The Workbook

1. \(\mu = \dfrac{\text{mass}}{\text{length}} = \dfrac{0.1}{2.5} = 0.04\) kg/m.

2. \(v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{16}{0.04}} = \sqrt{400} = 20\) m/s.

3. \(\lambda = \dfrac{v}{f} = \dfrac{20}{50} = 0.4\) m.

Tension is a force: \([T] = \mathrm{MLT^{-2}}\). Mass per unit length: \([\mu] = \mathrm{ML^{-1}}\).

\(\left[\dfrac{T}{\mu}\right] = \dfrac{\mathrm{MLT^{-2}}}{\mathrm{ML^{-1}}} = \mathrm{L^2T^{-2}}\).

The square root gives \(\mathrm{LT^{-1}}\), the dimensions of speed.

Convert first: \(\mu = 16\) g/m \(= 0.016\) kg/m.

\(v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{40}{0.016}} = \sqrt{2500} = 50\) m/s.

\(v = \sqrt{T/\mu}\), so \(v \propto \sqrt{T}\) and \(v \propto 1/\sqrt{\mu}\).

1. \(T \to 4T\): \(v \to \sqrt{4}\,v = 2v\).

2. \(T \to T/4\): \(v \to v/\sqrt{4} = v/2\).

3. \(\mu \to 4\mu\): \(v \to v/\sqrt{4} = v/2\).

\(v \propto \sqrt{T}\), so \(\dfrac{T_2}{T_1} = \left(\dfrac{v_2}{v_1}\right)^2\).

\(T_2 = 45 \times \left(\dfrac{25}{15}\right)^2 = 45 \times \dfrac{25}{9} = 125\) N.

1. The new tension is \(1.21\,T\). The new speed is \(\sqrt{1.21}\,v = 1.1\,v\).

   The wave speed increases by \(10\) percent.

2. \(\lambda = v/f\) with \(f\) unchanged. So the wavelength also increases by \(10\) percent.

1. Wave speed: \(v = \dfrac{40}{2} = 20\) m/s.

   From \(v = \sqrt{T/\mu}\): \(\mu = \dfrac{T}{v^2} = \dfrac{8}{400} = 0.02\) kg/m.

2. Mass \(= \mu L = 0.02 \times 4 = 0.08\) kg \(= 80\) g.

1. \(\omega = 60\pi\) rad/s and \(k = 3\pi\) rad/m. \(v = \dfrac{\omega}{k} = \dfrac{60\pi}{3\pi} = 20\) m/s.

2. \(T = \mu v^2 = 0.01 \times 400 = 4\) N.

1. A length \(\ell\) of wire has volume \(A\ell\) and mass \(\rho A \ell\). So \(\mu = \rho A\).

   \(A = 1\) mm\(^2 = 10^{-6}\) m\(^2\). \(\mu = 8000 \times 10^{-6} = 0.008\) kg/m.

2. \(v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{80}{0.008}} = \sqrt{10000} = 100\) m/s.

3. \(v = \sqrt{\dfrac{T}{\rho A}} = \sqrt{\dfrac{T/A}{\rho}} = \sqrt{\dfrac{\sigma}{\rho}}\), because the stress is \(\sigma = T/A\).

   Check: \(\sigma = 80/10^{-6} = 8 \times 10^{7}\) Pa, and \(\sqrt{8 \times 10^{7}/8000} = 100\) m/s.

1. In string 1 the crests are \(2\) m apart: \(\lambda_1 = 2\) m. In string 2 they are \(1\) m apart: \(\lambda_2 = 1\) m.

   The frequency is the same in both strings, and \(v = f\lambda\). The longer wavelength means the faster wave. String 1 carries the faster wave.

2. \(\dfrac{v_1}{v_2} = \dfrac{f\lambda_1}{f\lambda_2} = \dfrac{2}{1} = 2\).

3. \(v = \sqrt{T/\mu}\) with the same \(T\), so \(\dfrac{v_1}{v_2} = \sqrt{\dfrac{\mu_2}{\mu_1}}\).

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

Wave speed: \(v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{40}{0.1}} = 20\) m/s.

