The Workbook

1. (a) The wall cannot move. The string pulls up on the wall, and the wall pulls down on the string with an equal force. This downward pull sends back a trough. The reflected pulse is an inverted crest, that is, a trough moving left.

   (b) The ring is free to slide. The pulse lifts the ring to twice the crest height, and the ring then falls back, sending out a crest. The reflected pulse is a crest moving left.

2. Fixed end: the pulse is inverted, a phase change of \(\pi\). Free end: the pulse is not inverted, no phase change.

3. Neither changes. The speed is fixed by the tension and the mass per unit length of the same string. The shape is the same, only turned over in case (a).

The reflected wave travels along \(-x\), so its phase is \((100\pi t + 2\pi x)\). The amplitude stays \(2\) cm.

1. A fixed end inverts the wave: \(y_r = -2\sin(100\pi t + 2\pi x)\). This equals \(2\sin(100\pi t + 2\pi x + \pi)\), a phase change of \(\pi\).

2. A free end does not invert the wave: \(y_r = 2\sin(100\pi t + 2\pi x)\).

3. At \(x = 0\): \(y_i + y_r = 2\sin(100\pi t) - 2\sin(100\pi t) = 0\) for every \(t\).

   The fixed end is a node. The inversion is exactly what keeps the end at rest.

Phase difference \(= \dfrac{2\pi}{\lambda} \times\) path difference \(= \dfrac{2\pi}{4}\,\Delta x = \dfrac{\pi}{2}\,\Delta x\).

1. \(\Delta x = 1\) m: \(\Delta\phi = \dfrac{\pi}{2}\).

2. \(\Delta x = 2\) m \(= \dfrac{\lambda}{2}\): \(\Delta\phi = \pi\). The waves are in opposite phase and cancel (destructive interference).

3. \(\Delta x = 4\) m \(= \lambda\): \(\Delta\phi = 2\pi\). The waves are in phase and add fully (constructive interference).

Rule: a path difference of \(n\lambda\) gives full addition; a path difference of \((2n + 1)\dfrac{\lambda}{2}\) gives cancellation.

Resultant amplitude: \(R = \sqrt{A^2 + A^2 + 2A^2\cos\phi} = 2A\cos\dfrac{\phi}{2}\).

1. \(\phi = 0\): \(R = 2A\cos 0 = 2A\).

2. \(\phi = \dfrac{\pi}{2}\): \(R = 2A\cos\dfrac{\pi}{4} = \sqrt{2}A\).

3. \(\phi = \dfrac{2\pi}{3}\): \(R = 2A\cos\dfrac{\pi}{3} = A\).

4. \(\phi = \pi\): \(R = 2A\cos\dfrac{\pi}{2} = 0\).

\(R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi}\) with \(A_1 = 3\) cm and \(A_2 = 4\) cm.

1. Largest when \(\phi = 0\): \(R = 3 + 4 = 7\) cm. Smallest when \(\phi = \pi\): \(R = 4 - 3 = 1\) cm.

2. \(\phi = \dfrac{\pi}{2}\): \(\cos\phi = 0\), so \(R = \sqrt{9 + 16} = 5\) cm.

3. Intensity is proportional to the square of the amplitude: \(\dfrac{I_{\max}}{I_{\min}} = \left(\dfrac{7}{1}\right)^2 = 49\).

Both waves have the same \(\omega\) and \(k\), so they add like two phasors at \(90^\circ\).

\(R = \sqrt{3^2 + 4^2} = 5\) cm. The resultant leads \(y_1\) by \(\delta\) with \(\tan\delta = \dfrac{4\sin(\pi/2)}{3 + 4\cos(\pi/2)} = \dfrac{4}{3}\), so \(\delta = 53^\circ\).

\(y = 5\sin(\omega t - kx + 53^\circ)\), with \(y\) in centimetres.

