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Mechanical Waves - Class 12 Physics

A wave that is an oscillation of matter and is responsible for the transfer of energy through a medium is Mechanical Waves . Notes and Question Solutions for Class 12 NEB Physics Mechanical wave

A wave that is an oscillation of matter and is responsible for the transfer of energy through a medium.

Topics to be covered on Mechanical Waves:

Speed of Sound wave in Any Medium
Speed of Sound in Solid
Speed of Sound in Liquid
Speed of Sound in Air or Gas
Factors Affecting the speed of Sound in Gases or Air

Velocity of Waves

Sound is a form of energy which produces the sensation of hearing. It propagates through the medium, so it is a mechanical wave. When a sound wave travels through a medium, the particles of the medium vibrate along the direction of wave travel. This produces a series of higher and lower pressure regions called compressions and rarefactions respectively. So, there is a change in volume at different parts of the medium. Therefore, the longitudinal waves can only travel through a medium if the medium travel through a medium if the medium suffers a resistance to change in volume. In other words, the medium should posses volume of elasticity.

Velocity of Waves

Velocity of longitudinal wave on the medium:
$$ V = \sqrt {\frac {E}{\rho }} \text {where}\: E = \text {elasticity of medium} \: \rho = \text {density of medium} $$
Velocity of longitudinal wave on liquid or gas medium:
$$ V = \sqrt {\frac {B}{\rho }} \text {where}\: B = \text {bulk modulus of gas or liquid} \: \rho = \text {density of medium} $$
Velocity of longitudinal wave on solid medium:
$$ V = \sqrt {\frac {Y }{\rho }} \text {where} \: Y = \text {young’s modulus of elasticity } \: \rho = \text {density of medium} $$
Velocity of longitudinal wave on stretched string:
$$ V = \sqrt {\frac {T}{\mu }} \text {where}\: T = \text {tension on string} \: \mu = \text {mass per unit length} $$
Velocity of longitudinal wave on electromagnetic wave:
$$ V = \sqrt {\frac {1}{\mu \epsilon }} \text {where}\: \mu = \text {permeability of medium or free space} \: \rho = \text {permittivity of medium or free space } $$

Velocity of Sound Wave in Liquid and Solid

When a sound wave propagates through a medium, different regions of the medium undergo varying stress and strain. So, the velocity of a sound is governed the modulus of elasticity, E of the medium. It also depends on inertia of the medium and so, on the density, ρ of the medium. It can be shown that the velocity of sound,

$$ V = \sqrt {\frac {E}{\rho }} $$

In a liquid, modulus of elasticity is the bulk modulus of elasticity, B. so the velocity of the sound in liquid

$$ V = \sqrt {\frac {B}{\rho }} $$

When a longitudinal wave propagates in a solid rod or bar, the rod expands sideways slightly when it is compressed longitudinally while a fluid in a pipe with constant cross-section cannot move sideways. Using same method, we can show that the speed of longitudinal wave in rod is given by

$$ V = \sqrt {\frac {Y}{\rho }} $$

Where Y is Young’s modulus and ρ is the density of solid.

Velocity of Sound in a Gas

Dimensional Method

It is observed that the velocity of sound wave in a medium depends on the elasticity of the medium and its density. If v is velocity of the sound, E is the modulus of elasticity and ρ the density of the medium, and then we have,

$$ \begin{align*} v &\propto E^x \rho ^{y} \\ \text {or} \: V &= k E^x \rho ^{y} \dots (i) \end{align*} $$

where k is a proportionality constant, and x and y are indices to be determined.

$$ \begin{align*} \text {The dimension of velocity, v} &= [LT^{-1} ], \\ \text {Dimensional of modulus of elasticity,} \\ \text {E} = [ML^{-1}T^{-2}] \: \text {and} \\ \text {Dimension of density,} \rho = [ML^{-3}] \\ \text {From dimensional analysis,} \\ \text {from equation} \: (i), \text {we get} \\ [LT^{-1}] &= [ML^{-1} T^{-2}]^x. [ML^{-3} ] ^y \\ \text {or,} \: [LT^{-1}] &= [M]^{x+y} . [L] ^{-x-3y}. [T]^{-2x} \dots (ii) \\ \end{align*} For the equation to be correct, the indices of$ [M],\: [L] \text {and} \: [T]$ on both sides of equation (ii) must be equal.\begin{align*} 0 &= x+ y \dots (iii) \\ 1 &= -x-3y \dots (iv) \\ -1 &= -2x \dots (v) \\ \text {Solving these equations, we have} \\ x = 1/2 , y= -1/2. \text {Substituting these values in equation} \: (i), \text {we get} \\ v &= kE^{1/2} \rho ^{-1/2} = K \sqrt {\frac {E}{\rho } } \\ \text {From mathematical analysis,}\: k =1 \: \text {and so the velocity of sound is} \\ v &= K \sqrt {\frac {E}{\rho } }\\ \end{align*} $$

