Mechanical Waves

Newton's formula: Newton took the propagation of sound as isothermal, so the bulk modulus is $E = P$ and

$$v = \sqrt{\frac{P}{\rho}} \approx 280 \text{ m/s (air at NTP)}$$

which is about 16% lower than the measured 332 m/s.

Laplace's correction: Compressions and rarefactions are fast and air is a bad conductor of heat, so the process is adiabatic.

1. Adiabatic condition: $PV^\gamma =$ constant.
2. Differentiate: $\gamma PV^{\gamma-1}dV + V^\gamma dP = 0 \Rightarrow \gamma P = -\dfrac{dP}{dV/V} = E_{\text{adiabatic}}$
3. Substitute $E = \gamma P$ in $v = \sqrt{E/\rho}$:

$$\boxed{v = \sqrt{\frac{\gamma P}{\rho}}}$$

With $\gamma = 1.41$ for air, $v \approx 332$ m/s, which matches experiment.

The speed of sound in air, $v = \sqrt{\dfrac{\gamma P}{\rho}} = \sqrt{\dfrac{\gamma RT}{M}}$, depends on:

1. Temperature: $v \propto \sqrt{T}$. Speed increases as temperature rises (about 0.61 m/s per $^\circ$C near room temperature).
2. Humidity: moist air is less dense than dry air, so sound travels faster in humid air.
3. Density (nature of the gas): $v \propto \dfrac{1}{\sqrt{\rho}}$ at the same pressure; sound is faster in lighter gases.
4. Wind: sound moving with the wind travels faster ($v + w$) and against it slower ($v - w$).

Pressure has no effect when temperature is constant, because $P/\rho$ stays constant (Boyle's law).

A medium must have:

1. Elasticity, so that compressed or displaced particles return to their original position, and
2. Inertia (mass/density), so that particles can store and pass on energy to the next particles.

(Also, the friction or resistance in the medium should be low.)

(C) decreases

A P-wave is longitudinal, so it can travel in liquids, but its speed $v = \sqrt{E/\rho}$ is smaller there because liquids are much less elastic than solids. The speed does not become zero; it is the S-wave (transverse) that cannot pass through a liquid.

a) Laplace's correction

Newton assumed that sound travels through a gas under isothermal conditions, so he used the isothermal bulk modulus $E = P$:

$$v = \sqrt{\frac{P}{\rho}}$$

For air at NTP ($P = 1.013\times10^5$ Pa, $\rho = 1.293$ kg m$^{-3}$) this gives $v \approx 280$ m/s, but the measured value is about 332 m/s.

Laplace pointed out that compressions and rarefactions are very rapid and air is a poor conductor of heat, so heat cannot flow in or out. Hence the changes are adiabatic, not isothermal.

Corrected formula:

1. For an adiabatic change,

$$PV^\gamma = \text{constant}$$

2. Differentiating,

$$\gamma P V^{\gamma-1}\,dV + V^\gamma\,dP = 0 \;\Rightarrow\; \gamma P = -\frac{dP}{dV/V}$$

3. The right side is the bulk modulus, so the adiabatic bulk modulus is $E = \gamma P$.
4. Putting this in $v = \sqrt{E/\rho}$:

$$\boxed{v = \sqrt{\frac{\gamma P}{\rho}}}$$

For air, $\gamma = 1.41$:

$$v = \sqrt{\frac{1.41 \times 1.013\times10^5}{1.293}} \approx 332 \text{ m/s}$$

which agrees with experiment.

b) Numerical

Given: $v_0 = 330$ m s$^{-1}$ at $T_0 = 273$ K, $T = 25 + 273 = 298$ K, $f = 256$ Hz.

Since $v \propto \sqrt{T}$,

$$v = v_0\sqrt{\frac{T}{T_0}} = 330\sqrt{\frac{298}{273}} = 330 \times 1.0448 = 344.8 \text{ m s}^{-1}$$

Wavelength:

$$\lambda = \frac{v}{f} = \frac{344.8}{256} = 1.35 \text{ m}$$

Wavelength $\approx$ 1.35 m.

(D) Wave produced in the stretched string

In a stretched string the particles vibrate perpendicular to the length of the string (the direction in which the wave travels), so the wave is transverse. Sound in air, sound inside water and the wave in the stem of a tuning fork are all longitudinal.

Speed of sound in a gas (Laplace):

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

Effect of pressure: At constant temperature, Boyle's law gives $PV =$ constant, i.e. $P/\rho =$ constant. If pressure is doubled, density is also doubled, so $v$ does not change. Speed of sound is independent of pressure (at constant temperature).

Effect of temperature: For one mole of gas, $PV = RT$ and $\rho = M/V$, so

$$\frac{P}{\rho} = \frac{RT}{M} \;\Rightarrow\; v = \sqrt{\frac{\gamma RT}{M}} \;\Rightarrow\; v \propto \sqrt{T}$$

So speed increases with temperature: $\dfrac{v_1}{v_2} = \sqrt{\dfrac{T_1}{T_2}}$. Near $0^\circ$C, $v_t \approx v_0 + 0.61\,t$ m/s.

(i) Newton's formula and its discrepancy

Newton assumed that the compressions and rarefactions in sound occur isothermally, so the bulk modulus equals the pressure $P$, giving

$$v = \sqrt{\frac{P}{\rho}}$$

For air at STP, $P = 1.013\times10^5$ Pa and $\rho = 1.293\ \text{kg/m}^3$, so $v \approx 280$ m/s. The experimental value is about $332$ m/s, so the formula disagrees with experiment.

(ii) Laplace's correction

Sound travels so fast and air is such a poor conductor that the changes are rapid and adiabatic, not isothermal. For an adiabatic change $PV^\gamma = \text{constant}$. Differentiating,

$$V^\gamma\,dP + \gamma P V^{\gamma-1}dV = 0 \;\Rightarrow\; \gamma P = -\frac{dP}{dV/V} = B_{\text{ad}}$$

So the adiabatic bulk modulus is $\gamma P$, and

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

For air, $\gamma = 1.4$, which gives $v = \sqrt{1.4}\times280 \approx 331$ m/s, in agreement with experiment.

(iii) Pressure waves

As a sound wave travels in a gas, the particles produce alternate compressions (pressure above normal) and rarefactions (pressure below normal) at every point. The wave is thus a propagation of pressure variation, so it is called a pressure wave.

(a) Newton assumed that the compressions and rarefactions in a sound wave occur slowly enough that the temperature stays constant (an isothermal process), so the bulk modulus is $E = P$ and

$v = \sqrt{\frac{P}{\rho}} \approx 280\ \text{m/s at } 0^\circ\text{C}$

This disagrees with the measured value of about 332 m/s.

Laplace argued that air is a poor conductor of heat and the changes in a sound wave are very rapid, so there is no time for heat to flow between the compressed and rarefied regions. The process is therefore adiabatic, for which the bulk modulus is $E = \gamma P$, where $\gamma = C_p/C_v = 1.4$ for air. The corrected formula is

$v = \sqrt{\frac{\gamma P}{\rho}} \approx 332\ \text{m/s}$

which agrees with experiment.

(b) The pressure amplitude is $\Delta p_{max} = \rho v\omega s_0$, so

$s_0 = \frac{\Delta p_{max}}{\rho v\,(2\pi f)} = \frac{10}{1.29\times340\times2\pi\times100}$

$s_0 \approx 3.6\times10^{-5}\ \text{m}$

(C) wavelength

Frequency is fixed by the source, but the speed of sound changes with temperature ($v \propto \sqrt{T}$). Since $\lambda = v/f$, the wavelength changes.