The Workbook

1. The sound travels to the cliff and back, a distance \(2d\), in \(2\) s. \(2d = 340 \times 2 = 680\) m, so \(d = 340\) m.

2. Each echo travels \(2 \times 17 = 34\) m and takes \(\dfrac{34}{340} = 0.1\) s.

   The echo arrives with the next clap when the claps are \(0.1\) s apart: \(10\) claps per second.

\(\lambda = \dfrac{v}{f}\). Longest: \(\dfrac{340}{20} = 17\) m. Shortest: \(\dfrac{340}{20000} = 0.017\) m \(= 1.7\) cm.

Longitudinal. The air particles move back and forth along the direction in which the sound travels, making compressions and rarefactions.

Air has no rigidity, so it cannot support a sideways (shear) displacement. A transverse wave needs rigidity; a longitudinal wave needs only a bulk modulus.

Sound cannot travel through a vacuum. There are no particles to compress.

1. Compare with \(s = s_0\sin(\omega t - kx)\): \(s_0 = 4 \times 10^{-6}\) m \(= 4\ \mu\)m, \(\omega = 680\pi\) rad/s, \(k = 2\pi\) rad/m.

   \(f = \dfrac{\omega}{2\pi} = 340\) Hz. \(\lambda = \dfrac{2\pi}{k} = 1\) m. \(v = f\lambda = 340\) m/s.

2. \(\Delta P_0 = Bks_0 = 1.4 \times 10^{5} \times 2\pi \times 4 \times 10^{-6} = 1.12\pi \approx 3.5\) Pa.

   A displacement of a few micrometres gives a pressure swing of a few pascals, far smaller than the atmospheric pressure of \(10^{5}\) Pa.

\(k = \dfrac{2\pi}{\lambda} = 2\pi\) rad/m. From \(\Delta P_0 = Bks_0\): \(s_0 = \dfrac{\Delta P_0}{Bk} = \dfrac{8.8}{1.4 \times 10^{5} \times 2 \times \frac{22}{7}}\).

\(1.4 \times 2 \times \dfrac{22}{7} = 8.8\), so \(s_0 = \dfrac{8.8}{8.8 \times 10^{5}} = 10^{-5}\) m.

The pressure change is \(\Delta P = -B\dfrac{\partial s}{\partial x}\). Pressure is highest where the slope of the \(s\)-\(x\) curve is most negative.

1. At \(A\) the slope is negative. Particles to the left of \(A\) are shifted to the right, and particles to the right of \(A\) are shifted to the left. Both move towards \(A\), so the air at \(A\) is most compressed.

2. At \(B\) the slope is positive. Particles on both sides move away from \(B\), so the air at \(B\) is most rarefied.

3. At \(C\) the slope is zero, so \(\Delta P = 0\). The particles at \(C\) are displaced the most, but their neighbours are displaced by the same amount. So the air there is neither compressed nor rarefied.

1. The closed end is a wall. The air there cannot move: a displacement node. The air is squeezed against the wall and released, so the pressure swing is largest: a pressure antinode.

2. The open end is free to move: a displacement antinode. The pressure there stays at the outside pressure: a pressure node.

3. The open end is in contact with the open air. Any extra pressure is released at once into the surroundings, so the pressure there cannot rise or fall.

1. \(v = \sqrt{\dfrac{P}{\rho}} = \sqrt{\dfrac{1.0 \times 10^{5}}{1.25}} = \sqrt{80000} \approx 283\) m/s.

2. \(v = \sqrt{\dfrac{\gamma P}{\rho}} = \sqrt{1.4 \times 80000} = \sqrt{112000} \approx 335\) m/s.

3. Newton assumed that the compressions and rarefactions are isothermal. They are far too quick for heat to flow in or out, so the changes are adiabatic.

   The adiabatic bulk modulus is \(\gamma P\), not \(P\). The measured speed, about \(332\) m/s at \(0^\circ\)C, agrees with Laplace's formula.

\(v = \sqrt{\gamma RT/M}\), so \(v \propto \sqrt{T}\) for the same gas.

1. \(v = 330 \times \sqrt{\dfrac{1200}{300}} = 330 \times 2 = 660\) m/s.

2. \(\dfrac{T}{300} = \left(\dfrac{495}{330}\right)^2 = 1.5^2 = 2.25\), so \(T = 675\) K (about \(402^\circ\)C).

\(v \propto \sqrt{T}\) with \(T = 273 + t\): \(v = v_0\sqrt{\dfrac{273 + t}{273}} = v_0\left(1 + \dfrac{t}{273}\right)^{1/2}\).

For small \(t\), \((1 + x)^{1/2} \approx 1 + \dfrac{x}{2}\). So \(v \approx v_0\left(1 + \dfrac{t}{546}\right)\).

Rise per degree: \(\dfrac{v_0}{546} = \dfrac{332}{546} \approx 0.61\) m/s. The speed of sound rises by about \(0.6\) m/s for every degree of warming.

No change. At a fixed temperature, Boyle's law gives \(\rho \propto P\): doubling the pressure doubles the density.

