Wave in Gases

The speed of a longitudinal wave in a gas, such as the speed of sound, depends on the elastic properties and physical conditions of the gas. For an ideal gas, the Newton–Laplace equation provides the fundamental expression for the speed of sound.

\[ v=\sqrt{\frac{\gamma P}{\rho}} \]

Here, v is the speed of the longitudinal wave, γ is the ratio of specific heats of the gas, P is the pressure of the gas and ρ is its density.

Effect of Temperature

For an ideal gas, the pressure-to-density ratio can be expressed in terms of the absolute temperature and molar mass of the gas.

\[ \frac{P}{\rho}=\frac{RT}{M} \]

Substituting this relation in the Newton–Laplace equation:

\[ v=\sqrt{\frac{\gamma RT}{M}} \]

For a particular gas, γ, R and M remain constant. Therefore:

\[ \boxed{v\propto\sqrt{T}} \]

Thus, the speed of a longitudinal wave in a given gas increases with the square root of its absolute temperature.

For air, the speed of sound can approximately be written as:

\[ v\approx331\sqrt{\frac{T}{273}} \]

At 0°C, the absolute temperature is approximately 273 K. Therefore:

\[ v\approx331\,\mathrm{m\,s^{-1}} \]

At 20°C, the absolute temperature is approximately 293 K.

\[ v\approx331\sqrt{\frac{293}{273}} \]
\[ \boxed{v\approx343\,\mathrm{m\,s^{-1}}} \]

Therefore, the speed of sound in air increases from approximately 331 m s−1 at 0°C to approximately 343 m s−1 at 20°C.

Effect of Density

From the Newton–Laplace equation:

\[ v=\sqrt{\frac{\gamma P}{\rho}} \]

If the pressure and ratio of specific heats are kept constant, the speed of the longitudinal wave varies inversely as the square root of the density.

\[ \boxed{v\propto\frac{1}{\sqrt{\rho}}} \]

Therefore, at the same pressure, a gas with greater density has a lower wave speed, provided the other relevant properties remain unchanged.

For example, if the density of a gas is increased four times while its pressure and γ remain unchanged:

\[ \rho_2=4\rho_1 \]

Then:

\[ \frac{v_2}{v_1}=\sqrt{\frac{\rho_1}{\rho_2}} \]
\[ \frac{v_2}{v_1}=\frac{1}{2} \]

Thus, under these specified conditions, increasing the density four times reduces the wave speed to one-half of its original value.

Effect of Pressure

The Newton–Laplace equation appears to show that the speed of sound depends directly on the square root of pressure.

\[ v=\sqrt{\frac{\gamma P}{\rho}} \]

However, for an ideal gas at constant temperature, pressure and density change in the same proportion. From the ideal gas equation:

\[ \frac{P}{\rho}=\frac{RT}{M} \]

Therefore, at constant temperature and for the same gas:

\[ v=\sqrt{\frac{\gamma RT}{M}} \]

Since pressure does not appear in this expression:

\[ \boxed{\text{Speed of sound is independent of pressure at constant temperature}} \]

For example, if the pressure of an ideal gas is doubled at constant temperature, its density also doubles. Therefore:

\[ P_2=2P_1 \]
\[ \rho_2=2\rho_1 \]

Hence:

\[ \frac{v_2}{v_1} = \sqrt{\frac{P_2/\rho_2}{P_1/\rho_1}} \]
\[ \frac{v_2}{v_1}=1 \]

Therefore, the speed remains unchanged. Thus, for an ideal gas at constant temperature, changing the pressure alone does not change the speed of sound.

Effect of Molecular Mass

From the ideal-gas form of the Newton–Laplace equation:

\[ v=\sqrt{\frac{\gamma RT}{M}} \]

For gases at the same temperature, the speed depends on the molar mass of the gas as:

\[ \boxed{v\propto\frac{1}{\sqrt{M}}} \]

Therefore, at the same temperature, a gas with lower molar mass generally allows sound to travel faster, provided the ratio of specific heats is also taken into account.

For example, at approximately 0°C, the speed of sound is about 331 m s−1 in air, whereas it is about 1.3 × 103 m s−1 in hydrogen. The much lower molar mass of hydrogen is a major reason for its much greater sound speed.

