The Workbook

Compare \(y = 0.05\sin(20\pi t - 4\pi x)\) with the standard form \(y = A\sin(\omega t - kx)\).

1. \(A = 0.05\) m, \(\omega = 20\pi\) rad/s, \(k = 4\pi\) rad/m.

2. Frequency: \(f = \dfrac{\omega}{2\pi} = \dfrac{20\pi}{2\pi} = 10\) Hz.

   Time period: \(T = \dfrac{1}{f} = 0.1\) s.

   Wavelength: \(\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{4\pi} = 0.5\) m.

3. Wave speed: \(v = f\lambda = 10 \times 0.5 = 5\) m/s. (Check: \(v = \omega/k = 20\pi/4\pi = 5\) m/s.)

   The phase is \((\omega t - kx)\). The signs of \(t\) and \(x\) are opposite. So the wave travels along \(+x\).

Compare \(y = 2\sin(0.4\pi x + 100\pi t)\) with \(y = A\sin(kx + \omega t)\).

\(k = 0.4\pi\) rad/cm, so \(\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{0.4\pi} = 5\) cm.

\(\omega = 100\pi\) rad/s, so \(f = \dfrac{\omega}{2\pi} = 50\) Hz.

\(v = f\lambda = 50 \times 5 = 250\) cm/s \(= 2.5\) m/s.

The signs of \(x\) and \(t\) in the phase are the same. So the wave travels along \(-x\).

Rule: if \(x\) and \(t\) have opposite signs in the phase, the wave travels along \(+x\). If they have the same sign, the wave travels along \(-x\).

1. \((kx - \omega t)\): opposite signs. Along \(+x\).

2. \((\omega t + kx)\): same signs. Along \(-x\).

3. \((\omega t - kx)\): opposite signs. Along \(+x\). A cosine is only a sine with a shifted phase; the direction rule is the same.

The standard form \(y = A\sin 2\pi\left(\dfrac{t}{T} - \dfrac{x}{\lambda}\right)\) shows \(T\) and \(\lambda\) directly.

1. \(T = 0.2\) s and \(\lambda = 4\) m.

2. \(v = \dfrac{\lambda}{T} = \dfrac{4}{0.2} = 20\) m/s.

3. Phase at \(x = 1\) m, \(t = 0.1\) s: \(2\pi\left(\dfrac{0.1}{0.2} - \dfrac{1}{4}\right) = 2\pi(0.5 - 0.25) = \dfrac{\pi}{2}\).

   \(y = 0.1\sin\dfrac{\pi}{2} = 0.1\) m. The particle is at its highest point at this moment.

Wavelength: \(\lambda = \dfrac{v}{f} = \dfrac{10}{25} = 0.4\) m.

Wave number: \(k = \dfrac{2\pi}{\lambda} = \dfrac{2\pi}{0.4} = 5\pi\) rad/m.

Angular frequency: \(\omega = 2\pi f = 50\pi\) rad/s.

The wave travels along \(+x\), so the phase is \((\omega t - kx)\). Write \(y = 0.03\sin(50\pi t - 5\pi x)\) with \(y\) in metres.

Check the starting condition. At \(x = 0\), \(t = 0\): \(y = 0.03\sin 0 = 0\). The particle velocity is \(v_p = \dfrac{\partial y}{\partial t} = 0.03 \times 50\pi\cos(0) > 0\). So the particle moves upward. The equation is correct.

\(v = f\lambda = 4 \times 0.5 = 2\) m/s.

\(T = \dfrac{1}{f} = 0.25\) s.

\(k = \dfrac{2\pi}{\lambda} = \dfrac{2\pi}{0.5} = 4\pi\) rad/m.

\(\omega = 2\pi f = 8\pi\) rad/s.

1. Use \(\lambda = \dfrac{v}{f}\) in each string.

   Thin string: \(\lambda_1 = \dfrac{300}{50} = 6\) m. Thick string: \(\lambda_2 = \dfrac{150}{50} = 3\) m.

2. The frequency \(f\) is decided by the source that shakes the string. The frequency stays the same when the wave enters a new string.

   The wave speed \(v\) is decided by the string (its tension and its mass per unit length).

   The wavelength then follows from \(\lambda = v/f\). When \(v\) becomes half, \(\lambda\) also becomes half.

1. Crests are at \(x = 0.5\) m and \(x = 2.5\) m. The distance between neighbouring crests is the wavelength: \(\lambda = 2\) m.

   The highest displacement is \(2\) cm, so \(A = 2\) cm \(= 0.02\) m.

2. \(f = \dfrac{v}{\lambda} = \dfrac{4}{2} = 2\) Hz. \(T = \dfrac{1}{f} = 0.5\) s.

3. \(k = \dfrac{2\pi}{\lambda} = \pi\) rad/m and \(\omega = 2\pi f = 4\pi\) rad/s.

