The Workbook

1. Beat frequency \(= f_1 - f_2 = 260 - 256 = 4\) Hz: four loud moments every second.

2. Time between loud moments \(= \dfrac{1}{4} = 0.25\) s.

3. The ear hears the average, \(\dfrac{256 + 260}{2} = 258\) Hz, rising and falling in loudness \(4\) times a second.

1. \(3\) beats per second means \(|f_B - 512| = 3\): \(f_B = 509\) Hz or \(515\) Hz.

2. Wax adds mass to the prong. A heavier prong vibrates more slowly, so the frequency falls.

3. If \(f_B\) were \(515\) Hz, lowering it would bring it closer to \(512\) Hz and the beats would fall. The beats rose, so \(f_B\) moved away from \(512\) Hz.

   That happens when \(f_B\) was already below \(512\) Hz: \(f_B = 509\) Hz. Check: \(509 - 2 = 507\) gives \(512 - 507 = 5\) beats.

1. Filing removes mass from the prong, so the frequency rises.

   \(X\) was \(396\) Hz or \(404\) Hz. Raising \(396\) Hz would reduce the beats; raising \(404\) Hz increases them. So \(X = 404\) Hz. Check: \(406 - 400 = 6\).

2. Falling beats mean the frequency moved towards \(400\) Hz. Raising it did that, so \(X\) was below \(400\) Hz: \(X = 396\) Hz.

1. \(2\pi f_1 = 512\pi\) gives \(f_1 = 256\) Hz; \(2\pi f_2 = 520\pi\) gives \(f_2 = 260\) Hz.

   \(y_1 + y_2 = 2a\cos\left(\dfrac{520\pi - 512\pi}{2}t\right)\sin\left(\dfrac{520\pi + 512\pi}{2}t\right) = 2a\cos(4\pi t)\sin(516\pi t)\).

2. The fast factor \(\sin(516\pi t)\) has frequency \(\dfrac{516}{2} = 258\) Hz. This is the note heard.

3. The amplitude \(2a\cos(4\pi t)\) has frequency \(2\) Hz. But the ear responds to loudness, which depends on \(A^2\).

   \(A^2 = 4a^2\cos^2(4\pi t)\) is largest twice in every cycle of \(\cos(4\pi t)\), at \(\cos = +1\) and at \(\cos = -1\). So there are \(2 \times 2 = 4\) loud moments per second.

1. The envelope is widest at \(t = 0\), \(0.25\) s and \(0.5\) s. The loud moments are \(0.25\) s apart, so the beat frequency is \(\dfrac{1}{0.25} = 4\) Hz.

2. The other fork differs from \(256\) Hz by \(4\) Hz: \(252\) Hz or \(260\) Hz. The graph alone cannot tell which.

1. \(f_1 = \dfrac{500}{2} = 250\) Hz and \(f_2 = \dfrac{506}{2} = 253\) Hz. Beat frequency \(= 3\) Hz.

2. When the two waves are in phase the amplitudes add: \(5 + 5 = 10\) mm.

3. Loud moments are \(\dfrac{1}{3}\) s apart. A silent moment lies halfway between two loud ones, so the time is \(\dfrac{1}{6}\) s.

The wire is \(252\) Hz or \(260\) Hz. Raising the tension raises the frequency of the wire (\(f \propto \sqrt{T}\)).

The beats became fewer, so the wire's frequency moved towards \(256\) Hz. A rising frequency moves towards \(256\) Hz only if it started below: the wire was \(252\) Hz.

1. \(f = \dfrac{v}{\lambda}\): \(\dfrac{330}{5} = 66\) Hz and \(\dfrac{330}{5.5} = 60\) Hz. Beats \(= 66 - 60 = 6\) per second.

2. \(\dfrac{330}{1} = 330\) Hz and \(\dfrac{330}{1.1} = 300\) Hz. Beat frequency \(= 30\) Hz.

