The Workbook

1. Both wires hang from the same support and are at the same temperature. If the room warms, or the support yields, both wires lengthen equally.

   Only the extra extension of the test wire, caused by the added load, moves the spirit level. So the errors cancel.

2. The extension of the test wire relative to the reference wire. The screw is turned until the spirit level is horizontal again, and the change in the screw reading is the extension.

3. The fixed load keeps the reference wire straight and taut, so that its kinks do not change during the experiment.

\(F = 2 \times 10 = 20\) N, \(A = 0.2\) mm\(^2 = 2 \times 10^{-7}\) m\(^2\), \(\Delta L = 10^{-3}\) m.

\(Y = \dfrac{FL}{A\,\Delta L} = \dfrac{20 \times 2}{2 \times 10^{-7} \times 10^{-3}} = 2 \times 10^{11}\) Pa.

1. The line passes through \(80\) N at \(4\) mm: slope \(= \dfrac{80}{4} = 20\) N/mm \(= 2 \times 10^{4}\) N/m.

2. \(F = \dfrac{YA}{L}\Delta L\), so a graph of \(F\) against \(\Delta L\) is a straight line of slope \(\dfrac{YA}{L}\).

   \(Y = \dfrac{\text{slope} \times L}{A} = \dfrac{2 \times 10^{4} \times 2}{2 \times 10^{-7}} = 2 \times 10^{11}\) Pa.

3. A line drawn through many points averages the random errors of the single readings. A curved graph would also show at once that the elastic limit had been crossed.

1. If the loading and unloading readings agree, the wire has stayed elastic. Averaging the two readings also removes the backlash of the screw.

2. The wire takes a moment to reach its new length, and the added load sets the frame swinging. The student waits for the wire to settle.

\(\Delta L = \dfrac{FL}{AY}\). A long wire (large \(L\)) and a thin wire (small \(A\)) give a larger extension for the same load. A larger extension is easier to measure accurately.

Beyond the elastic limit the wire takes a permanent set and Hooke's law no longer holds, so \(Y\) cannot be found from the readings.

1. The tension at height \(x\) holds the rod below it, of mass \(\rho A x\). So \(T(x) = \rho g A x\).

2. A small element of length \(dx\) at height \(x\) stretches by \(d(\Delta L) = \dfrac{T\,dx}{AY} = \dfrac{\rho g x\,dx}{Y}\).

   \(\Delta L = \displaystyle\int_0^L \dfrac{\rho g x}{Y}\,dx = \dfrac{\rho g L^2}{2Y}\).

3. \(W = \rho g A L\), so \(\rho g = \dfrac{W}{AL}\) and \(\Delta L = \dfrac{WL}{2AY}\).

   A rod with the same \(W\) hung at its end stretches by \(\dfrac{WL}{AY}\), twice as much. The own weight acts on average at half the length.

\(\Delta L = \dfrac{\rho g L^2}{2Y} = \dfrac{8000 \times 10 \times 100}{2 \times 2 \times 10^{11}} = \dfrac{8 \times 10^{6}}{4 \times 10^{11}} = 2 \times 10^{-5}\) m \(= 0.02\) mm.

1. \(a = \dfrac{F}{m}\). The part of length \(x\) from the free end has mass \(\dfrac{mx}{L}\) and is pulled only by the tension at its right face: \(T(x) = \dfrac{mx}{L}a = \dfrac{Fx}{L}\).

2. At \(x = \dfrac{L}{2}\): \(T = \dfrac{F}{2}\). The tension grows from \(0\) at the free end to \(F\) at the pulled end.

3. \(\Delta L = \displaystyle\int_0^L \dfrac{T\,dx}{AY} = \dfrac{F}{ALY}\int_0^L x\,dx = \dfrac{FL}{2AY}\), half the value for a rod stretched by \(F\) at both ends.

1. \(a = \dfrac{30 - 10}{m}\). The tension at distance \(x\) from the \(10\) N end is \(T = 10 + (30 - 10)\dfrac{x}{L}\), growing linearly from \(10\) N to \(30\) N.

   At the middle: \(T = 10 + 10 = 20\) N.

2. The tension varies linearly, so the average tension \(20\) N gives the elongation: \(\Delta L = \dfrac{20 \times 1}{10^{-6} \times 2 \times 10^{11}} = 10^{-4}\) m \(= 0.1\) mm.

   In general \(\Delta L = \dfrac{(F_1 + F_2)L}{2AY}\).

1. Strain \(= \dfrac{\text{stress}}{Y} = \dfrac{2 \times 10^{8}}{2 \times 10^{11}} = 10^{-3}\).

2. \(u = \dfrac{1}{2}\,\text{stress} \times \text{strain} = \dfrac{1}{2} \times 2 \times 10^{8} \times 10^{-3} = 10^{5}\) J/m\(^3\).