1. The peaks close at \(2 + 2 = 4\) m/s. They meet after \(\dfrac{4}{4} = 1\) s.

2. By superposition the displacements add: crest plus equal trough gives zero everywhere. The string is straight.

   The particles are still moving. The energy is entirely kinetic at that instant.

3. Each pulse passes through the other unchanged. Later the crest is far to the right and the trough far to the left, each with its original shape and speed.

1. Use \(\sin C + \sin D = 2\sin\dfrac{C + D}{2}\cos\dfrac{C - D}{2}\).

   \(y = y_1 + y_2 = 2 \times 2\sin(\pi x)\cos(50\pi t) = 4\sin(\pi x)\cos(50\pi t)\).

   The \(x\) part and the \(t\) part are separate factors. Each particle oscillates with its own fixed amplitude \(4\sin(\pi x)\). The shape does not travel. This is a standing wave.

2. Nodes: \(\sin(\pi x) = 0\), so \(\pi x = 0, \pi, 2\pi, \ldots\), that is, \(x = 0, 1, 2, \ldots\) m.

   Antinodes: \(\sin(\pi x) = \pm 1\), so \(x = 0.5, 1.5, 2.5, \ldots\) m.

3. At an antinode the amplitude is \(4\) cm, twice the amplitude of each travelling wave.

   Neighbouring nodes are \(\dfrac{\lambda}{2}\) apart. Here \(k = \pi\), so \(\lambda = 2\) m and the node spacing is \(1\) m.

Compare with \(y = 2A\sin(kx)\cos(\omega t)\): \(2A = 3\) cm, \(k = 0.5\pi\) rad/cm, \(\omega = 100\pi\) rad/s.

1. \(\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{0.5\pi} = 4\) cm. \(f = \dfrac{\omega}{2\pi} = 50\) Hz.

   \(v = f\lambda = 50 \times 4 = 200\) cm/s \(= 2\) m/s.

2. Nodes where \(\sin(0.5\pi x) = 0\): \(x = 0, 2, 4, 6, 8\) cm. Antinodes halfway between: \(x = 1, 3, 5, 7\) cm.

3. Amplitude at \(x\): \(3\sin(0.5\pi x)\). At \(x = \dfrac{1}{3}\) cm: \(3\sin\dfrac{\pi}{6} = 3 \times \dfrac{1}{2} = 1.5\) cm.

1. No. The amplitude is \(2A\sin(kx)\), which depends on the position. Particles at the nodes never move; particles at the antinodes move the most.

2. All particles in one loop have the same sign of \(\sin(kx)\). They reach their extreme positions together and pass through the mean position together: they are in phase.

   In the next loop \(\sin(kx)\) has the opposite sign. Those particles are in opposite phase, a phase difference of \(\pi\).

3. No. A node never moves, so no energy can pass through a node. The energy of each loop stays in that loop and changes between kinetic and potential.

1. Node spacing is \(\dfrac{\lambda}{2}\). So \(\lambda = 2 \times 25 = 50\) cm.

2. \(f = \dfrac{v}{\lambda} = \dfrac{100}{0.5} = 200\) Hz.

3. An antinode lies halfway between two nodes: \(\dfrac{\lambda}{4} = 12.5\) cm from a node.

1. Three loops fill \(1.8\) m. One loop is \(\dfrac{\lambda}{2}\), so \(\dfrac{3\lambda}{2} = 1.8\) m and \(\lambda = 1.2\) m.

   \(f = \dfrac{v}{\lambda} = \dfrac{60}{1.2} = 50\) Hz.

2. Amplitude at \(x\): \(4\sin(kx)\) with \(k = \dfrac{2\pi}{1.2}\) rad/m. At \(x = 0.1\) m: \(kx = \dfrac{2\pi \times 0.1}{1.2} = \dfrac{\pi}{6}\).

   Amplitude at \(Q\): \(4\sin\dfrac{\pi}{6} = 2\) cm.