   No. The ear cannot separate more than about \(10\) loud moments per second; the sensation of one sound lasts about \(0.1\) s. At \(30\) Hz the beats blur into a rough continuous sound.

Open pipe: \(f = \dfrac{v}{2L} = \dfrac{330}{2 \times 0.5} = 330\) Hz. The fork is \(325\) Hz or \(335\) Hz.

Warming the air raises the speed of sound, so the pipe's frequency rises above \(330\) Hz. The beats increased, so the pipe moved away from the fork: the fork is below \(330\) Hz. Fork \(= 325\) Hz.

Each fork is \(4\) Hz above the one before. From the first to the last there are \(24\) steps, so \(f_{25} = f_1 + 24 \times 4 = f_1 + 96\).

Also \(f_{25} = 2f_1\). So \(2f_1 = f_1 + 96\), giving \(f_1 = 96\) Hz and \(f_{25} = 192\) Hz.

\(A\) is \(378\) Hz or \(390\) Hz. Wax lowers the frequency of \(A\).

The beats fell, so \(A\) moved towards \(384\) Hz. A falling frequency moves towards \(384\) Hz only from above: \(A = 390\) Hz. Check: \(390 - 4 = 386\) gives \(2\) beats.

1. The intensity peaks every \(0.5\) s, so the beat frequency is \(\dfrac{1}{0.5} = 2\) Hz.

2. The other fork differs by \(2\) Hz: \(398\) Hz or \(402\) Hz.

3. The beat frequency would be \(4\) Hz, so the peaks would come every \(0.25\) s, twice as close together.

1. The sound in the lower path goes down one arm of the U-piece and up the other. Moving the U-piece down by \(x\) lengthens both arms by \(x\).

   So the lower path grows by \(2x\), and the path difference between the two routes changes by \(2x\).

2. Loud (maximum): path difference \(= n\lambda\), the two waves arrive in phase.

   Faint (minimum): path difference \(= (2n + 1)\dfrac{\lambda}{2}\), the two waves arrive in opposite phase.

1. From one minimum to the next the path difference changes by \(\lambda\). So \(2x = \lambda\): \(\lambda = 2 \times 8.25 = 16.5\) cm.

2. \(v = f\lambda = 2000 \times 0.165 = 330\) m/s.

Successive maxima are also \(\lambda\) apart in path difference: \(\lambda = 2 \times 10 = 20\) cm.

\(f = \dfrac{v}{\lambda} = \dfrac{330}{0.2} = 1650\) Hz.

From a maximum to the next minimum the path difference changes by \(\dfrac{\lambda}{2}\). So \(2x = \dfrac{\lambda}{2}\): \(\lambda = 4x = 20\) cm.

\(f = \dfrac{330}{0.2} = 1650\) Hz.

1. \(\lambda = \dfrac{v}{f} = \dfrac{330}{1000} = 0.33\) m \(= 33\) cm. Path difference \(= 116.5 - 100 = 16.5\) cm \(= \dfrac{\lambda}{2}\).

   The waves arrive in opposite phase: a minimum.

2. The nearest maxima are at path differences \(0\) and \(\lambda = 33\) cm, each \(16.5\) cm away from the present value.

   A change of \(16.5\) cm needs a movement of \(x = \dfrac{16.5}{2} = 8.25\) cm, in either direction.

1. Maximum amplitude \(= a + 2a = 3a\); minimum amplitude \(= 2a - a = a\).

   Intensity is proportional to the square of the amplitude: \(\dfrac{I_{\max}}{I_{\min}} = \dfrac{(3a)^2}{a^2} = 9\).

2. The two waves lose different amounts of energy on their two paths, so they reach \(D\) with different amplitudes. Unequal amplitudes cannot cancel completely.

Successive minima are \(\dfrac{\lambda}{2}\) apart in tube movement. Doubling the frequency halves the wavelength.

New distance \(= \dfrac{8.25}{2} = 4.125\) cm.

Path difference at \(P\): \(4 - 3 = 1\) m. The speakers are in phase, so only the path difference matters.

