A thick climbing rope hangs from the ceiling of a school gymnasium. Its bottom end hangs free just above the floor. Give the bottom end a quick sideways flick. A kink runs up the rope towards the ceiling. Watch closely and the kink is slow at first, then faster and faster as it climbs.
Using the formula
The formula \(v = \sqrt{T/\mu}\) needs two numbers. The tension \(T\) is usually given. The mass per unit length \(\mu\) is found from the mass \(m\) and the length \(L\) of a uniform string, \(\mu = m/L\). Both must be in SI units, kilograms and metres, before they enter the formula.
A guitar string has a length of \(64\,\text{cm}\) and a mass of \(2.0\,\text{g}\). It is stretched with a tension of \(80\,\text{N}\). What is the speed of a wave on this string?
First find the linear mass density \(\mu\), in SI units. Length \(L = 64\,\text{cm} = 0.64\,\text{m}\). Mass \(m = 2.0\,\text{g} = 0.0020\,\text{kg}\). Tension \(T = 80\,\text{N}\).
\(\mu = \dfrac{m}{L} = \dfrac{0.0020\,\text{kg}}{0.64\,\text{m}} = 0.003125\,\text{kg/m}\).
\(v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{80\,\text{N}}{0.003125\,\text{kg/m}}} = \sqrt{25600\,\text{m}^2/\text{s}^2} = 160\,\text{m/s}\).
The speed of the wave on the guitar string is \(160\,\text{m/s}\).
A sitar string \(1.0\,\text{m}\) long has a mass of \(2.5\,\text{g}\). It is stretched with a tension of \(100\,\text{N}\). What is the speed of a wave on it, in m/s?
A rope whose tension changes
So far the tension was the same all along the string. A heavy rope hanging from a support is different. Each piece of the rope holds up the weight of all the rope below it. So the tension is zero at the free bottom end and largest at the support.
Measure the height \(y\) upward from the bottom end. The rope below the height \(y\) has length \(y\), so its mass is \(\mu y\) and its weight is \(\mu g y\). This weight is the tension at that height, \(T(y) = \mu g y\). Here \(g\) is the acceleration due to gravity.
A long, heavy rope of uniform mass per unit length \(\mu\) hangs vertically from a support. The rope has length \(L\) and mass \(M\). You flick the bottom end, and a pulse is made there. (a) How does the pulse travel up the rope? (b) What is the speed of the pulse as it moves upward? (c) What is its acceleration? (d) How long does it take to reach the top?
(a) At a height \(y\) above the bottom, the tension comes from the weight of the rope below that point, \(T(y) = \mu g y\). The tension is smallest at the bottom and grows upward. So the pulse starts slow and speeds up as it climbs.
(b) The speed on a string is \(v = \sqrt{T/\mu}\). With the tension at the height \(y\), \(v(y) = \sqrt{\mu g y / \mu} = \sqrt{g y}\).
(c) By the chain rule, \(a = dv/dt = (dv/dy)(dy/dt) = v\,dv/dy\). Differentiating, \(dv/dy = g/(2\sqrt{g y})\). So \(a = \sqrt{g y} \times g/(2\sqrt{g y}) = g/2\). The acceleration of the pulse is constant, \(g/2\).
(d) In a small time \(dt\) the pulse climbs \(dy = v\,dt\), so \(dt = dy/\sqrt{g y}\). Adding from \(y = 0\) to \(y = L\), \(t = \int_0^L dy/\sqrt{g y} = 2\sqrt{L/g}\).
The second equation of motion gives the same time, because the acceleration is constant. The pulse starts from rest with acceleration \(g/2\), so \(L = \tfrac12 (g/2)\,t^2\), and \(t = 2\sqrt{L/g}\).
The answers hold a surprise. The speed \(\sqrt{g y}\) does not contain \(\mu\) or \(M\). A heavier rope has more tension at every height, but also more mass to move, in the same proportion. The two effects cancel. Every hanging rope of the same length carries a pulse to its top in the same time.
Watch the dots in Fig 2.1. They are dropped at equal times, \(0.4\,\text{s}\) apart. The gaps between them are \(0.4\), \(1.2\), \(2.0\) and \(2.8\,\text{m}\), and they keep growing in the ratio \(1 : 3 : 5 : 7\). That odd-number pattern is the mark of motion from rest with constant acceleration. Here the acceleration is \(g/2 = 5\,\text{m/s}^2\), and the pulse reaches the top after \(2\sqrt{20/10} = 2.83\,\text{s}\), moving at \(\sqrt{200} = 14.1\,\text{m/s}\).
The pulse speeds up, but no piece of the rope moves upward. Each piece only swings sideways as the pulse passes it. What accelerates is the place where the disturbance is, because the rope higher up is tighter.
A heavy rope of length \(L\) and mass \(M\) hangs from a hook. What is the tension in the rope at its midpoint?
A heavy rope \(10\,\text{m}\) long hangs from the ceiling of a tall hall. A pulse is made at its free bottom end. Take \(g = 10\,\text{m/s}^2\). How long does the pulse take to reach the top, in seconds?
Two ropes of the same length hang side by side. Rope B has twice the mass per unit length of rope A. Pulses are made at both bottom ends at the same instant. Which pulse reaches the top first?
Now return to the rope in the school gym. The kink climbs at \(\sqrt{g y}\), so it is slow near the floor and faster near the ceiling. It gains speed at a steady \(g/2\). A rope \(5\,\text{m}\) long carries it to the ceiling in \(2\sqrt{5/10} = 1.4\,\text{s}\), whatever the rope is made of.
Tension and mass side by side
The speed \(v = \sqrt{T/\mu}\) grows as the square root of the tension and falls as the square root of the mass per unit length. So a change in either one changes the speed by its square root.
The tension in a string is quadrupled while its mass per unit length stays the same. What happens to the wave speed?
Three strings carry identical pulses, released at the same instant from the same start. The first has tension \(T\) and mass per unit length \(\mu\). The second has \(4T\) and \(\mu\), the third \(T\) and \(4\mu\). When the first pulse has gone a distance \(d\), how far has the third gone?
Play the three strings in the figure above. All three pulses start together from the same line. Four times the tension doubles the speed, so that pulse has gone twice as far, \(2d\). Four times the mass per unit length halves the speed, so that pulse has gone half as far, \(d/2\).
Which changes make a wave travel faster along a stretched string?
Marin Mersenne was a French friar who wrote letters to nearly every scientist in Europe. He also measured musical strings with great care. In his book Harmonie universelle of 1636 he stated the laws of a vibrating string. The pitch rises as the square root of the tension. It falls with the length, and it falls as the square root of the string's heaviness. The square roots in his laws are the square root in \(v = \sqrt{T/\mu}\), found by measurement decades before anyone could derive them.
- Putting grams and centimetres into \(\mu = m/L\).Convert to kilograms and metres first; \(\mu\) is in kg/m.
- Using one tension for a heavy hanging rope.The tension at a height \(y\) above the free end is \(\mu g y\), zero at the bottom.
- Thinking a heavier hanging rope carries a pulse faster.\(v = \sqrt{g y}\) has no \(\mu\). Every rope of the same length takes \(2\sqrt{L/g}\).
- Taking four times \(\mu\) to give a quarter of the speed.The speed goes as \(1/\sqrt{\mu}\), so four times \(\mu\) halves it.