Speed of a Transverse Wave on Stretched String

The Speed of a Transverse Wave on a Stretched String

A wave travelling along a stretched string is a classic example of a transverse mechanical wave. The speed of such a wave is not arbitrary — it is determined entirely by two properties of the string: how tightly it is stretched (the tension) and how heavy it is per unit length (the linear mass density). The derivation that follows shows exactly how these two factors combine to give the wave speed.

Consider a small element of the string of length $\Delta l$ and mass $\Delta m$. If the string has a linear mass density $\mu$ (mass per unit length), then $\Delta m = \mu \, \Delta l$. The string is under a constant tension $T$, which is the same at every point along it.

When a wave pulse travels along the string, each small element of the string is displaced from its equilibrium position. The tension forces at the two ends of the element are not exactly opposite — they have slightly different directions because the string is curved. The net force on the element comes from the vector sum of these two tension forces.

Deriving the Wave Speed

Take a small element of the string of length $\Delta l$ that is part of a wave travelling to the right. The string makes a small angle $\theta$ with the horizontal at the left end of the element, and a slightly different angle $\theta + \Delta \theta$ at the right end. For small displacements (which is the usual case for wave motion on a string), these angles are small, so $\sin \theta \approx \theta$ and $\cos \theta \approx 1$.

The net vertical force on the element is the difference between the vertical components of tension at the two ends:

For small angles, $\sin \theta \approx \theta$, so:

Now, the angle $\theta$ at any point on the string is related to the slope of the string at that point. For small displacements, $\theta \approx \tan \theta = \frac{\partial y}{\partial x}$, where $y(x,t)$ is the vertical displacement of the string at position $x$ and time $t$. Therefore, $\Delta \theta$ is the change in slope over the length of the element:

Since the element is nearly horizontal for small displacements, its length $\Delta l$ is approximately equal to $\Delta x$. So:

This net force causes the element to accelerate vertically. Using Newton's second law, $F_{\text{net}} = \Delta m \, a$, where $\Delta m = \mu \, \Delta x$ and the vertical acceleration is $a = \frac{\partial^2 y}{\partial t^2}$:

Cancelling $\Delta x$ from both sides gives the wave equation for the string:

This is a standard wave equation of the form $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$, where $v$ is the wave speed. Comparing the two forms:

Therefore, the speed of a transverse wave on a stretched string is:

This is the central result of the section. The speed depends only on the tension $T$ and the linear mass density $\mu$, not on the shape or amplitude of the wave.

Properties of the Wave Speed

The textbook lists three important properties that follow directly from the formula $v = \sqrt{T/\mu}$.

Property (I): The speed of a transverse wave on a stretched string is independent of the shape of the wave and of its amplitude.

This is evident from the formula — neither the shape nor the amplitude appears in $v = \sqrt{T/\mu}$. The wave speed is a property of the medium (the string under tension), not of the particular disturbance travelling through it.

Property (II): The speed is directly proportional to the square root of the tension in the string.

If you double the tension, the speed increases by a factor of $\sqrt{2}$, not by a factor of 2. This is because tension provides the restoring force — more tension means a stronger restoring force for a given curvature, so the wave can travel faster. But the relationship is not linear because the inertia of the string also plays a role.

Property (III): The speed is inversely proportional to the square root of the linear mass density of the string.

A heavier string (larger $\mu$) has more inertia per unit length, so it responds more sluggishly to the restoring force, and the wave travels more slowly. If you double the mass per unit length (say by using a thicker string of the same material), the speed decreases by a factor of $\sqrt{2}$.

Dimensional Analysis Check

The formula $v = \sqrt{T/\mu}$ is dimensionally consistent. Tension $T$ has units of force: $[T] = \text{MLT}^{-2}$. Linear mass density $\mu$ has units of mass per length: $[\mu] = \text{ML}^{-1}$. Therefore:

Taking the square root gives $\text{LT}^{-1}$, which is the dimension of speed. This dimensional check confirms that the formula is physically plausible. …