\(\omega = 2\pi f = 10\pi\) rad/s and \(A = 0.1\) m.

\(P = \dfrac{1}{2}\mu\omega^2 A^2 v = \dfrac{1}{2} \times 0.1 \times (10\pi)^2 \times (0.1)^2 \times 20 = \pi^2 \approx 9.87\) W.

\(P = \dfrac{1}{2}\mu\omega^2 A^2 v\), with \(\omega = 2\pi f\) and \(v = \sqrt{T/\mu}\).

1. \(P \propto A^2\). Doubling \(A\) makes the power \(4P\).

2. \(P \propto f^2\). Doubling \(f\) makes the power \(4P\).

3. \(P \propto v \propto \sqrt{T}\). Four times the tension doubles \(v\), so the power is \(2P\).

1. Energy per unit length: \(u = \dfrac{1}{2}\mu\omega^2 A^2 = \dfrac{1}{2} \times 0.1 \times (10\pi)^2 \times (0.1)^2 = 0.05\pi^2 \approx 0.49\) J/m.

2. \(\lambda = \dfrac{v}{f} = \dfrac{20}{5} = 4\) m. Energy in one wavelength: \(E = u\lambda = 0.05\pi^2 \times 4 = 0.2\pi^2 \approx 1.97\) J.

3. One wavelength passes a point in one time period: \(T = \dfrac{1}{f} = 0.2\) s.

   \(\dfrac{E}{T} = \dfrac{0.2\pi^2}{0.2} = \pi^2\) W. This equals the power of Question 11, because \(P = u\,v = u\lambda f\).

1. At \(P\). A particle at the mean position moves with its maximum speed \(A\omega\). So the kinetic energy per unit length is largest at \(P\).

   At \(Q\) and \(R\) the particles are momentarily at rest, so the kinetic energy there is zero.

2. At \(P\). The slope of the string is steepest at the mean position. A steep piece of string is stretched more than a flat piece.

   At a crest or a trough the slope is zero, and the string there is not stretched at all.

3. Zero. At \(Q\) the kinetic energy is zero (the particle is at rest) and the potential energy is zero (the string is unstretched).

   In a travelling wave both energies are largest at the same place, unlike in simple harmonic motion.

\(P = \dfrac{1}{2}\mu\omega^2 A^2 v\) and \(v = \sqrt{T/\mu}\). So \(P = \dfrac{1}{2}\omega^2 A^2\sqrt{\mu T}\).

With the same \(\omega\), \(A\) and \(T\): \(P \propto \sqrt{\mu}\).

\(\dfrac{P_2}{P_1} = \sqrt{\dfrac{\mu_2}{\mu_1}} = \sqrt{4} = 2\).

1. The tension at height \(y\) holds up the rope below that point. The mass below is \(\mu y\), so \(T = \mu g y\).

   At \(y = 2.5\) m: \(T = 0.2 \times 10 \times 2.5 = 5\) N. At the top (\(y = 10\) m): \(T = 0.2 \times 10 \times 10 = 20\) N.

2. \(v = \sqrt{T/\mu}\). At \(y = 2.5\) m: \(v = \sqrt{5/0.2} = \sqrt{25} = 5\) m/s. At the top: \(v = \sqrt{20/0.2} = \sqrt{100} = 10\) m/s.

3. The tension is not the same everywhere. Higher points carry the weight of more rope, so the tension grows with \(y\).

   The pulse speed depends on the tension, so the speed grows with \(y\) as well.

1. \(T = \mu g y\), so \(v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{\mu g y}{\mu}} = \sqrt{gy}\).

   The mass per unit length cancels. A thick rope and a thin rope of the same length give the same pulse speed at the same height.

2. In a small time \(dt\) the pulse climbs \(dy = v\,dt = \sqrt{gy}\,dt\). So \(dt = \dfrac{dy}{\sqrt{gy}}\).

   \(t = \displaystyle\int_0^L \dfrac{dy}{\sqrt{g}\,\sqrt{y}} = \dfrac{1}{\sqrt{g}}\left[2\sqrt{y}\right]_0^L = 2\sqrt{\dfrac{L}{g}}\).