3. \(P\) and \(Q\) are in the same loop: phase difference \(0\). \(P\) and \(R\) are in neighbouring loops: phase difference \(\pi\).

   When \(P\) is at its highest point, \(Q\) is at its highest point (only \(2\) cm up) and \(R\) is at its lowest point.

1. \(\omega = 50\pi\) rad/s, so \(T = \dfrac{2\pi}{\omega} = 0.04\) s.

   The string is straight when \(\cos(50\pi t) = 0\): \(50\pi t = \dfrac{\pi}{2}, \dfrac{3\pi}{2}\), that is, \(t = 0.01\) s and \(t = 0.03\) s.

   The string is straight twice in every period, at \(\dfrac{T}{4}\) and \(\dfrac{3T}{4}\).

2. At those instants every particle is passing through its mean position with its maximum speed. The energy is entirely kinetic.

1. In the fundamental mode the string is one loop: \(L = \dfrac{\lambda}{2}\), so \(\lambda = 2L = 2\) m.

   \(f_1 = \dfrac{v}{\lambda} = \dfrac{200}{2} = 100\) Hz. In general \(f_1 = \dfrac{v}{2L}\).

2. \(f_n = n f_1\): \(f_2 = 200\) Hz and \(f_3 = 300\) Hz.

1. Three loops: \(\dfrac{3\lambda}{2} = 60\) cm, so \(\lambda = 40\) cm \(= 0.4\) m.

   \(v = f\lambda = 300 \times 0.4 = 120\) m/s.

2. \(f_1 = \dfrac{v}{2L} = \dfrac{120}{2 \times 0.6} = 100\) Hz. Check: \(300 = 3 \times 100\).

3. \(300\) Hz \(= 3f_1\): the third harmonic. Overtones are counted above the fundamental, so this is the second overtone.

1. For a string fixed at both ends, consecutive resonant frequencies differ by \(f_1\).

   \(f_1 = 360 - 300 = 60\) Hz.

2. \(300 = 5 \times 60\) and \(360 = 6 \times 60\): the fifth and the sixth harmonics.

3. \(f_1 = \dfrac{v}{2L}\), so \(L = \dfrac{v}{2f_1} = \dfrac{120}{120} = 1\) m.

\(f_1 = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}\), so \(f_1 \propto \sqrt{T}\) and \(f_1 \propto \dfrac{1}{L}\).

1. \(T \to 4T\): \(f_1 \to \sqrt{4} \times 100 = 200\) Hz.

2. \(L \to L/2\): \(f_1 \to 2 \times 100 = 200\) Hz.

For \(150\) Hz: \(\dfrac{T'}{T} = \left(\dfrac{150}{100}\right)^2 = 2.25\). The tension must become \(2.25\) times.

1. Fourth harmonic: four loops, \(L = 4 \times \dfrac{\lambda}{2}\), so \(\lambda = \dfrac{2L}{4} = 1\) m.

   \(f = \dfrac{v}{\lambda} = \dfrac{80}{1} = 80\) Hz.

2. \(k = \dfrac{2\pi}{\lambda} = 2\pi\) rad/m and \(\omega = 2\pi f = 160\pi\) rad/s. The antinode amplitude is \(0.02\) m.

   \(y = 0.02\sin(2\pi x)\cos(160\pi t)\). Check: \(x = 0\) and \(x = 2\) m give \(\sin(2\pi x) = 0\), nodes at both fixed ends.

3. Nodes every \(\dfrac{\lambda}{2} = 0.5\) m: \(x = 0, 0.5, 1, 1.5, 2\) m.

Compare with \(y = 2A\sin(kx)\cos(\omega t)\): \(k = 3\pi\) rad/m and \(\omega = 600\pi\) rad/s.

1. \(\lambda = \dfrac{2\pi}{k} = \dfrac{2}{3}\) m. Number of loops \(= \dfrac{L}{\lambda/2} = \dfrac{1}{1/3} = 3\). This is the third harmonic.