For two gases at the same temperature:

\[ \frac{v_1}{v_2} = \sqrt{\frac{\gamma_1M_2}{\gamma_2M_1}} \]

If the two gases also have the same value of γ, this becomes:

\[ \boxed{\frac{v_1}{v_2}=\sqrt{\frac{M_2}{M_1}}} \]

Effect of Ratio of Specific Heats

The Newton–Laplace equation shows that the speed of sound also depends on the ratio of specific heats, γ.

\[ v=\sqrt{\frac{\gamma RT}{M}} \]

For a given temperature and molar mass:

\[ \boxed{v\propto\sqrt{\gamma}} \]

Thus, a gas having a larger ratio of specific heats has a greater speed of sound when its temperature and molar mass are the same.

For example, consider two gases having the same temperature and molar mass, with γ values 1.40 and 1.67 respectively.

\[ \frac{v_2}{v_1}=\sqrt{\frac{1.67}{1.40}} \]
\[ \frac{v_2}{v_1}\approx1.09 \]

Thus, under these hypothetical conditions, the gas with γ = 1.67 would have a wave speed approximately 9% greater than the gas with γ = 1.40.

Effect of Humidity

Humidity also affects the speed of sound in air. Moist air contains water vapour, whose effective molar mass is lower than that of the main components of dry air. At the same temperature and pressure, increasing humidity therefore generally decreases the average molar mass of the air mixture.

From the ideal-gas form:

\[ v=\sqrt{\frac{\gamma RT}{M}} \]

A decrease in the effective molar mass tends to increase the speed of sound. Therefore, at the same temperature and pressure, sound generally travels slightly faster in humid air than in dry air.

\[ \boxed{\text{Speed of sound increases with increase in humidity}} \]

The effect of humidity is smaller than the effect of temperature under ordinary atmospheric conditions, but it is important in accurate measurements of the speed of sound.

Summary of the Dependence

For an ideal gas, the most useful form of the Newton–Laplace equation is:

\[ \boxed{v=\sqrt{\frac{\gamma RT}{M}}} \]

Therefore, for a given gas:

\[ \boxed{v\propto\sqrt{T}} \]

For gases at the same temperature, the dependence on molar mass is:

\[ \boxed{v\propto\frac{1}{\sqrt{M}}} \]

For a given temperature and molar mass, the dependence on the ratio of specific heats is:

\[ \boxed{v\propto\sqrt{\gamma}} \]

Thus, the speed of a longitudinal wave in a gas is primarily determined by the temperature, molar mass and ratio of specific heats of the gas. For an ideal gas at constant temperature, the speed of sound is independent of pressure, while humidity causes a small increase in the speed of sound in air.

Formation of Stationary Waves on a Stretched String

A stationary wave is formed when two waves of the same amplitude, frequency and wavelength travel through the same medium in opposite directions and superpose on each other. The resulting wave pattern appears to remain stationary, with certain points of the medium permanently at rest and other points vibrating with maximum amplitude.

Stationary waves are commonly produced on a stretched string when a progressive wave travelling along the string is reflected from a fixed end and travels back along the string. The incident and reflected waves then travel in opposite directions and interfere with each other.

Formation of a Stationary Wave by Superposition

Consider two progressive waves of equal amplitude A, equal angular frequency ω and equal wave number k, travelling through a stretched string in opposite directions.

Let the first wave travel in the positive x-direction. Its displacement is:

\[ y_1=A\sin(\omega t-kx) \]

The second wave travels in the negative x-direction. Its displacement is:

\[ y_2=A\sin(\omega t+kx) \]

According to the principle of superposition, the resultant displacement at any point is equal to the algebraic sum of the displacements produced by the two waves.

\[ y=y_1+y_2 \]

Substituting the expressions for the two waves:

\[ y=A\sin(\omega t-kx)+A\sin(\omega t+kx) \]

Taking A common:

\[ y=A\left[\sin(\omega t-kx)+\sin(\omega t+kx)\right] \]

Using the trigonometric identity:

\[ \sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right) \]

Here:

\[ C=\omega t-kx \]
\[ D=\omega t+kx \]

Therefore:

\[ \frac{C+D}{2}=\omega t \]
\[ \frac{C-D}{2}=-kx \]

Since:

\[ \cos(-kx)=\cos(kx) \]

Substituting these results:

\[ y=2A\sin\omega t\cos kx \]

Hence, the equation of the stationary wave is:

\[ \boxed{y=2A\sin\omega t\cos kx} \]

This equation shows that the displacement of every particle depends on time through \(\sin\omega t\), while its amplitude depends on position through \(\cos kx\).