1. Minimum when \(1\) m \(= \dfrac{\lambda}{2}\), so \(\lambda = 2\) m. Lowest frequency \(f = \dfrac{330}{2} = 165\) Hz.

2. Maximum when \(1\) m \(= \lambda\) (the case \(n = 1\); \(n = 0\) would need infinite wavelength). \(f = \dfrac{330}{1} = 330\) Hz.

3. Next minimum when \(1\) m \(= \dfrac{3\lambda}{2}\), so \(\lambda = \dfrac{2}{3}\) m and \(f = 330 \times \dfrac{3}{2} = 495\) Hz.

\(\lambda = \dfrac{330}{330} = 1\) m.

1. Every point on the perpendicular bisector is equally far from both speakers. The path difference is zero everywhere, so the listener hears a steady loud sound.

2. At a distance \(x\) from \(S_1\) the path difference is \(|x - (4 - x)| = |2x - 4|\).

   Maxima at \(|2x - 4| = 0, 1, 2\) m: \(x = 2\) m, \(x = 1.5\) and \(2.5\) m, \(x = 1\) and \(3\) m. The maxima are \(\dfrac{\lambda}{2} = 0.5\) m apart, like the antinodes of a standing wave.

\(\lambda = \dfrac{v}{f} = \dfrac{1320}{2000} = 0.66\) m \(= 66\) cm.

Successive minima need a tube movement of \(\dfrac{\lambda}{2} = 33\) cm. In air the same frequency needed \(8.25\) cm, four times less, because the speed of sound in hydrogen is four times larger.

Moving source, listener at rest: \(f' = f\,\dfrac{v}{v \mp v_s}\), minus for approach.

1. \(f' = 600 \times \dfrac{330}{330 - 30} = 600 \times \dfrac{330}{300} = 660\) Hz.

2. \(f' = 600 \times \dfrac{330}{330 + 30} = 600 \times \dfrac{330}{360} = 550\) Hz.

Moving listener, source at rest: \(f' = f\,\dfrac{v \pm v_o}{v}\), plus for approach.

1. \(f' = 600 \times \dfrac{330 + 33}{330} = 600 \times 1.1 = 660\) Hz.

2. \(f' = 600 \times \dfrac{330 - 33}{330} = 600 \times 0.9 = 540\) Hz.

1. \(\lambda = \dfrac{v}{f} = \dfrac{330}{600} = 0.55\) m.

2. In one period \(T = \dfrac{1}{600}\) s the source moves \(v_sT\), so the crests ahead are squeezed and those behind are stretched.

   Ahead: \(\lambda' = \dfrac{v - v_s}{f} = \dfrac{300}{600} = 0.5\) m. Behind: \(\lambda' = \dfrac{v + v_s}{f} = \dfrac{360}{600} = 0.6\) m.

3. No. The wavelength is fixed by the source and the air. A moving listener simply meets the crests more often (or less often), so the frequency changes although the wavelength does not.

General formula: \(f' = f\,\dfrac{v \pm v_o}{v \mp v_s}\), upper signs for approach.

1. \(f' = 600 \times \dfrac{330 + 33}{330 - 30} = 600 \times \dfrac{363}{300} = 726\) Hz.

2. \(f' = 600 \times \dfrac{330 - 33}{330 + 30} = 600 \times \dfrac{297}{360} = 495\) Hz.

Approaching: \(660 = f\,\dfrac{330}{330 - u}\). Receding: \(550 = f\,\dfrac{330}{330 + u}\).

Divide: \(\dfrac{660}{550} = \dfrac{330 + u}{330 - u} = 1.2\). So \(330 + u = 396 - 1.2u\), \(2.2u = 66\), \(u = 30\) m/s.

Then \(f = 660 \times \dfrac{330 - 30}{330} = 660 \times \dfrac{300}{330} = 600\) Hz.

