A sitar player turns a wooden peg before a concert. The string grows tighter, and the note it gives rises. On a guitar the thick bass strings give low notes and the thin ones high notes. The player changes nothing but the tightness and the thickness of the string. Somehow these two alone decide how the string sings.
What decides the speed of a wave?
The speed at which a wave travels is one of its most basic properties. It is decided entirely by the medium that carries the wave. For a transverse wave on a stretched string, the medium offers two properties. One is how tightly the string is stretched, its tension \(T\). The other is how much mass it has in each metre of length, its linear mass density \(\mu\).
In this lesson we find how these two combine. The method is Newton's second law, applied to one tiny piece of the string while the wave passes through it. Nothing else is needed.
In this chapter \(T\) always means the tension in the string, in newtons. It is never a time period. When a period is needed, the chapter says "one period" in words.
Two assumptions
Imagine watching a tiny piece of the string while the wave ripples through it. To keep the analysis clear, we make two standard assumptions. The first is small amplitude. The sideways displacement \(y\) of the string is much smaller than the wavelength \(\lambda\). As a result, the angle \(\theta\) between the string and the horizontal is always small.
The second assumption is purely transverse motion. Each bit of the string moves only up and down, across the string. No bit moves sideways along the string. So the horizontal position \(x\) of a bit never changes, and only its displacement \(y\) does.
The segment and its mass
Now fix attention on a short segment of the string between the points \(x\) and \(x + dx\). As the wave passes, the segment is slightly curved. Its true length along the curve is \(ds\), a little more than \(dx\). The mass of the segment is \(\mu\,ds\), since \(\mu\) is the mass in each unit of length.
For a small amplitude, the curve is nearly flat, so \(ds\) is very nearly \(dx\). The mass of the segment can therefore be written \(dm = \mu\,dx\). This one step uses the first assumption.
The curved segment in Fig 1.1 has length \(ds\) along the string. Why may its mass be written as \(\mu\,dx\) instead of \(\mu\,ds\)?
The forces on the segment
The segment moves only up and down. Its vertical acceleration is \(a_y\), the second derivative of \(y\) with respect to time at a fixed \(x\). We write it \(a_y = \partial^2 y / \partial t^2\). The curly \(\partial\) marks a partial derivative: \(y\) depends on both \(x\) and \(t\), and here only \(t\) changes.
Newton's second law for the vertical motion of the segment reads \(dF_y = (dm)\,a_y\). Here \(dF_y\) is the net vertical force on the segment, and \(dm = \mu\,dx\) is its mass. So we need the forces. The string on each side pulls the segment with the tension \(T\). Each pull acts along the tangent to the string at that end.
In Fig 1.2 the tangent makes an angle \(\theta\) with the horizontal at \(x\). At \(x + dx\) the string has turned a little, and the angle is \(\theta + d\theta\). The pull at the left end points down and to the left. The pull at the right end points up and to the right.
All the motion is vertical, so the segment has no net horizontal force. The horizontal components of the two pulls cancel. Only the difference of the vertical components is left, and it is the net vertical force.
\[ F_{\text{up}} = T\sin(\theta + d\theta) \]
Why. each pull has size \(T\) along its tangent. Its vertical part is \(T\) times the sine of the tangent's angle with the horizontal. At the right end it points up.
\[ F_{\text{down}} = T\sin\theta \]
Why. the vertical part of the pull at the left end, which points down.
\[ T\cos(\theta + d\theta) \approx T\cos\theta \approx T \]
Why. the horizontal parts. For small angles each cosine is very nearly \(1\), so the two horizontal parts are equal and opposite, as purely transverse motion requires.
\[ \sin\theta \approx \tan\theta \]
Why. for a small angle the sine and the tangent are nearly equal, because the cosine is nearly \(1\). The same holds for \(\theta + d\theta\).
\[ dF_y = T\left(\tan(\theta + d\theta) - \tan\theta\right) \]
Read it. the net vertical force is the upward pull minus the downward pull.
Why do the horizontal components of the tension at the two ends of the segment cancel?