2. \(v = \dfrac{\omega}{k} = \dfrac{600\pi}{3\pi} = 200\) m/s.

3. \(f = \dfrac{\omega}{2\pi} = 300\) Hz is the third harmonic, so \(f_1 = \dfrac{300}{3} = 100\) Hz. Check: \(\dfrac{v}{2L} = \dfrac{200}{2} = 100\) Hz.

1. The fixed end is a node and the free end is an antinode. The simplest mode is a quarter wavelength: \(L = \dfrac{\lambda}{4}\), so \(\lambda = 4\) m.

   \(f_1 = \dfrac{v}{4L} = \dfrac{200}{4} = 50\) Hz.

2. Only odd multiples are allowed: \(f = (2n - 1)f_1\). The next two are \(3f_1 = 150\) Hz and \(5f_1 = 250\) Hz.

3. A node at one end and an antinode at the other need an odd number of quarter wavelengths: \(L = (2n - 1)\dfrac{\lambda}{4}\).

   An even harmonic would need a whole number of half wavelengths, which puts a node at both ends. The free end cannot be a node.

1. Count the quarter wavelengths from the node at the wall to the antinode at the ring: there are five. \(L = \dfrac{5\lambda}{4}\), so \(\lambda = \dfrac{4 \times 50}{5} = 40\) cm.

2. \(f = \dfrac{v}{\lambda} = \dfrac{80}{0.4} = 200\) Hz.

   \(f_1 = \dfrac{v}{4L} = \dfrac{80}{4 \times 0.5} = 40\) Hz. Check: \(200 = 5 \times 40\).

3. \(200\) Hz \(= 5f_1\): the fifth harmonic. The allowed modes are \(f_1, 3f_1, 5f_1, \ldots\), so this is the second overtone.

1. For one end free, the resonant frequencies are \(f_1, 3f_1, 5f_1, \ldots\). Consecutive ones differ by \(2f_1\).

   \(2f_1 = 250 - 150 = 100\) Hz, so \(f_1 = 50\) Hz.

2. \(150 = 3 \times 50\) and \(250 = 5 \times 50\): the third and the fifth harmonics.

3. \(f_1 = \dfrac{v}{4L}\), so \(L = \dfrac{v}{4f_1} = \dfrac{100}{200} = 0.5\) m.

\(f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}\), from \(f = \dfrac{v}{2L}\) and \(v = \sqrt{T/\mu}\).

Law of length: \(f \propto \dfrac{1}{L}\) when \(T\) and \(\mu\) are constant.

Law of tension: \(f \propto \sqrt{T}\) when \(L\) and \(\mu\) are constant.

Law of mass: \(f \propto \dfrac{1}{\sqrt{\mu}}\) when \(L\) and \(T\) are constant.

1. The wire carries the weight in the pan: \(T = 4 \times 10 = 40\) N.

2. \(\mu = 1\) g/m \(= 0.001\) kg/m. \(v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{40}{0.001}} = \sqrt{40000} = 200\) m/s.

3. \(f = \dfrac{v}{2L} = \dfrac{200}{2 \times 0.5} = 200\) Hz.

1. At the same tension, \(f \propto \dfrac{1}{L}\), so \(fL\) is constant.

   \(L_2 = \dfrac{f_1 L_1}{f_2} = \dfrac{256 \times 40}{320} = 32\) cm.

2. The law of length. A higher frequency needs a shorter wire.

At fixed length, \(f \propto \sqrt{T}\), and \(T = mg\). So \(f \propto \sqrt{m}\).

\(\dfrac{m_2}{m_1} = \left(\dfrac{f_2}{f_1}\right)^2 = \left(\dfrac{300}{200}\right)^2 = \dfrac{9}{4}\), so \(m_2 = 4 \times \dfrac{9}{4} = 9\) kg.