1. Ahead: \(f' = 600 \times \dfrac{330}{330 - 165} = 600 \times 2 = 1200\) Hz.

   Behind: \(f' = 600 \times \dfrac{330}{330 + 165} = 600 \times \dfrac{2}{3} = 400\) Hz.

2. With \(v_s = v\) the denominator \(v - v_s\) is zero and the formula gives an infinite frequency.

   Physically, the source keeps up with its own wavefronts. All the crests it emits pile up on top of one another into a single strong pulse, a shock wave. The formula no longer applies.

1. Listener behind a receding source: \(f' = f\,\dfrac{v + v_o}{v + v_s} = 600 \times \dfrac{360}{360} = 600\) Hz.

   The distance between source and listener does not change, so the listener receives crests at the same rate as they are sent out.

2. \(f' = f\,\dfrac{v - v_o}{v} = f \times \dfrac{330 - 330}{330} = 0\). The sound moves at the same speed as the listener and never catches up. The listener hears nothing.

1. Wind carries the whole sound pattern with it. Replace \(v\) by \(v + w\) when the wind blows from source to listener.

   \(f' = f\,\dfrac{v + w}{v + w - v_s} = 600 \times \dfrac{340}{340 - 40} = 600 \times \dfrac{340}{300} = 680\) Hz.

2. No. With both at rest, \(f' = f\,\dfrac{v + w}{v + w} = f\). The wind changes the speed and the wavelength of the sound, but every crest still reaches the listener once per period.

1. The wall is a stationary listener and the car is an approaching source: \(f_1 = 600 \times \dfrac{330}{330 - 30} = 660\) Hz.

2. The wall now acts as a stationary source of \(660\) Hz. The driver is a listener approaching it at \(30\) m/s.

   \(f_2 = 660 \times \dfrac{330 + 30}{330} = 660 \times \dfrac{360}{330} = 720\) Hz.

1. Approaching listener: \(f_1 = 550 \times \dfrac{330 + 30}{330} = 550 \times \dfrac{12}{11} = 600\) Hz.

2. The car re-emits \(600\) Hz as a source approaching the siren at \(30\) m/s: \(f_2 = 600 \times \dfrac{330}{330 - 30} = 660\) Hz.

1. Source and listener approach at \(30\) m/s each: \(f' = 600 \times \dfrac{330 + 30}{330 - 30} = 600 \times \dfrac{360}{300} = 720\) Hz.

2. After crossing they move apart: \(f' = 600 \times \dfrac{330 - 30}{330 + 30} = 600 \times \dfrac{300}{360} = 500\) Hz.

1. Direct sound, source receding: \(f_1 = 600 \times \dfrac{330}{360} = 550\) Hz.

   The wall receives \(600 \times \dfrac{330}{300} = 660\) Hz from the approaching source and reflects it unchanged to the stationary listener: \(f_2 = 660\) Hz.

2. Beats \(= 660 - 550 = 110\) per second. No: the ear cannot count more than about \(10\) beats per second, so the listener hears two separate notes.

1. \(v_s = r\omega = 1.5 \times 20 = 30\) m/s.

2. Only the velocity component along the line to \(O\) matters. At \(A\) the whistle moves straight towards \(O\): \(f' = 600 \times \dfrac{330}{300} = 660\) Hz, the highest.

   At \(B\) the whistle moves straight away from \(O\): \(f' = 600 \times \dfrac{330}{360} = 550\) Hz, the lowest. At \(C\) the velocity is perpendicular to the line to \(O\), so the frequency is \(600\) Hz.

3. The whistle's velocity is always perpendicular to the line to the centre. The distance to the centre never changes, so the observer there hears a steady \(600\) Hz.

1. To the right. Each wavefront is a circle centred on the point where the source was when it was emitted. The smallest, newest circle is centred farthest to the right, so the source has been moving right.

2. \(P\). The wavefronts are crowded ahead of the source, so the wavelength is shorter and the frequency higher at \(P\).