Amplitude of a Particle in a Stationary Wave

The stationary-wave equation is:

\[ y=2A\sin\omega t\cos kx \]

Comparing this with the general equation of simple harmonic motion, the amplitude of a particle at position x is:

\[ \boxed{A_x=2A|\cos kx|} \]

Therefore, unlike a progressive wave, the amplitude of a particle in a stationary wave depends on its position along the string.

Nodes

A node is a point on a stationary wave where the amplitude of vibration is always zero. Therefore, a particle situated at a node remains permanently at rest.

For a stationary wave:

\[ A_x=2A|\cos kx| \]

At a node:

\[ A_x=0 \]

Therefore:

\[ \cos kx=0 \]

Hence:

\[ kx=\frac{(2n+1)\pi}{2} \]

Therefore, the positions of the nodes are:

\[ \boxed{x=\frac{(2n+1)\lambda}{4}} \]

where \(n=0,1,2,3,\ldots\).

Antinodes

An antinode is a point on a stationary wave where the amplitude of vibration is maximum. The particles at antinodes vibrate with maximum amplitude.

For a stationary wave, the amplitude is:

\[ A_x=2A|\cos kx| \]

The amplitude is maximum when:

\[ |\cos kx|=1 \]

Therefore:

\[ kx=n\pi \]

Hence, the positions of the antinodes are:

\[ \boxed{x=\frac{n\lambda}{2}} \]

where \(n=0,1,2,3,\ldots\).

The maximum amplitude at an antinode is:

\[ \boxed{A_{\max}=2A} \]

Diagram of Nodes and Antinodes

The following diagram represents the characteristic pattern of a stationary wave on a stretched string. The points marked N are nodes and the points marked A are antinodes.

N N N N N NA A A A A λ/2 λ/4Stationary Wave

The distance between two successive nodes is \(\lambda/2\), and the distance between two successive antinodes is also \(\lambda/2\). The distance between a node and the nearest antinode is \(\lambda/4\).

Distance Between Nodes and Antinodes

The positions of successive nodes differ by half a wavelength. Therefore, the distance between two successive nodes is:

\[ \boxed{\text{Distance between successive nodes}=\frac{\lambda}{2}} \]

Similarly, the distance between two successive antinodes is:

\[ \boxed{\text{Distance between successive antinodes}=\frac{\lambda}{2}} \]

The distance between a node and its nearest antinode is:

\[ \boxed{\text{Distance between a node and nearest antinode}=\frac{\lambda}{4}} \]

Phase of Particles in a Stationary Wave

The equation of a stationary wave is:

\[ y=2A\sin\omega t\cos kx \]

The factor \(\cos kx\) determines the amplitude and sign of the displacement of a particle at position x. The sign of this factor determines whether the particle vibrates in phase or opposite phase with another particle.

Particles Vibrating in the Same Phase

All particles lying between two consecutive nodes vibrate in the same phase. They reach their extreme positions simultaneously, pass through their mean positions simultaneously and move in the same direction at the same instant.

Therefore, the phase difference between any two particles within the same segment between consecutive nodes is:

\[ \boxed{\Delta\phi=0} \]

Such a region between two successive nodes is called a loop or segment of the stationary wave.

Particles Vibrating in Opposite Phase

Particles belonging to two adjacent loops, separated by a node, vibrate in opposite phases. When particles in one loop move upward from their mean positions, particles in the adjacent loop move downward from their mean positions.

Therefore, the phase difference between particles belonging to adjacent loops is:

\[ \boxed{\Delta\phi=\pi} \]

Thus, the phase difference between two vibrating particles in a stationary wave is either 0 or π, depending on whether they belong to the same loop or adjacent loops.

Phase Difference Between Two Particles

For two particles situated within the same loop of a stationary wave, the particles vibrate in the same phase, even though their amplitudes may be different.

\[ \boxed{\Delta\phi=0} \]

For two particles situated in adjacent loops, the particles vibrate in opposite phases.

\[ \boxed{\Delta\phi=\pi} \]

A node itself has zero amplitude and therefore does not undergo vibration. Consequently, assigning a vibration phase to a perfectly stationary node is not physically meaningful. The statements about phase refer to the vibrating particles on either side of the node.

Meaning of Adjacent Loops

The region between two successive nodes is called a loop. Two loops separated by a common node are called adjacent loops.