\(\mu = \rho A = \rho \dfrac{\pi d^2}{4}\), so \(\mu \propto d^2\) for the same material.

At the same \(L\) and \(T\), \(f \propto \dfrac{1}{\sqrt{\mu}} \propto \dfrac{1}{d}\).

\(\dfrac{f_1}{f_2} = \dfrac{2d}{d} = 2\). The thinner wire has twice the frequency: the ratio is \(2 : 1\).

1. \(fL\) is constant: \(fL = 200 \times 90 = 18000\) Hz cm.

   \(30\) cm part: \(f = \dfrac{18000}{30} = 600\) Hz. \(60\) cm part: \(f = \dfrac{18000}{60} = 300\) Hz.

2. \(f \propto \dfrac{1}{L}\), so frequencies in the ratio \(3 : 2\) need lengths in the ratio \(2 : 3\).

   \(L_1 = \dfrac{2}{5} \times 90 = 36\) cm and \(L_2 = 54\) cm. Place the bridge \(36\) cm from one end.

1. The fork forces the wire at the fork frequency. At one length the natural frequency of the wire equals the fork frequency.

   This is resonance: the wire vibrates with a large amplitude and throws the rider off.

2. At the middle of the vibrating length. In the fundamental mode the middle is the antinode, where the amplitude is largest.

   Near a bridge the wire is almost at rest (a node), and the rider would stay on even at resonance.

1. In the transverse arrangement the prong moves across the string. Each swing of the prong is one swing of the string end.

   The string vibrates at the fork frequency: \(100\) Hz.

2. In the longitudinal arrangement the prong moves along the string. The string is pulled tight and then slackened once in every fork vibration.

   The string end goes up in one tightening and down in the next. One string vibration takes two fork vibrations, so the string frequency is \(\dfrac{100}{2} = 50\) Hz.

1. Four loops in \(1\) m: \(\lambda = \dfrac{2 \times 1}{4} = 0.5\) m. In the transverse arrangement \(f = 50\) Hz, so \(v = f\lambda = 25\) m/s.

2. \(T = \mu v^2 = 0.008 \times 625 = 5\) N.

3. \(T = mg\), so \(m = \dfrac{5}{10} = 0.5\) kg.

With \(f\), \(L\) and \(\mu\) fixed, \(p^2 T\) is constant (Question 12). From Question 9: \(p^2 T = 16 \times 5 = 80\) N.

1. \(p = 2\): \(T = \dfrac{80}{4} = 20\) N. Fewer loops need a larger tension.

2. \(p = 8\): \(T = \dfrac{80}{64} = 1.25\) N.

In the longitudinal arrangement the string frequency is half the fork frequency: \(f = 25\) Hz.

Four loops again means \(\lambda = 0.5\) m, so \(v = f\lambda = 25 \times 0.5 = 12.5\) m/s.

\(T = \mu v^2 = 0.008 \times 156.25 = 1.25\) N. This is one quarter of the \(5\) N of Question 9, because \(v\) is halved and \(T \propto v^2\).

1. With \(p\) loops, \(L = p\dfrac{\lambda}{2}\), so \(\lambda = \dfrac{2L}{p}\). Also \(v = f\lambda = \sqrt{T/\mu}\).

   \(f \times \dfrac{2L}{p} = \sqrt{\dfrac{T}{\mu}}\). Square both sides: \(\dfrac{4L^2 f^2}{p^2} = \dfrac{T}{\mu}\), so \(p^2 T = 4L^2 f^2 \mu\), a constant.

2. \(p_1^2 T_1 = p_2^2 T_2\): \(9 \times 4 = 4 \times T_2\), so \(T_2 = 9\) N.

1. Three loops in \(90\) cm: \(\lambda = \dfrac{2 \times 90}{3} = 60\) cm. \(v = f\lambda = 60 \times 0.6 = 36\) m/s.

2. \(p^2 T\) is constant, so \(p \propto \dfrac{1}{\sqrt{T}}\). Nine times the tension gives \(p = \dfrac{3}{\sqrt{9}} = 1\) loop.