From the force to the wave equation
Set the net vertical force equal to mass times acceleration, and the rest follows step by step. The key idea is that \(\tan\theta\) is the slope of the string, \(\partial y / \partial x\). The difference of two slopes over a distance \(dx\) is how fast the slope changes along the string.
\[ \begin{aligned} &T\left(\tan(\theta + d\theta) - \tan\theta\right) \\ &\quad = (\mu\,dx)\dfrac{\partial^2 y}{\partial t^2} \end{aligned} \]
Why. Newton's second law, \(dF_y = (dm)\,a_y\), with the force just found and \(dm = \mu\,dx\).
\[ \begin{aligned} &T\,\dfrac{\tan(\theta + d\theta) - \tan\theta}{dx} \\ &\quad = \mu\dfrac{\partial^2 y}{\partial t^2} \end{aligned} \]
Why. divide both sides by \(dx\), the length of the segment.
\[ \dfrac{\tan(\theta + d\theta) - \tan\theta}{dx} = \dfrac{\partial(\tan\theta)}{\partial x} \]
Why. the change in \(\tan\theta\) over a tiny distance \(dx\), divided by \(dx\), is the rate of change of \(\tan\theta\) along the string.
\[ \tan\theta = \dfrac{\partial y}{\partial x} \]
Why. \(\tan\theta\) is the slope of the string at an instant.
\[ \dfrac{\partial(\tan\theta)}{\partial x} = \dfrac{\partial^2 y}{\partial x^2} \]
Why. the rate of change of the slope along the string is the second derivative of \(y\) with respect to \(x\).
\[ T\dfrac{\partial^2 y}{\partial x^2} = \mu\dfrac{\partial^2 y}{\partial t^2} \]
Read it. put steps 3 and 5 into step 2. This is the wave equation for a string.
The speed, read from the equation
The chapter Introduction to Waves found the general wave equation for any wave that travels at speed \(v\) without changing shape: \(\partial^2 y / \partial x^2 = (1/v^2)\,\partial^2 y / \partial t^2\). The string's equation has exactly this form. So the string carries waves, and the two equations can be compared term by term.
\[ \dfrac{\partial^2 y}{\partial x^2} = \dfrac{\mu}{T}\dfrac{\partial^2 y}{\partial t^2} \]
Why. divide the string's wave equation by \(T\), so that it has the same layout as the general wave equation.
\[ \dfrac{1}{v^2} = \dfrac{\mu}{T} \]
Why. the two equations match term by term, so the factors in front of \(\partial^2 y / \partial t^2\) are equal.
\[ v = \sqrt{\dfrac{T}{\mu}} \]
Read it. turn both sides upside down and take the positive square root. The wave speed is a positive number.
The speed of a wave on a string depends only on the tension and the mass per unit length. It does not depend on the frequency or the amplitude of the wave. A higher tension raises the speed, and a higher mass per unit length lowers it.
The result makes physical sense. A tighter string pulls each displaced piece back harder, so a disturbance is handed on more quickly. A heavier string has more inertia in each piece, so each piece responds more slowly. The square root means that the speed grows more slowly than the tension itself.
Two identical strings are under different tensions. String A has four times the tension of string B. How many times faster do waves travel on string A than on string B?
A string with a mass per unit length of \(0.01\,\text{kg/m}\) is stretched with a tension of \(36\,\text{N}\). What is the speed of a transverse wave on it, in m/s?
The tension is in newtons and the mass per unit length in kg/m. Why does \(\sqrt{T/\mu}\) come out in m/s?
A student shakes the end of a stretched rope twice as often per second as before. The tension and the rope are unchanged. What happens to the speed of the wave on the rope?
Jean le Rond d'Alembert was left as a baby on the steps of a Paris church. He grew up to be a leading mathematician of France. In 1747 he sent the Berlin Academy a paper on the shape of a stretched string set vibrating. He started from the force on each tiny piece of the string, as this lesson did. He obtained the wave equation of the string and showed that its solutions are two shapes travelling in opposite directions at the same speed. It was one of the first partial differential equations of physics, and it is the equation of this lesson.
Now return to the sitar peg. Turning the peg raises the tension \(T\), so the waves on the string travel faster. The thick bass strings of a guitar have a large \(\mu\), so waves on them travel slower. The speed of the waves, together with the length of the string, sets the note. The chapter Standing Waves on a String shows exactly how.
- Taking \(T\) in \(v = \sqrt{T/\mu}\) to be a time period.Here \(T\) is the tension in newtons. The formula has no period in it.
- Using the mass of the whole string for \(\mu\).\(\mu\) is the mass per unit length, \(m/L\), in kg/m.
- Thinking that shaking harder or faster makes the wave travel faster.The speed depends only on \(T\) and \(\mu\). Amplitude and frequency change the wave, not its speed.
- Forgetting the square root.Four times the tension gives twice the speed, not four times.