Particles in the same loop vibrate in the same phase, whereas particles in adjacent loops vibrate in opposite phases.

\[ \boxed{\text{Same loop}\rightarrow\Delta\phi=0} \]
\[ \boxed{\text{Adjacent loops}\rightarrow\Delta\phi=\pi} \]

Why the Wave is Called Stationary

In a progressive wave, the wave pattern travels continuously through the medium. In a stationary wave, however, the positions of nodes and antinodes remain fixed in space.

The particles of the medium continue to vibrate about their mean positions, but the overall pattern of nodes and antinodes does not travel along the string.

For this reason, the wave is called a stationary wave.

Stationary Wave and Standing Wave

A stationary wave is also called a standing wave. The two terms refer to the same physical phenomenon: a wave pattern in which fixed nodes and antinodes are formed due to the superposition of two waves of the same frequency, wavelength and amplitude travelling in opposite directions.

Thus:

\[ \boxed{\text{Stationary wave}=\text{Standing wave}} \]

Important Characteristics of a Stationary Wave

The amplitude of vibration is different at different positions along the string. It is zero at nodes and maximum at antinodes.

There is no net transfer of energy along the string by an ideal stationary wave. Energy oscillates locally between kinetic and potential forms within the different segments of the string.

The distance between two successive nodes or two successive antinodes is \(\lambda/2\), while the distance between a node and its nearest antinode is \(\lambda/4\).

All particles between two successive nodes vibrate with the same phase, whereas particles in adjacent loops vibrate in opposite phases.

The nodes and antinodes remain fixed in their positions and therefore the wave pattern does not propagate through the medium.

Therefore, a stationary wave on a stretched string is formed by the superposition of two identical waves travelling in opposite directions, and its displacement is represented by y = 2A sin(ωt) cos(kx). The formation of fixed nodes and antinodes is the most important characteristic of a stationary wave.

Formation of Stationary Waves When a String is Fixed at Both Ends

When a stretched string is fixed at both ends, a transverse progressive wave travelling along the string is reflected at the fixed ends. The incident and reflected waves travel in opposite directions and superpose with each other, producing a stationary wave.

Since both ends of the string are fixed, the displacement of the particles at the two ends must always be zero. Therefore, both ends of the string necessarily become nodes of the stationary wave.

Conditions for Formation of Stationary Waves

Consider a stretched string of length L, fixed at both ends. Let the wavelength of the waves travelling along the string be λ.

For a stationary wave to be formed on the string, the two fixed ends must be nodes. Therefore, the length of the string must contain an integral number of half-wavelengths.

\[ L=n\frac{\lambda}{2} \]

where n = 1, 2, 3, … represents the harmonic number.

Therefore, the allowed wavelengths are:

\[ \boxed{\lambda_n=\frac{2L}{n}} \]

The corresponding frequencies are called the natural frequencies or harmonics of the stretched string.

Fundamental Mode or First Harmonic

The simplest mode of vibration is obtained when there is one loop between the two fixed ends. There are nodes at both ends and one antinode at the centre of the string.

N NAFirst harmonic (n = 1)

For the fundamental mode, the length of the string contains one-half of the wavelength.

\[ L=\frac{\lambda_1}{2} \]

Therefore:

\[ \boxed{\lambda_1=2L} \]

The frequency of a wave is related to its velocity and wavelength by:

\[ f=\frac{v}{\lambda} \]

Therefore, the fundamental frequency is:

\[ f_1=\frac{v}{2L} \]

For a stretched string, the transverse wave velocity is:

\[ v=\sqrt{\frac{T}{\mu}} \]

Substituting the wave velocity:

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

This is called the fundamental frequency or first harmonic frequency of the stretched string.

Second Harmonic

In the second harmonic, the string vibrates with two loops. There are three nodes, including the two fixed ends, and two antinodes between them.

N N NA ASecond harmonic (n = 2)

For the second harmonic, the length of the string contains one complete wavelength.

\[ L=\lambda_2 \]

Therefore:

\[ \boxed{\lambda_2=L} \]

The frequency of the second harmonic is:

\[ f_2=\frac{v}{\lambda_2} \]

Substituting \(\lambda_2=L\):

\[ \boxed{f_2=\frac{v}{L}} \]

Using the expression for the wave velocity:

\[ \boxed{f_2=\frac{1}{L}\sqrt{\frac{T}{\mu}}} \]

Comparing the second harmonic frequency with the fundamental frequency:

\[ \boxed{f_2=2f_1} \]

Third Harmonic

In the third harmonic, the string vibrates with three loops. There are four nodes, including the two fixed ends, and three antinodes.