3. In the longitudinal position the string frequency is halved, so \(v = f\lambda\) is halved for the same \(\lambda\).

   \(T \propto v^2\), so the tension needed is \(\dfrac{T}{4}\).

1. \(A_r = \dfrac{v_2 - v_1}{v_2 + v_1}A_i = \dfrac{10 - 30}{10 + 30} \times 4 = -2\) mm.

   \(A_t = \dfrac{2v_2}{v_1 + v_2}A_i = \dfrac{20}{40} \times 4 = 2\) mm.

   The minus sign means the reflected wave is inverted. A wave reflected from a heavier string always inverts.

2. \(v = \sqrt{T/\mu}\) with the same \(T\): \(\dfrac{\mu_2}{\mu_1} = \left(\dfrac{v_1}{v_2}\right)^2 = \left(\dfrac{30}{10}\right)^2 = 9\).

1. \(A_r = \dfrac{30 - 10}{30 + 10} \times 4 = +2\) mm. The reflected wave is not inverted: reflection from a lighter string keeps the sign.

   \(A_t = \dfrac{2 \times 30}{10 + 30} \times 4 = 6\) mm.

2. Yes. Power depends on the string as well as on the amplitude: \(P = \dfrac{1}{2}\mu\omega^2A^2v = \dfrac{T\omega^2}{2}\cdot\dfrac{A^2}{v}\) at equal tension.

   The light string carries a larger amplitude with a smaller \(\mu\). Check: \(\dfrac{P_r}{P_i} = \dfrac{2^2}{4^2} = \dfrac{1}{4}\) and \(\dfrac{P_t}{P_i} = \dfrac{6^2}{4^2} \times \dfrac{v_1}{v_2} = \dfrac{36}{16} \times \dfrac{1}{3} = \dfrac{3}{4}\). Sum \(= 1\).

At equal tension and frequency, \(P \propto \dfrac{A^2}{v}\) (see Question 15).

1. Reflected fraction \(= \dfrac{A_r^2}{A_i^2} = \left(\dfrac{v_2 - v_1}{v_2 + v_1}\right)^2 = \left(\dfrac{-20}{40}\right)^2 = \dfrac{1}{4}\).

2. Transmitted fraction \(= \dfrac{A_t^2}{A_i^2} \times \dfrac{v_1}{v_2} = \dfrac{4v_2^2}{(v_1 + v_2)^2} \times \dfrac{v_1}{v_2} = \dfrac{4v_1v_2}{(v_1 + v_2)^2} = \dfrac{4 \times 30 \times 10}{40^2} = \dfrac{3}{4}\).

\(\dfrac{1}{4} + \dfrac{3}{4} = 1\): the incident power is shared between the two waves, none is lost.

1. \(\lambda = \dfrac{v}{f}\). String 1: \(\dfrac{30}{50} = 0.6\) m. String 2: \(\dfrac{10}{50} = 0.2\) m.

2. No. The junction moves at the frequency of the incoming wave, and drives both the reflected and the transmitted wave at that frequency.

3. The reflected wave from a heavier string is inverted: phase change \(\pi\). The transmitted wave is never inverted: phase change \(0\).

1. Light to heavy is like meeting a nearly fixed end. The reflected pulse is a trough (inverted) and is smaller than the incident crest.

2. The transmitted pulse is a crest, smaller than the incident pulse. The heavy string has a smaller wave speed, so the pulse moves more slowly.

   The pulse takes the same time to pass any point, so a slower pulse is narrower.

3. Heavy to light is like meeting a nearly free end. The reflected pulse is a crest (not inverted) and smaller than the incident pulse.

   The transmitted pulse is a crest, larger than the incident pulse, faster and wider.

\(v_2 = \sqrt{T/\mu_2}\).

