The Workbook

1. Open pipe: \(f_1 = \dfrac{v}{2L} = \dfrac{330}{2 \times 0.5} = 330\) Hz.

2. \(f_n = nf_1\): \(f_2 = 660\) Hz and \(f_3 = 990\) Hz.

1. Three nodes: the pipe holds three half wavelengths. \(\dfrac{3\lambda}{2} = 75\) cm, so \(\lambda = 50\) cm.

   \(f = \dfrac{v}{\lambda} = \dfrac{330}{0.5} = 660\) Hz.

2. \(f_1 = \dfrac{v}{2L} = \dfrac{330}{1.5} = 220\) Hz. Check: \(660 = 3 \times 220\).

3. \(660 = 3f_1\): the third harmonic. The overtones are counted above the fundamental, so this is the second overtone.

1. In an open pipe consecutive resonances differ by \(f_1\): \(f_1 = 880 - 660 = 220\) Hz.

2. \(660 = 3 \times 220\) and \(880 = 4 \times 220\): the third and the fourth harmonics.

3. \(L = \dfrac{v}{2f_1} = \dfrac{330}{440} = 0.75\) m.

\(L = \dfrac{v}{2f_1} = \dfrac{330}{660} = 0.5\) m.

The harmonics of an open pipe are \(f_1, 2f_1, 3f_1\): the ratio is \(1 : 2 : 3\).

1. (a) has one node, (b) two, (c) three. The number of nodes is the harmonic number: (a) first, (b) second, (c) third harmonic.

2. Each mode holds \(n\) half wavelengths: \(L = \dfrac{n\lambda}{2}\), so \(\lambda = \dfrac{2L}{n}\).

   (a) \(\lambda = 2L\), (b) \(\lambda = L\), (c) \(\lambda = \dfrac{2L}{3}\).

3. An open end is in contact with the outside air. The air there is free to move, so the displacement is largest: an antinode.

   Antinodes at both ends fit whenever the length is a whole number of half wavelengths, \(L = n\lambda/2\), for every \(n\). So every harmonic is allowed.

\(f_1 = \dfrac{v}{2L} \propto v\) for the same pipe. The speed becomes \(4\) times, so \(f_1 = 4 \times 330 = 1320\) Hz.

1. Closed pipe: \(f_1 = \dfrac{v}{4L} = \dfrac{330}{4 \times 0.25} = 330\) Hz.

2. Only odd harmonics: \(3f_1 = 990\) Hz and \(5f_1 = 1650\) Hz.

1. From the open end (antinode) to the closed end (node) the curve covers three quarter wavelengths: \(L = \dfrac{3\lambda}{4}\).

   \(\lambda = \dfrac{4 \times 75}{3} = 100\) cm \(= 1\) m. \(f = \dfrac{v}{\lambda} = 330\) Hz.

2. \(f_1 = \dfrac{v}{4L} = \dfrac{330}{3} = 110\) Hz. Check: \(330 = 3 \times 110\).

3. \(330 = 3f_1\): the third harmonic. A closed pipe has only odd harmonics, so this is its first overtone.

1. In a closed pipe consecutive resonances differ by \(2f_1\): \(2f_1 = 770 - 550 = 220\) Hz, so \(f_1 = 110\) Hz.

2. \(550 = 5 \times 110\) and \(770 = 7 \times 110\): the fifth and the seventh harmonics.

3. \(L = \dfrac{v}{4f_1} = \dfrac{330}{440} = 0.75\) m.

1. \(\dfrac{f_{\text{open}}}{f_{\text{closed}}} = \dfrac{v/2L}{v/4L} = 2\). The open pipe sounds an octave higher.

2. The open pipe. It gives all the harmonics, \(f_1, 2f_1, 3f_1, \ldots\), while the closed pipe gives only the odd ones, \(f_1, 3f_1, 5f_1, \ldots\).

3. \(\dfrac{v}{4L_c} = \dfrac{v}{2L_o}\), so \(L_o = 2L_c\). The ratio \(L_c : L_o = 1 : 2\).

Closed pipe: second overtone \(= 5f_1 = \dfrac{5v}{4L_c}\). Open pipe: second overtone \(= 3f_1 = \dfrac{3v}{2L_o}\).

Equal frequencies: \(\dfrac{5}{4L_c} = \dfrac{3}{2L_o}\), so \(L_o = \dfrac{6}{5}L_c = \dfrac{6}{5} \times 50 = 60\) cm.

1. Curve 1 is the displacement: it is largest at the open end and zero at the closed wall, where the air cannot move.

   Curve 2 is the pressure: it is zero at the open end (atmospheric pressure) and largest at the closed end. There the air is squeezed against the wall.