Velocity of Sound in a Gas

Consider a single pulse in which air is compressed, travel from left to right with a velocity v through the air in a long tube as shown in the figure. Let us run along with the pulse at that velocity so that the pulse appears to stand still and air is moving at velocity v through it from right to left.

Let the pressure of the undisturbed air be P and the pressure inside the pulse be $ P + \Delta P $ , where $\Delta P$ is positive due to compression. Consider a slice of air of thickness $\Delta $ and cross-sectional area A moving toward the pulse at velocity v. As this element of air enters the pulse, its front face encounters a region of the higher pressure, which slow its speed $ v + \Delta v$, where $ \Delta v$ \: is negative. This slowing is complete where the back face reaches the pulse, which requires time interval $\Delta t = \Delta x / \Delta v $. During $\Delta t$ the average force on the front face is $ (P + \Delta P)$ A toward the right. Therefore, average net force on the element during $\Delta t$ is

$$ \begin{align*} F &= PA – (P+\Delta P)A = -\Delta PA \dots (i) \\\end{align*} The minus sign indicates that the net force on the air element is directed to the right in figure.The volume of the element is $A\Delta x,$ so mass of the element is \begin{align*} \Delta m &= \rho A\Delta x = \rho Av\: \Delta t \\ \text {Then the average acceleration of the during} \Delta t \: \text {is} \\ a &= \frac {\Delta v}{\Delta t} \text {Newton’s second law of motion, we have} \\ -\Delta PA &= \rho A v\: \Delta t .\: \frac {\Delta v}{\Delta t} \\ \text {or,} \: \rho v^2 &= - {\Delta P}{\Delta v/v} \\ \end{align*} The air that occupies a volume $V = A v\Delta t \:$ outside the pulse is compressed by an amount $ \Delta V = A\Delta v \Delta $ As it enters the pulse. Hence, \begin{align*} \frac {\Delta V}{V} = \frac {A \Delta v\Delta t}{A v\Delta t}\\ &= \frac {\Delta v}{v} \\ \text {Combining these equations, we have} \\ pv^2 &= - {\Delta P}{\Delta v/v} = - {\Delta P}{\Delta V/V} = B \\ \text {or,} \: V = \sqrt {\frac {B}{\rho }} \\ \end{align*} $$

Newton’s Formula and Factors Affecting for Velocity of Sound in a Gas

Newton’s Formula for Velocity of Sound in a Gas

When a sound wave travels through the medium, there alternate rarefactions and compression and rarefactions are produced.

Newton assumed that in compression heat is produced by vibrating particles which transfer into surrounding. The rarefaction heat is lost by particles so heat is taken from surrounding. As a whole heat lost is equal to heat gain i.e. sound wave travels on air by isothermal process.

In isothermal process,

$$ \begin{align*} PV &= \text {constant} \\ \text {Differentiating equation} \: (i)\: \text {we get}, \\ PdV + VdP &= 0 \\ \text {or,} \: PdV & = -VdP \\ \text {or,} \: P &= - V\frac {dP}{dV} \\ \text {or,} \: P &= \frac {dP}{(-dV/V)} = B\\ \therefore B = \text {Bulk modulus of gas air}\\ \therefore B &= \frac {dP}{(-dV/V)} \\ \text {or,} \: P &= B \\\text {where, P} = \text {atmospheric pressure} \\ \text {So, the velocity of sound in air is written by} \\ V &= \sqrt {\frac {E}{\rho }} = \sqrt {\frac {B}{\rho }} \\ \text {or,} \: \sqrt {\frac {P}{\rho }} \: [\therefore P = B]\\ \text {At NTP, Atmospheric pressure (p)} &= 760\: \text {mm of Hg} \\ &= 760\: \text {mm of Hg} \\ &= 1.013\times 10^5 N/m^2 \\ \text {density of air at NTP,} = \rho = 1.293 \: kg/m^3 \\ \text {Velocity of sound in air at NTP becomes,} \\ V &= \sqrt {\frac {1.013 \times 10^5}{1.293}} \\ \therefore V &= 280 \: m/s \\ \end{align*} $$

Laplace Correction for Velocity of Sound

Newton’s formula for velocity is wrong as it gives value 280 m/s which is wrong because velocity of sound calculated by modern experiment is 332 m/s.