3. \(t = 2\sqrt{\dfrac{10}{10}} = 2\) s.

1. \(\lambda = v/f\). At \(y = 2.5\) m: \(v = 5\) m/s, so \(\lambda = 5/5 = 1\) m. At \(y = 10\) m: \(v = 10\) m/s, so \(\lambda = 10/5 = 2\) m.

2. The frequency is fixed by the source and stays \(5\) Hz all the way up.

   The speed grows with height, so the wavelength grows with height. The crests spread apart as the wave climbs.

\(\mu = \dfrac{3}{7.5} = 0.4\) kg/m.

1. At the lower end the rope holds only the block: \(T = 1 \times 10 = 10\) N.

   At the top the rope holds the block and the whole rope: \(T = (1 + 3) \times 10 = 40\) N.

2. Lower end: \(v = \sqrt{10/0.4} = \sqrt{25} = 5\) m/s. Top: \(v = \sqrt{40/0.4} = \sqrt{100} = 10\) m/s.

From the working of Question 17, the time to climb from \(y_1\) to \(y_2\) is \(t = \dfrac{2}{\sqrt{g}}\left(\sqrt{y_2} - \sqrt{y_1}\right)\).

\(t = \dfrac{2}{\sqrt{10}}\left(\sqrt{10} - \sqrt{2.5}\right) = 2\left(1 - \sqrt{\dfrac{2.5}{10}}\right) = 2(1 - 0.5) = 1\) s.

The pulse takes \(1\) s for the lower \(2.5\) m and only \(1\) s more for the upper \(7.5\) m.

The time to reach height \(y\) from the lower end is \(t = 2\sqrt{y/g}\). So \(y = \dfrac{g t^2}{4}\): the height grows as the square of the time.

The whole rope takes \(t_0 = 2\sqrt{L/g}\). At \(t = t_0/2\): \(y = \dfrac{g}{4} \times \dfrac{t_0^2}{4} = \dfrac{1}{4} \times \dfrac{g t_0^2}{4} = \dfrac{L}{4}\).

Half the time takes the pulse only a quarter of the way up. For the \(10\) m rope this is \(y = 2.5\) m at \(t = 1\) s, as in Question 20.

\(v = \sqrt{gy}\), so \(v^2 = gy\). Differentiate with respect to time: \(2v\dfrac{dv}{dt} = g\dfrac{dy}{dt}\).

But \(\dfrac{dy}{dt} = v\), the pulse speed. So \(2v\,a = g\,v\), which gives \(a = \dfrac{g}{2}\).

The acceleration is constant and does not depend on \(y\) or on the rope. A pulse climbing a hanging rope moves like a body with acceleration \(g/2\).

1. \(v = \sqrt{gy}\). \(P\): \(\sqrt{10 \times 0.4} = 2\) m/s. \(Q\): \(\sqrt{10 \times 1.6} = 4\) m/s. \(R\): \(\sqrt{10 \times 3.6} = 6\) m/s. Top: \(\sqrt{10 \times 6.4} = 8\) m/s.

2. Going down, the pulse passes through smaller tension. The speed falls from \(8\) m/s at the top to zero at the free end. The pulse slows down.

3. The time down equals the time up: \(t = 2\sqrt{\dfrac{L}{g}} = 2\sqrt{\dfrac{6.4}{10}} = 2 \times 0.8 = 1.6\) s.

The pulse from the lower end reaches height \(y\) after \(t_1 = 2\sqrt{y/g}\).

Coming down from the top to \(y\) takes the pulse the same time as climbing from \(y\) to the top. So \(t_2 = 2\sqrt{L/g} - 2\sqrt{y/g}\).

They meet when \(t_1 = t_2\): \(2\sqrt{y/g} = 2\sqrt{L/g} - 2\sqrt{y/g}\), so \(2\sqrt{y} = \sqrt{L}\) and \(y = \dfrac{L}{4} = 2.5\) m.

Time: \(t = 2\sqrt{2.5/10} = 1\) s. The pulses meet \(2.5\) m above the lower end, one second after they start.