N N N NA A AThird harmonic (n = 3)

For the third harmonic, the length of the string contains three half-wavelengths.

\[ L=\frac{3\lambda_3}{2} \]

Therefore:

\[ \boxed{\lambda_3=\frac{2L}{3}} \]

The frequency of the third harmonic is:

\[ f_3=\frac{v}{\lambda_3} \]

Substituting \(\lambda_3=2L/3\):

\[ f_3=\frac{3v}{2L} \]

Using the expression for the wave velocity:

\[ \boxed{f_3=\frac{3}{2L}\sqrt{\frac{T}{\mu}}} \]

Comparing the third harmonic frequency with the fundamental frequency:

\[ \boxed{f_3=3f_1} \]

General Expression for the nth Harmonic

For the nth harmonic, the string contains n loops. The length of the string therefore contains n half-wavelengths.

\[ L=n\frac{\lambda_n}{2} \]

Rearranging:

\[ \boxed{\lambda_n=\frac{2L}{n}} \]

The frequency of the nth harmonic is given by:

\[ f_n=\frac{v}{\lambda_n} \]

Substituting \(\lambda_n=2L/n\):

\[ f_n=\frac{nv}{2L} \]

For a stretched string:

\[ v=\sqrt{\frac{T}{\mu}} \]

Therefore:

\[ \boxed{f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu}}} \]

Since the fundamental frequency is:

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

The frequency of the nth harmonic can therefore be written as:

\[ \boxed{f_n=nf_1} \]

General Pattern of Nodes and Antinodes

N N N N N N N NA A A A A A An loops → n + 1 nodes → n antinodes

For the nth harmonic, there are n loops, n antinodes and n + 1 nodes, including the two fixed ends.

The fundamental mode corresponds to \(n=1\), the second harmonic corresponds to \(n=2\), the third harmonic corresponds to \(n=3\), and so on.

Harmonic Frequencies of a String Fixed at Both Ends

The first few harmonic frequencies are therefore:

\[ \boxed{f_1=\frac{1}{2L}\sqrt{\frac{T}{\mu}}} \]
\[ \boxed{f_2=\frac{2}{2L}\sqrt{\frac{T}{\mu}}} \]
\[ \boxed{f_3=\frac{3}{2L}\sqrt{\frac{T}{\mu}}} \]

Hence, the harmonic frequencies are integral multiples of the fundamental frequency.

\[ \boxed{f_1:f_2:f_3:f_4:\cdots=1:2:3:4:\cdots} \]

Therefore, a string fixed at both ends can vibrate only at certain discrete frequencies called its natural frequencies or harmonics.

The fundamental frequency is the lowest natural frequency of vibration, while the higher harmonics are integral multiples of the fundamental frequency.

Thus, when a string is fixed at both ends, stationary waves are formed only when the length of the string contains an integral number of half-wavelengths. The allowed wavelengths and frequencies are determined by the length of the string, tension and linear mass density.

Laws of Transverse Vibrations of a Stretched String

The frequency of transverse vibrations of a stretched string depends on the length of the string, the tension in the string and the mass per unit length of the string.

For a string fixed at both ends, the fundamental frequency is given by:

\[ f=\frac{1}{2L}\sqrt{\frac{T}{\mu}} \]

where \(f\) is the fundamental frequency, \(L\) is the length of the vibrating string, \(T\) is the tension in the string and \(\mu\) is the mass per unit length of the string.

Fundamental Formula for a Stretched String

The velocity of a transverse wave travelling along a stretched string is:

\[ v=\sqrt{\frac{T}{\mu}} \]

For the fundamental mode of a string fixed at both ends, the wavelength is:

\[ \lambda=2L \]

Using the relation between wave velocity, frequency and wavelength:

\[ v=f\lambda \]

Therefore:

\[ f=\frac{v}{\lambda} \]

Substituting the values of \(v\) and \(\lambda\):

\[ f=\frac{1}{2L}\sqrt{\frac{T}{\mu}} \]

Hence, the fundamental frequency of a stretched string is:

\[ \boxed{f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}} \]

First Law of Transverse Vibrations — Law of Length

The first law states that, when the tension and mass per unit length of a stretched string are kept constant, the frequency of transverse vibration is inversely proportional to the length of the vibrating string.