2. At the closed end, the pressure antinode.

3. At the open end, the displacement antinode, where the air moves the most.

1. \(e = 0.6r = 0.6 \times 2 = 1.2\) cm.

2. The antinode lies a little outside the open end. Effective length \(= L + e = 23.8 + 1.2 = 25\) cm.

3. \(f_1 = \dfrac{v}{4(L + e)} = \dfrac{330}{4 \times 0.25} = 330\) Hz.

Each open end adds \(e = 0.6 \times 1 = 0.6\) cm. Effective length \(= L + 2e = 48.8 + 1.2 = 50\) cm.

\(f_1 = \dfrac{v}{2(L + 2e)} = \dfrac{330}{2 \times 0.5} = 330\) Hz.

The antinode is a distance \(e\) above the open end. So the effective air columns are \(l_1 + e\) and \(l_2 + e\).

1. First resonance: \(l_1 + e = \dfrac{\lambda}{4}\). Second resonance: \(l_2 + e = \dfrac{3\lambda}{4}\).

   Subtract: \(l_2 - l_1 = \dfrac{\lambda}{2}\), so \(\lambda = 2(l_2 - l_1)\). The unknown \(e\) cancels.

2. From the two equations, \(l_2 + e = 3(l_1 + e)\), so \(l_2 - 3l_1 = 2e\) and \(e = \dfrac{l_2 - 3l_1}{2}\).

Lower. The true air column is \(L + e\), longer than \(L\), so the true fundamental \(\dfrac{v}{4(L + e)}\) is smaller than \(\dfrac{v}{4L}\).

\(e = 0.6r\) grows with the radius. The pipe of radius \(3\) cm needs the larger correction, \(1.8\) cm against \(0.6\) cm.

1. \(\lambda = 2(l_2 - l_1) = 2 \times (74 - 24) = 100\) cm \(= 1\) m.

   \(v = f\lambda = 330 \times 1 = 330\) m/s.

2. \(e = \dfrac{l_2 - 3l_1}{2} = \dfrac{74 - 72}{2} = 1\) cm.

   \(e = 0.6r\), so \(r = \dfrac{1}{0.6} = \dfrac{5}{3} \approx 1.7\) cm.

3. The antinode sits \(e\) above the open end. The quarter and three-quarter wavelengths are \(l_1 + e\) and \(l_2 + e\), and \(l_2 + e = 3(l_1 + e)\) gives \(l_2 = 3l_1 + 2e\), a little more than \(3l_1\).

\(\lambda = 2(l_2 - l_1) = 2 \times (49.5 - 16) = 67\) cm \(= 0.67\) m.

\(v = f\lambda = 512 \times 0.67 \approx 343\) m/s.

\(e = \dfrac{l_2 - 3l_1}{2} = \dfrac{49.5 - 48}{2} = 0.75\) cm.

1. \(l_1 = \dfrac{\lambda}{4}\), so \(\lambda = 100\) cm \(= 1\) m. \(v = f\lambda = 330\) m/s.

2. Resonances at \(\dfrac{\lambda}{4}, \dfrac{3\lambda}{4}, \dfrac{5\lambda}{4}\): the next two are \(75\) cm and \(125\) cm.

1. The air cannot move into the water. The air particles at the surface stay at rest, which is exactly what a closed end does: a displacement node.

2. The open top is an antinode and the water surface is a node. The shortest column with a node at one end and an antinode at the other is a quarter wavelength.

3. At resonance the air column vibrates with a large amplitude and the sound becomes suddenly loud. The student lowers the water until the loudness is greatest.

Resonance lengths are \(\dfrac{\lambda}{4}, \dfrac{3\lambda}{4}, \dfrac{5\lambda}{4}, \ldots\), as long as they fit in the \(100\) cm tube.

1. \(\lambda = \dfrac{330}{330} = 1\) m: lengths \(25\) cm, \(75\) cm (\(125\) cm does not fit). Two resonances.

2. \(\lambda = \dfrac{330}{660} = 0.5\) m: lengths \(12.5\), \(37.5\), \(62.5\), \(87.5\) cm. Four resonances.

\(l_1 = \dfrac{\lambda}{4} = \dfrac{v}{4f}\), so \(v = 4fl_1\).

Cold day: \(v = 4 \times 330 \times 0.25 = 330\) m/s. Hot day: \(v = 4 \times 330 \times 0.26 = 343.2\) m/s.

The speed of sound rises with temperature, so the wavelength of the same fork is longer on the hot day. A longer quarter wavelength needs a longer air column.