   At \(t = 0\) the figure is \(y = 0.02\sin(\pi x)\): check \(x = 0.5\) m gives \(\sin(\pi/2) = 1\), a crest.

   The wave moves along \(+x\), so the phase is \((kx - \omega t)\). Therefore \(y = 0.02\sin(\pi x - 4\pi t)\), in SI units.

1. The graph reaches \(3\) cm, so \(A = 3\) cm \(= 0.03\) m.

   The graph repeats after \(0.2\) s, so \(T = 0.2\) s.

2. \(f = \dfrac{1}{T} = 5\) Hz. \(\omega = 2\pi f = 10\pi\) rad/s.

   \(\lambda = \dfrac{v}{f} = \dfrac{10}{5} = 2\) m. \(k = \dfrac{2\pi}{\lambda} = \pi\) rad/m.

3. At \(x = 0\) the graph is \(y = 0.03\sin(10\pi t)\): it starts at zero and rises.

   The wave travels along \(+x\), so \(y = 0.03\sin(10\pi t - \pi x)\), in SI units. Put \(x = 0\) to check that the equation gives the graph.

1. Every crest of the dashed curve is to the right of the matching crest of the solid curve. The whole shape has moved towards \(+x\).

   So the wave travels along \(+x\).

2. At \(x = 1\) m the solid curve is at \(y = 0\) and is going down to the right: the slope is negative.

   Slope rule for a wave along \(+x\): \(v_p = -v\dfrac{\partial y}{\partial x}\). A negative slope gives a positive \(v_p\). The particle moves up.

   Check with the figure: at \(x = 1\) m the dashed curve is above the axis. The particle has indeed moved up.

Two points a distance \(\Delta x\) apart differ in phase by \(\Delta\phi = \dfrac{2\pi}{\lambda}\Delta x\). Here \(\lambda = 2\) m, so \(\Delta\phi = \pi\,\Delta x\).

1. \(\Delta x = 0.5\) m: \(\Delta\phi = \dfrac{\pi}{2}\).

2. \(\Delta x = 1\) m: \(\Delta\phi = \pi\). The two particles always move in opposite directions.

3. \(\Delta x = 2\) m \(= \lambda\): \(\Delta\phi = 2\pi\). The two particles are in phase; they move exactly together.

The phase of one particle changes by \(\Delta\phi = \omega\,\Delta t = 2\pi f\,\Delta t\). Here \(f = 5\) Hz, so \(\Delta\phi = 10\pi\,\Delta t\).

1. \(\Delta t = 0.05\) s: \(\Delta\phi = \dfrac{\pi}{2}\).

2. \(\Delta t = 0.1\) s: \(\Delta\phi = \pi\).

3. \(\Delta t = 0.2\) s \(= T\): \(\Delta\phi = 2\pi\). After one full period the particle returns to the same state.

\(\Delta\phi = \dfrac{2\pi}{\lambda}\Delta x\), so \(\lambda = \dfrac{2\pi\,\Delta x}{\Delta\phi}\).

\(\lambda = \dfrac{2\pi \times 0.10}{\pi/3} = 2 \times 0.10 \times 3 = 0.6\) m.

From \(y = 0.02\sin(100\pi t - 2\pi x)\): \(\omega = 100\pi\) rad/s and \(k = 2\pi\) rad/m.

1. \(\Delta x = 0.75 - 0.25 = 0.5\) m. \(\Delta\phi = k\,\Delta x = 2\pi \times 0.5 = \pi\).

2. \(\Delta\phi = \omega\,\Delta t = 100\pi \times 0.005 = \dfrac{\pi}{2}\).

3. A phase difference of \(\pi\) means the two points are always in opposite states. When one is at a crest, the other is at a trough. They move in opposite directions.

From \(y = 0.05\sin(10\pi t - 2\pi x)\): \(A = 0.05\) m, \(\omega = 10\pi\) rad/s, \(k = 2\pi\) rad/m.

1. \(v_{p,\max} = A\omega = 0.05 \times 10\pi = 0.5\pi \approx 1.57\) m/s.

2. \(a_{p,\max} = A\omega^2 = 0.05 \times (10\pi)^2 = 5\pi^2 \approx 49.3\) m/s\(^2\).

3. \(v = \dfrac{\omega}{k} = \dfrac{10\pi}{2\pi} = 5\) m/s.

The maximum particle speed (\(1.57\) m/s) is smaller than the wave speed (\(5\) m/s). The particles only move up and down a small distance; the wave shape moves along the string.

Particle velocity: \(v_p = \dfrac{\partial y}{\partial t} = 0.05 \times 10\pi\cos(10\pi t - 2\pi x) = 0.5\pi\cos(10\pi t - 2\pi x)\).

At \(x = 0.25\) m, \(t = 0.05\) s the phase is \(10\pi(0.05) - 2\pi(0.25) = 0.5\pi - 0.5\pi = 0\).