From the fundamental formula:

\[ f=\frac{1}{2L}\sqrt{\frac{T}{\mu}} \]

For constant tension \(T\) and constant linear mass density \(\mu\), the quantity \(\sqrt{T/\mu}\) remains constant.

\[ f\propto\frac{1}{L} \]

Therefore:

\[ \boxed{f\propto\frac{1}{L}} \]

Thus, if the length of the string is increased, its frequency decreases. If the length is decreased, its frequency increases.

Length = L Length = L/2L decreases f increases

For example, if the length of the string is reduced to half while the tension and linear density remain unchanged:

\[ L’=\frac{L}{2} \]

Then the new frequency is:

\[ f’=\frac{1}{2L’}\sqrt{\frac{T}{\mu}} \]

Substituting \(L’=L/2\):

\[ f’=2f \]

Thus, halving the length doubles the frequency.

Second Law of Transverse Vibrations — Law of Tension

The second law states that, when the length and mass per unit length of a stretched string are kept constant, the frequency of transverse vibration is directly proportional to the square root of the tension in the string.

From the fundamental formula:

\[ f=\frac{1}{2L}\sqrt{\frac{T}{\mu}} \]

For constant length \(L\) and constant linear density \(\mu\), the remaining variable is the tension \(T\).

\[ f\propto\sqrt{T} \]

Therefore:

\[ \boxed{f\propto\sqrt{T}} \]

Thus, increasing the tension increases the frequency, but the frequency does not increase in the same ratio as the tension. It increases as the square root of the tension.

T Lower tension 4T Higher tensionTension increases → frequency increases

For example, if the tension is increased four times:

\[ T’=4T \]

Then:

\[ f’\propto\sqrt{4T} \]

Therefore:

\[ f’=2f \]

Thus, when the tension is increased four times, the frequency becomes twice its original value.

Third Law of Transverse Vibrations — Law of Linear Density

The third law states that, when the length and tension of a stretched string are kept constant, the frequency of transverse vibration is inversely proportional to the square root of the mass per unit length of the string.

The mass per unit length of the string is represented by \(\mu\).

\[ \mu=\frac{m}{L} \]

From the fundamental formula:

\[ f=\frac{1}{2L}\sqrt{\frac{T}{\mu}} \]

For constant length \(L\) and constant tension \(T\):

\[ f\propto\frac{1}{\sqrt{\mu}} \]

Therefore:

\[ \boxed{f\propto\frac{1}{\sqrt{\mu}}} \]

Thus, a heavier string with greater mass per unit length vibrates with a lower frequency, while a lighter string with smaller mass per unit length vibrates with a higher frequency, provided the length and tension remain unchanged.

Large μ Heavier string Small μ Lighter stringμ decreases → frequency increases

For example, if the mass per unit length becomes four times its original value:

\[ \mu’=4\mu \]

Then:

\[ f’\propto\frac{1}{\sqrt{4\mu}} \]

Therefore:

\[ f’=\frac{f}{2} \]

Thus, increasing the linear density four times reduces the frequency to half.

Combined Expression of the Three Laws

The three laws can be combined into a single proportionality relation. From the fundamental formula:

\[ f=\frac{1}{2L}\sqrt{\frac{T}{\mu}} \]

Therefore:

\[ \boxed{f\propto\frac{\sqrt{T}}{L\sqrt{\mu}}} \]

Hence, the frequency increases with the square root of tension and decreases with length and the square root of linear density.

Relation with Mass and Length of the String

Since the linear mass density is mass per unit length:

\[ \mu=\frac{m}{L} \]

Substituting this relation in the fundamental frequency formula:

\[ f=\frac{1}{2L}\sqrt{\frac{T}{m/L}} \]

Therefore:

\[ f=\frac{1}{2L}\sqrt{\frac{TL}{m}} \]

Thus:

\[ \boxed{f=\frac{1}{2}\sqrt{\frac{T}{mL}}} \]

This form shows that, when the total mass of the string is considered, the frequency is inversely proportional to the square root of the product of mass and length.

Effect of Changing the Material of the String

The linear density of a string also depends on its material and cross-sectional area. If the density of the material is \(\rho\) and the cross-sectional area is \(A\), then:

\[ \mu=\rho A \]

Substituting this into the fundamental frequency formula:

\[ f=\frac{1}{2L}\sqrt{\frac{T}{\rho A}} \]

Therefore, for constant length, tension and cross-sectional area:

\[ \boxed{f\propto\frac{1}{\sqrt{\rho}}} \]

Thus, a string made of a denser material has a lower frequency than a string of the same length and cross-sectional area made of a less dense material, when the tension is the same.