The ratio \(P/\rho\) stays the same, so \(v = \sqrt{\gamma P/\rho}\) stays the same. The speed of sound depends on the temperature, not on the pressure.

\(v = \sqrt{\gamma RT/M}\). With the same \(\gamma\) and \(T\): \(v \propto \dfrac{1}{\sqrt{M}}\).

\(\dfrac{v_H}{v_O} = \sqrt{\dfrac{M_O}{M_H}} = \sqrt{\dfrac{32}{2}} = 4\). So \(v_H = 4 \times 330 = 1320\) m/s.

Moist air. A water molecule (\(M = 18\) g/mol) is lighter than the nitrogen and oxygen molecules it replaces (\(28\) and \(32\) g/mol).

Moist air therefore has a smaller average molar mass and a smaller density at the same pressure. \(v \propto 1/\sqrt{M}\), so sound travels a little faster in moist air.

Convert the molar mass: \(M = 32\) g/mol \(= 0.032\) kg/mol.

\(v = \sqrt{\dfrac{1.4 \times 8.3 \times 300}{0.032}} = \sqrt{\dfrac{3486}{0.032}} = \sqrt{108937} \approx 330\) m/s.

1. For a point source, \(I = \dfrac{P}{4\pi r^2}\), so \(I \propto \dfrac{1}{r^2}\).

   \(\dfrac{I_P}{I_Q} = \left(\dfrac{r_Q}{r_P}\right)^2 = \left(\dfrac{6}{2}\right)^2 = 9\).

2. \(I_P = \dfrac{10}{4\pi \times 2^2} = \dfrac{10}{16\pi} = \dfrac{5}{8\pi} \approx 0.2\) W/m\(^2\).

3. The power spreads over a sphere of area \(4\pi r^2\). The same energy per second crosses a larger and larger area, so the energy per square metre falls.

\(L = 10\log_{10}\dfrac{I}{I_0}\) with \(I_0 = 10^{-12}\) W/m\(^2\).

1. \(L = 10\log 10^{6} = 60\) dB.

2. \(L = 10\log 10^{3} = 30\) dB.

3. \(L = 10\log 10^{12} = 120\) dB. This is the threshold of pain.

From \(L = 10\log(I/I_0)\): \(I = I_0 \times 10^{L/10}\).

1. \(I = 10^{-12} \times 10^{8} = 10^{-4}\) W/m\(^2\).

2. \(I = 10^{-12} \times 10^{2} = 10^{-10}\) W/m\(^2\).

3. \(I = 10^{-12} \times 10^{0} = 10^{-12}\) W/m\(^2\), the threshold itself.

If \(I\) becomes \(nI\), the level rises by \(10\log n\). The change does not depend on the starting level.

1. \(10\log 2 = 3\) dB. New level \(53\) dB.

2. \(10\log 10 = 10\) dB. New level \(60\) dB.

3. \(10\log 100 = 20\) dB. New level \(70\) dB.

1. Point source: \(I \propto \dfrac{1}{r^2}\). From \(10\) m to \(100\) m the intensity falls by a factor of \(100\).

   The level falls by \(10\log 100 = 20\) dB: \(60 - 20 = 40\) dB.

2. Line source: the power spreads over a cylinder of area \(2\pi r \ell\), so \(I \propto \dfrac{1}{r}\). The intensity falls by a factor of \(10\).

   The level falls by \(10\log 10 = 10\) dB: \(60 - 10 = 50\) dB. The road is still loud at \(100\) m, while the point source has become quiet.

\(I = \dfrac{1}{2}\rho v\omega^2 s_0^2\), so \(I \propto s_0^2\).

1. Doubling \(s_0\) makes \(I\) four times: the level rises by \(10\log 4 = 20\log 2 = 6\) dB. New level \(76\) dB.

2. Halving \(s_0\) makes \(I\) one quarter: the level falls by \(6\) dB. New level \(64\) dB.

1. \(\rho v = 1.25 \times 320 = 400\) kg/(m\(^2\) s). \(I = \dfrac{20^2}{2 \times 400} = \dfrac{400}{800} = 0.5\) W/m\(^2\).

2. \(L = 10\log\dfrac{0.5}{10^{-12}} = 10\log(5 \times 10^{11}) = 10(0.7 + 11) = 117\) dB.

   A pressure swing of only \(20\) Pa, one five-thousandth of atmospheric pressure, is already painfully loud.

\(\omega = 2\pi \times 1000 = 2000\pi\) rad/s and \(\rho v = 400\) kg/(m\(^2\) s).

\(s_0^2 = \dfrac{2I}{\rho v\omega^2} = \dfrac{2 \times 10^{-12}}{400 \times (2000\pi)^2} = \dfrac{2 \times 10^{-12}}{1.6 \times 10^{9}\pi^2} = \dfrac{1.25 \times 10^{-21}}{\pi^2}\).

\(s_0 = \dfrac{\sqrt{1.25 \times 10^{-21}}}{\pi} = \dfrac{3.5 \times 10^{-11}}{\pi} \approx 1.1 \times 10^{-11}\) m.

The air particles move by about a tenth of the size of an atom, and the ear still hears the sound.