After 100 years Laplace corrected the Newton’s assumption for velocity of sound. Newton assumed that the compression and rarefaction is slow process so sound waves propagate through an isothermal process in gas. According to Laplace, the processes of compression and rarefaction occur so rapidly that neither heat is transferred to the surrounding during rarefaction. Thus, the temperature in different region does not remain constant. So, the sound waves in a gas propagate through an adiabatic process. The equation of an adiabatic process is

\begin{align*} PV^{\gamma} &= \text {constant} \dots (iv) \end{align*}

Where γ is the ratio of molar heat capacity of the air at constant pressure to that at constant volume (C_p/C_v =γ). Differentiating equations (iv) on both sides, we get

$$ \begin{align*} V^{\gamma} dP + P(\gamma V^{\gamma - 1}dV) &= 0 \\ \text {Dividing this equation by } V^{\gamma - 1} \: \text {we get} \\ \text {or,} \: VdP + \gamma P\: dV = 0 \\ \text {or,} \: \gamma P &= - \frac {VdP}{dV} \\ \text {or,} \: \gamma P &= - \frac {dP}{dV/V} \\ \text {or,} \: \gamma P &= B \dots (v) \\ \text {On substituting this value of B in equation of the wave,} \\ v &= \sqrt {\frac {B}{\rho}}, \text {we get} \\ v &= \sqrt {\frac {\gamma P}{\rho}} \dots (vi) \\ \text {For air}, \gamma = 1.4 \text {and at NTP, velocity of sound in air is given by} \\ v &= \sqrt {\frac {\gamma P}{\rho}} \\ &= \sqrt {\frac {1.4 \times 1.013 \times 10^5}{1.293}} \\ &= 331.2 ms^{-1} \end{align*} $$

Thus result closely agrees with the experimental value. Thus, Laplace’s formula gives the correct velocity of the sound in air.

Factors Affecting the Velocity of Sound in a Gas

Effect of temperature
The ideal gas equation is given by
$$ \begin{align*} PV &= n RT \\ \text {or,} PV &= \frac {m}{M}RT \: \left ( m = \text {actual mass of gas molecules} \: M = \text {molar mass} \right ) \\ \text {or,} \: \frac {PV}{m} &= \frac {RT}{M} \\ \therefore \frac {P}{\rho } &= \frac {RT}{M} \\ \text {According to Laplace correction} \\ v &= \sqrt {\frac {\gamma P}{\rho }} \\ &= \sqrt {\frac {\gamma RT}{M }} \\ \text {For any gas medium}\\ \gamma , R, M \text {are constant} \\ \therefore v \propto \sqrt {T} \\ \end{align*} $$
For two different gas medium V₁ and V₂ be the velocities of sound at temperature T₁ and T₂ then,
$$ \frac {v_1}{v_2} = \sqrt {\frac {T_1}{T_2}} $$
Hence, velocity of sound is directly proportional to square root of temperature of gas. If the temperature increases then velocity of sound increases.
Effect of pressure
$$\text {Since,}\: v = \sqrt {\frac {\gamma P}{\rho }} = \sqrt {\frac {\gamma RT}{M}} $$
at constant temperature, the velocity of sound is independent to the pressure.
Effect of density
$$ \begin{align*} \text {Since,}\: v = \sqrt {\frac {\gamma P}{\rho }} \\ \text {at constant pressure,} \: v \propto \frac {1}{\sqrt {\rho }} \\ \end{align*} $$
i.e. velocity of sound in inversely proportional to density of a medium.
Effect of humidity
The density of humid air is low than that of dry air. Since, the velocity of sound is inversely proportional to the square root of density. So, a velocity of sound is more humid air than in dry air.
Effect of direction of wind
The velocity of sound increases with the same direction of wind but velocity of sound decreases in opposite direction of the wind.
$$ v = v_s \pm v_w $$
Effect of amplitude, frequency and wavelength
The velocity of sound is independent to the amplitude, frequency and wavelength of around but if amplitude becomes very high then the velocity of sound may change.