1. \(\mu_2 \to \infty\) gives \(v_2 \to 0\). Then \(A_r = \dfrac{0 - v_1}{0 + v_1}A_i = -A_i\) and \(A_t = \dfrac{0}{v_1}A_i = 0\).

   The wave is fully reflected and inverted, as at a fixed end.

2. \(\mu_2 \to 0\) gives \(v_2 \to \infty\). Divide numerator and denominator by \(v_2\): \(A_r = \dfrac{1 - v_1/v_2}{1 + v_1/v_2}A_i \to A_i\) and \(A_t = \dfrac{2}{v_1/v_2 + 1}A_i \to 2A_i\).

   The wave is fully reflected without inversion, and the junction moves with twice the amplitude, as at a free end.

1. \(\dfrac{v_1}{v_2} = \sqrt{\dfrac{\mu_2}{\mu_1}} = \sqrt{4} = 2\).

2. Put \(v_1 = 2v_2\): \(A_r = \dfrac{v_2 - 2v_2}{v_2 + 2v_2}A_i = -\dfrac{1}{3} \times 3 = -1\) mm. \(A_t = \dfrac{2v_2}{3v_2}A_i = \dfrac{2}{3} \times 3 = 2\) mm.

3. Reflected fraction \(= \left(\dfrac{A_r}{A_i}\right)^2 = \dfrac{1}{9}\). Transmitted fraction \(= \dfrac{4v_1v_2}{(v_1 + v_2)^2} = \dfrac{4 \times 2}{9} = \dfrac{8}{9}\). Sum \(= 1\).

Inverted and half: \(\dfrac{A_r}{A_i} = \dfrac{v_2 - v_1}{v_2 + v_1} = -\dfrac{1}{2}\).

\(2v_2 - 2v_1 = -v_2 - v_1\), so \(3v_2 = v_1\) and \(\dfrac{v_2}{v_1} = \dfrac{1}{3}\).

\(\dfrac{\mu_2}{\mu_1} = \left(\dfrac{v_1}{v_2}\right)^2 = 9\). String 2 is nine times heavier per metre.

1. \(v_1 = \dfrac{\omega}{k_1} = \dfrac{100\pi}{2\pi} = 50\) m/s. \(\mu_2 = 4\mu_1\) gives \(v_2 = \dfrac{v_1}{2} = 25\) m/s.

2. \(A_r = \dfrac{25 - 50}{25 + 50} \times 3 = -1\) mm. The reflected wave travels along \(-x\) in string 1 with the same \(k_1 = 2\pi\).

   \(y_r = -\sin(100\pi t + 2\pi x)\) mm.

3. \(A_t = \dfrac{2 \times 25}{75} \times 3 = 2\) mm. In string 2, \(k_2 = \dfrac{\omega}{v_2} = \dfrac{100\pi}{25} = 4\pi\) rad/m.

   \(y_t = 2\sin(100\pi t - 4\pi x)\) mm. Check at \(x = 0\): \(y_i + y_r = 3\sin(100\pi t) - \sin(100\pi t) = 2\sin(100\pi t) = y_t\). The string is continuous at the junction.

1. \(\dfrac{A_t}{A_i} = \dfrac{2v_2}{v_1 + v_2} = \dfrac{4}{6} = \dfrac{2}{3}\). So \(6v_2 = 2v_1 + 2v_2\), \(4v_2 = 2v_1\), and \(\dfrac{v_2}{v_1} = \dfrac{1}{2}\).

2. \(\dfrac{\mu_2}{\mu_1} = \left(\dfrac{v_1}{v_2}\right)^2 = 4\).

3. \(A_r = \dfrac{v_2 - v_1}{v_2 + v_1}A_i = \dfrac{1 - 2}{1 + 2} \times 6 = -2\) mm. Check: \(A_i + A_r = 6 - 2 = 4\) mm \(= A_t\).