3. Volume \(= 1 \times 10^{-6} = 10^{-6}\) m\(^3\). \(U = 10^{5} \times 10^{-6} = 0.1\) J.

1. Work \(= \displaystyle\int F\,dx\), and an integral of \(F\) over \(x\) is the area under the \(F\)-\(x\) graph.

2. Area of the triangle \(= \dfrac{1}{2} \times 200 \times 10^{-3} = 0.1\) J, the same as the strain energy of Question 10 (stress \(2 \times 10^{8}\) Pa on \(1\) mm\(^2\) is \(200\) N).

3. The force is not \(200\) N throughout. It grows from zero to \(200\) N as the wire stretches, so only the average force, \(100\) N, does the work.

\(U = \dfrac{1}{2}F\,\Delta L = \dfrac{F^2 L}{2AY}\). With the same \(F\), \(L\) and \(Y\): \(U \propto \dfrac{1}{A} \propto \dfrac{1}{r^2}\).

\(\dfrac{U_1}{U_2} = \dfrac{(2r)^2}{r^2} = 4\). The thin wire stores four times the energy; it stretches four times as much under the same pull.

\(U = \dfrac{1}{2}k(\Delta L)^2 \propto (\Delta L)^2\). Three times the extension gives \(9\) times the energy: \(0.9\) J.

Start from \(u = \dfrac{1}{2}\,\text{stress} \times \text{strain}\) and use \(\text{stress} = Y \times \text{strain}\).

Replace the stress: \(u = \dfrac{1}{2}Y(\text{strain})^2\). Replace the strain by \(\dfrac{\text{stress}}{Y}\) instead: \(u = \dfrac{(\text{stress})^2}{2Y}\).

1. Free, the rod would lengthen by \(\alpha L\,\Delta T\). The walls push it back by the same amount, a compressive strain of \(\dfrac{\alpha L\,\Delta T}{L} = \alpha\,\Delta T\).

   Stress \(= Y \times \text{strain} = Y\alpha\,\Delta T\). The length \(L\) has cancelled.

2. Stress \(= 2 \times 10^{11} \times 1.2 \times 10^{-5} \times 50 = 1.2 \times 10^{8}\) Pa.

   Force \(= \text{stress} \times A = 1.2 \times 10^{8} \times 2 \times 10^{-4} = 2.4 \times 10^{4}\) N, about \(2.4\) tonnes weight.

The natural length of the hot rod is longer than the gap between the walls. Keeping it at the old length means squeezing it from its new natural length, which is a real strain.

The walls push inward on the rod, so the stress is compressive.

1. \(\Delta L = \alpha L\,\Delta T = 1.2 \times 10^{-5} \times 10 \times 40 = 4.8 \times 10^{-3}\) m \(= 4.8\) mm.

2. The first \(2\) mm of expansion is free. The remaining \(2.8\) mm is prevented: strain \(= \dfrac{2.8 \times 10^{-3}}{10} = 2.8 \times 10^{-4}\).

   Stress \(= Y \times \text{strain} = 2 \times 10^{11} \times 2.8 \times 10^{-4} = 5.6 \times 10^{7}\) Pa.

1. \(F = \dfrac{YA}{L}\Delta L\) has the form \(F = k\,\Delta L\) with \(k = \dfrac{YA}{L}\).

   \(k = \dfrac{2 \times 10^{11} \times 10^{-6}}{2} = 10^{5}\) N/m.

2. \(\Delta L = \dfrac{F}{k} = \dfrac{100}{10^{5}} = 10^{-3}\) m \(= 1\) mm.

\(T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{1}{10^{5}}} = \dfrac{2\pi}{316} \approx 0.02\) s (with \(\pi^2 = 10\), \(2\pi = 2\sqrt{10}\) and \(T = 2\sqrt{10^{-4}} = 0.02\) s exactly).

\(f = \dfrac{1}{T} = 50\) Hz. A stiff wire makes a fast oscillation.

\(k = \dfrac{YA}{L}\), so \(k \propto \dfrac{1}{L}\) and \(k \propto A\).

1. Half the length: \(2k\) each.

2. Side by side the areas add (or: springs in parallel add): \(2k + 2k = 4k\).

3. End to end the length doubles (springs in series): \(\dfrac{k}{2}\).

1. The block falls until the wire stops it. At the lowest point the block is momentarily at rest, so the loss of gravitational energy equals the strain energy: \(mgx = \dfrac{1}{2}kx^2\).

   \(x = \dfrac{2mg}{k} = \dfrac{2 \times 20}{10^{5}} = 4 \times 10^{-4}\) m \(= 0.4\) mm.

2. At rest, \(kx_0 = mg\) gives \(x_0 = \dfrac{20}{10^{5}} = 0.2\) mm. A suddenly applied load stretches the wire twice as far, and the stress is momentarily doubled.