Important Results of the Three Laws

Law of Length: When tension and linear density are constant, the frequency is inversely proportional to the length of the string.

\[ \boxed{f\propto\frac{1}{L}} \]

Law of Tension: When length and linear density are constant, the frequency is proportional to the square root of tension.

\[ \boxed{f\propto\sqrt{T}} \]

Law of Linear Density: When length and tension are constant, the frequency is inversely proportional to the square root of linear density.

\[ \boxed{f\propto\frac{1}{\sqrt{\mu}}} \]

Combining all three laws gives:

\[ \boxed{f\propto\frac{\sqrt{T}}{L\sqrt{\mu}}} \]

For a string fixed at both ends, the complete expression for the fundamental frequency is:

\[ \boxed{f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}} \]

These three laws explain how the pitch of a vibrating string can be changed by changing its length, tension or mass per unit length. They form the fundamental basis of the working of musical string instruments such as the violin, guitar and sitar.

Beats and Applications of Beats

When two sound waves of slightly different frequencies travel through the same medium and reach an observer simultaneously, the sound heard alternately becomes loud and faint at regular intervals. This phenomenon is called beats.

Beats are produced because of the periodic reinforcement and cancellation of the two sound waves due to their slightly different frequencies.

Formation of Beats

Consider two sound waves having equal amplitudes and slightly different frequencies \(f_1\) and \(f_2\).

Let their angular frequencies be \(\omega_1\) and \(\omega_2\).

\[ \omega_1=2\pi f_1 \]
\[ \omega_2=2\pi f_2 \]

The two waves may be represented by:

\[ y_1=A\sin\omega_1t \]
\[ y_2=A\sin\omega_2t \]

According to the principle of superposition, the resultant displacement is the algebraic sum of the individual displacements.

\[ y=y_1+y_2 \]
\[ y=A\sin\omega_1t+A\sin\omega_2t \]

Using the trigonometric identity:

\[ \sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right) \]

The resultant displacement becomes:

\[ y=2A\cos\left(\frac{\omega_1-\omega_2}{2}t\right) \sin\left(\frac{\omega_1+\omega_2}{2}t\right) \]

Therefore, the resultant wave has a rapidly oscillating factor and a slowly varying amplitude.

\[ \boxed{A_R=2A\left|\cos\left(\frac{\omega_1-\omega_2}{2}t\right)\right|} \]

Thus, the amplitude of the resultant sound periodically increases and decreases with time. This periodic variation in amplitude produces beats.

Beat Frequency

The time interval between two successive maxima or two successive minima of sound intensity is called the time interval between beats.

Let the angular frequencies of the two waves be \(\omega_1\) and \(\omega_2\).

The amplitude variation is governed by:

\[ \cos\left(\frac{\omega_1-\omega_2}{2}t\right) \]

Successive maxima of amplitude occur when the cosine factor changes from one maximum to the next maximum. Therefore, the angular frequency of the amplitude variation is:

\[ \omega_b=|\omega_1-\omega_2| \]

Since:

\[ \omega=2\pi f \]

We obtain:

\[ 2\pi f_b=2\pi|f_1-f_2| \]

Therefore:

\[ \boxed{f_b=|f_1-f_2|} \]

Thus, the beat frequency is equal to the absolute difference between the frequencies of the two sound waves.

Number of Beats per Second

The number of beats heard per second is called the beat frequency.

\[ \boxed{n_b=|f_1-f_2|} \]

For example, if two sound sources have frequencies \(256\,Hz\) and \(260\,Hz\), the number of beats produced per second is:

\[ n_b=|260-256| \]
\[ \boxed{n_b=4\,Hz} \]

Therefore, four beats are heard every second.

Maximum Loudness in Beats

A maximum in loudness occurs when the two waves arrive at the point in the same phase. Their displacements then reinforce each other.

For two waves of equal amplitude \(A\), the maximum resultant amplitude is:

\[ \boxed{A_{\max}=2A} \]

Therefore, the sound becomes loud when the waves interfere constructively.