3. \(\lambda_P = \dfrac{v - v_s}{f}\) and \(\lambda_Q = \dfrac{v + v_s}{f}\). \(\dfrac{\lambda_P}{\lambda_Q} = \dfrac{330 - 30}{330 + 30} = \dfrac{300}{360} = \dfrac{5}{6}\).

1. Component along \(SP\): \(v_s\cos 60^\circ = 60 \times 0.5 = 30\) m/s, towards \(P\).

2. Use this component as the source speed: \(f' = 600 \times \dfrac{330}{330 - 30} = 660\) Hz.

3. At \(N\) the velocity is perpendicular to \(NP\). The component along the line to \(P\) is zero, so \(f' = 600\) Hz.

Component of the listener's velocity along the line to the source: \(33\cos 60^\circ = 16.5\) m/s, towards the source.

\(f' = 600 \times \dfrac{330 + 16.5}{330} = 600 \times 1.05 = 630\) Hz.

1. Far away and approaching the graph is flat at \(660\) Hz. Far away and receding it is flat at \(550\) Hz. Far away the motion is almost along the line to the observer, so these are the full approach and recession values.

2. \(\dfrac{f_{\max}}{f_{\min}} = \dfrac{v + u}{v - u} = \dfrac{660}{550} = 1.2\). So \(330 + u = 1.2(330 - u)\), giving \(u = 30\) m/s.

3. \(f = f_{\max}\,\dfrac{v - u}{v} = 660 \times \dfrac{300}{330} = 600\) Hz.

   The sound heard at \(t_0\) left the source when it moved perpendicular to the line to the observer, so the true frequency is heard then. The graph passes through \(600\) Hz at \(t_0\): it checks.

1. \(720 = 600 \times \dfrac{330 + v_o}{330}\), so \(330 + v_o = 396\) and \(v_o = 66\) m/s.

2. \(720 = 600 \times \dfrac{330}{330 - v_s}\), so \(330 - v_s = 275\) and \(v_s = 55\) m/s. A smaller source speed gives the same rise.

The source moves away from the listener and the listener moves towards the source.

\(f' = f\,\dfrac{v + v_o}{v + v_s} = 600 \times \dfrac{330 + 33}{330 + 30} = 600 \times \dfrac{363}{360} = 605\) Hz.

The listener gains on the source slowly, so the frequency is only a little above \(600\) Hz.

1. Step 1, car as approaching listener: \(f_1 = f\,\dfrac{v + u}{v}\).

   Step 2, car as approaching source of \(f_1\): \(f' = f_1\,\dfrac{v}{v - u} = f\,\dfrac{v + u}{v}\cdot\dfrac{v}{v - u} = f\,\dfrac{v + u}{v - u}\).

2. \(\dfrac{v + u}{v - u} = \dfrac{1 + u/v}{1 - u/v} \approx \left(1 + \dfrac{u}{v}\right)\left(1 + \dfrac{u}{v}\right) \approx 1 + \dfrac{2u}{v}\) for small \(u/v\).

   So \(f' \approx f + \dfrac{2uf}{v}\), that is, \(\Delta f \approx \dfrac{2uf}{v}\). This is how a speed gun measures the speed of a car.

3. Exact: \(f' = 550 \times \dfrac{360}{300} = 660\) Hz. Approximate: \(\Delta f = \dfrac{2 \times 30 \times 550}{330} = 100\) Hz, so \(f' \approx 650\) Hz.

   The approximation is \(10\) Hz low because \(u/v = 0.09\) is not very small.

1. \(f' = f\,\dfrac{v}{v - 0.1v} = \dfrac{f}{0.9} = 1.111f\). The rise is \(11.1\) percent.

2. \(f' = f\,\dfrac{v + 0.1v}{v} = 1.1f\). The rise is \(10\) percent.

3. The two cases are physically different. A moving source shortens the wavelength in the air; a moving listener meets unchanged wavelengths more often.

   The factors \(\dfrac{1}{1 - 0.1}\) and \(1 + 0.1\) are equal only to first order. The difference grows as the speed approaches the speed of sound.