\(v_p = 0.5\pi\cos 0 = 0.5\pi \approx 1.57\) m/s. The sign is positive, so the particle moves upward.

Particle acceleration: \(a_p = -\omega^2 y\). Here \(y = 0.05\sin 0 = 0\), so \(a_p = 0\). The particle is passing through its mean position with maximum speed.

1. For a wave along \(+x\): \(v_p = -v\dfrac{\partial y}{\partial x} = -20 \times 0.02 = -0.4\) m/s.

   The negative sign means the particle is moving down.

2. For a wave along \(-x\), \(y = f(x + vt)\), so \(\dfrac{\partial y}{\partial t} = +v\dfrac{\partial y}{\partial x}\).

   \(v_p = +20 \times 0.02 = +0.4\) m/s. The particle is moving up.

Use the slope rule for a wave along \(+x\): \(v_p = -v\dfrac{\partial y}{\partial x}\).

1. \(P\): the string rises to the right, the slope is positive, so \(v_p\) is negative. \(P\) moves down.

   \(Q\): at a crest the slope is zero, so \(v_p = 0\). \(Q\) is momentarily at rest.

   \(R\): the string falls to the right, the slope is negative, so \(v_p\) is positive. \(R\) moves up.

   \(S\): at a trough the slope is zero. \(S\) is momentarily at rest.

2. The speed is largest where the slope is steepest, at the mean position. \(P\) and \(R\) have the maximum speed.

3. For a wave along \(-x\) the sign of \(v_p\) reverses: \(P\) moves up, \(Q\) is at rest, \(R\) moves down, \(S\) is at rest.

1. Maximum particle speed: \(v_{p,\max} = A\omega\). Wave speed: \(v = \dfrac{\omega}{k}\).

   Ratio: \(\dfrac{v_{p,\max}}{v} = \dfrac{A\omega}{\omega/k} = Ak = A \times \dfrac{2\pi}{\lambda} = \dfrac{2\pi A}{\lambda}\).

2. The two speeds are equal when \(\dfrac{2\pi A}{\lambda} = 1\), that is, \(\lambda = 2\pi A\).

   \(\lambda = 2\pi \times 0.5\) cm \(= \pi\) cm \(\approx 3.14\) cm.

1. The displacement depends on \(x\) and \(t\) only through the combination \((x - 3t)\). Any function of \((x - vt)\) is a travelling wave.

2. Compare \((x - 3t)\) with \((x - vt)\): \(v = 3\) cm/s. The sign is minus, so the wave travels along \(+x\).

3. The denominator \((x - 3t)^2 + 1\) is smallest when \(x - 3t = 0\). Then \(y_{\max} = \dfrac{2}{1} = 2\) cm.

   At \(t = 2\) s the maximum is at \(x = 3 \times 2 = 6\) cm.

Take out the factor of \(x\): \(2x + 6t = 2(x + 3t)\).

So \(y = \dfrac{4}{1 + 4(x + 3t)^2}\), a function of \((x + 3t)\).

Compare with \((x + vt)\): the speed is \(3\) m/s and the plus sign means the pulse travels along \(-x\).

A travelling wave must be a function of \((ax \pm bt)\) only, with \(x\) and \(t\) appearing in this single combination.

1. \(A\sin(ax)\cos(bt)\) is a product of a function of \(x\) and a function of \(t\). It is a standing wave, not a travelling wave.

2. \(A\,e^{-(ax - bt)^2}\) depends only on \((ax - bt) = a\left(x - \dfrac{b}{a}t\right)\). It is a travelling wave with speed \(v = \dfrac{b}{a}\) along \(+x\).

3. \(A\sin(ax^2 - bt^2)\) contains \(x^2\) and \(t^2\) separately, not the combination \((ax \pm bt)\). It is not a travelling wave.

A pulse keeps its shape and moves with the wave speed. In \(3\) s the whole pulse shifts by \(2 \times 3 = 6\) m along \(+x\).

1. The peak was at \(x = 1\) m. At \(t = 3\) s the peak is at \(x = 1 + 6 = 7\) m.

2. The pulse occupied \(0 \le x \le 2\) m at \(t = 0\). At \(t = 3\) s the string is disturbed for \(6\) m \(\le x \le 8\) m.

3. For a shape moving along \(+x\) with speed \(v\): \(y(x, t) = f(x - vt) = f(x - 2t)\).

1. The pulse moves along \(-x\) with speed \(5\) m/s, so replace \(x\) by \((x + 5t)\): \(y(x, t) = \dfrac{3}{(x + 5t)^2 + 4}\).

2. At \(x = -10\) m, \(t = 2\) s: \(x + 5t = -10 + 10 = 0\).

   \(y = \dfrac{3}{0 + 4} = 0.75\) m. This is the peak of the pulse, which has reached \(x = -10\) m at this time.