Formation of BeatsLoud Faint LoudAmplitude periodically increases and decreases

Minimum Loudness in Beats

A minimum in loudness occurs when the two waves arrive in opposite phases. Their displacements then partially or completely cancel each other.

For two waves of equal amplitude, complete cancellation occurs at the instant of minimum amplitude.

\[ \boxed{A_{\min}=0} \]

Therefore, the sound becomes faint when destructive interference occurs.

Time Interval Between Successive Beats

The time interval between two successive maxima of loudness is called the beat period.

The beat frequency is:

\[ f_b=|f_1-f_2| \]

Since frequency is the reciprocal of time period:

\[ T_b=\frac{1}{f_b} \]

Therefore:

\[ \boxed{T_b=\frac{1}{|f_1-f_2|}} \]

Thus, the smaller the difference between the two frequencies, the longer the interval between successive beats.

Conditions for Hearing Beats

For clearly audible beats to be produced, the two sound waves should have frequencies that are close to each other but not exactly equal.

The two waves should have appreciable amplitudes so that the variations in loudness can be detected by the ear.

The frequencies should be sufficiently close that the beats are distinctly audible. If the frequency difference is too large, the rapid variations in loudness are difficult for the human ear to distinguish as separate beats.

For musical applications, a small difference in frequency is generally preferred for producing clearly distinguishable beats.

Effect of Unequal Amplitudes

If the two waves have unequal amplitudes \(A_1\) and \(A_2\), the resultant amplitude varies between a maximum and a minimum value.

The maximum amplitude is:

\[ \boxed{A_{\max}=A_1+A_2} \]

The minimum amplitude is:

\[ \boxed{A_{\min}=|A_1-A_2|} \]

Complete cancellation occurs only when the two amplitudes are equal.

\[ A_1=A_2 \]

Beats and Intensity

The intensity of a wave is proportional to the square of its amplitude.

\[ I\propto A^2 \]

Therefore, when the resultant amplitude increases, the sound intensity increases and the sound is perceived as louder.

When the resultant amplitude decreases, the sound intensity decreases and the sound is perceived as fainter.

Thus, the periodic variation in amplitude produces a periodic variation in sound intensity, which is perceived as beats.

Applications of Beats

Beats have several important applications in physics and music.

1. Tuning Musical Instruments: Beats are used to tune musical instruments. A vibrating instrument string is compared with a standard frequency source. When the two frequencies are nearly equal, beats are produced.

As the instrument is tuned closer to the correct frequency, the number of beats per second decreases.

When the frequencies become exactly equal, the beats disappear.

\[ \boxed{f_1=f_2\Rightarrow f_b=0} \]

2. Determination of Unknown Frequency: Beats can be used to determine the unknown frequency of a sound source when the frequency of another source is known.

If the known frequency is \(f_1\) and the beat frequency is \(f_b\), then the unknown frequency \(f_2\) can be:

\[ \boxed{f_2=f_1+f_b} \]

or:

\[ \boxed{f_2=f_1-f_b} \]

The correct value is selected depending on whether the unknown frequency is higher or lower than the known frequency.

3. Adjustment of Frequencies: Beats can be used to compare two frequencies and determine whether they are equal or slightly different.

This makes the beat phenomenon useful wherever two frequencies need to be matched accurately.

Beats as a Consequence of Superposition

Beats are a direct consequence of the principle of superposition. When two waves of slightly different frequencies overlap, their individual displacements combine to produce a resultant wave whose amplitude varies periodically with time.

Therefore, the formation of beats demonstrates an important application of wave superposition.

\[ \boxed{f_b=|f_1-f_2|} \]

Important Results

The resultant displacement of two sound waves of equal amplitude and slightly different frequencies is:

\[ \boxed{y=2A\cos\left(\frac{\omega_1-\omega_2}{2}t\right) \sin\left(\frac{\omega_1+\omega_2}{2}t\right)} \]

The beat frequency is:

\[ \boxed{f_b=|f_1-f_2|} \]

The time interval between successive beats is:

\[ \boxed{T_b=\frac{1}{|f_1-f_2|}} \]

For equal-amplitude waves, the maximum resultant amplitude is:

\[ \boxed{A_{\max}=2A} \]

For equal-amplitude waves, the minimum resultant amplitude is:

\[ \boxed{A_{\min}=0} \]

Thus, beats are produced by the superposition of two waves of slightly different frequencies. The periodic variation in loudness provides a useful method for comparing frequencies, tuning musical instruments and determining unknown frequencies.