Speed of Waves on a String

Pick up a sitar and pluck the open baajh taar — the main playing string. A note rings out, bright and steady. Now turn the tuning peg to increase the tension; pluck again. The pitch rises, sharp and distinct. You have just changed the wave speed on the string, and the pitch shifted accordingly. Loosen it the other way and the pitch falls.

Now switch instruments. A sarangi is played at the same mehfil. Its gut strings are thicker and heavier than the sitar's steel strings, and tuned to a comparable tension. The sarangi's drone strings give a deeper, throatier note than the sitar. Walk over to a rudra veena and its great melody strings — thick, heavy, long — produce an even lower fundamental.

Three instruments, three pitches, and the single physical quantity responsible is the wave speed on the string. The relationship, one of the simplest and most important formulas in wave physics, is

where T is the tension in the string (in newtons) and \mu is the linear mass density — the mass per unit length of the string (in kg/m). Tighter string pulls the wave speed up; heavier string pulls the wave speed down. Pitch and wave speed rise together because, on a string of fixed length, the fundamental frequency is f_1 = v/(2L) — you will meet that relation in the chapter on standing waves.

This article does three things. First, it derives v = \sqrt{T/\mu} from Newton's second law applied to a small curved segment of the string — a derivation that looks daunting the first time but is mostly careful bookkeeping. Second, it works through examples that turn the formula into numbers (a sitar string, a bungee cord, a clothesline). Third, it computes the energy and power carried by a sinusoidal wave on the string, reaching the universal scaling P \propto A^2 f^2 that generalises to every kind of wave in every medium.

You should have met introduction-to-waves and the wave equation before this article; you will use the sinusoidal form y(x,t) = A\sin(kx - \omega t) throughout.

Setting up — a small element of string

Consider a long, uniform, stretched string laid along the x-axis in equilibrium. Its tension is T (newtons), its mass per unit length is \mu (kg/m). A small disturbance is passing, so the string's shape at time t is described by a function y(x, t) — the vertical displacement of each point from the equilibrium y = 0 line.

Assumptions: The string has no thickness (it is a perfect curve); it has uniform density \mu; it is perfectly flexible (it can curve but cannot resist bending stiffly, unlike a steel rod); the tension T is the same everywhere along the string; the transverse displacements are small (|\partial y/\partial x| \ll 1), so that the slope angles everywhere are tiny and their sines can be approximated by the slopes themselves. Air drag is neglected. Gravity on a horizontal string either acts perpendicular to the wave motion (does not couple to the transverse dynamics) or is small compared to T.

Zoom in on a tiny segment of the string between x and x + \Delta x. In equilibrium this segment has length \Delta x and mass \mu \, \Delta x. When a wave passes, the segment deforms slightly; since the displacements are small, its length is still very nearly \Delta x — the deformation of length is second-order small and can be ignored.

The tension acts at both ends of the segment, pulling along the string's local direction. At x + \Delta x, the tension pulls forward along the tangent at that point; at x, the tension pulls backward along the tangent at that point. The two tangent directions differ slightly because the string is curved. That small difference in angle is what produces the net vertical force on the segment.

Deriving v = \sqrt{T/\mu}

The plan: write Newton's second law for the small segment's vertical motion, relate the small tangent angles to the slope \partial y/\partial x, expand in small quantities, and identify the resulting equation as the wave equation with wave speed \sqrt{T/\mu}.

Step 1. Write down the vertical components of the tension at both ends.

At x + \Delta x, the tension pulls outward along the local tangent. If the slope of the string there is \tan\theta_2 = \partial y/\partial x \big|_{x+\Delta x}, and the angle \theta_2 is small, then the vertical component of this tension is

At x, similarly,

Why: for small angles, \sin\theta \approx \tan\theta \approx \theta. The slope of the string \partial y/\partial x is \tan\theta (rise over run). This is the only small-angle approximation the whole derivation makes.

Step 2. Write the net vertical force on the segment.

The right end pulls the segment upward (when \partial y/\partial x|_{x+\Delta x} > 0); the left end pulls the segment downward (when \partial y/\partial x|_x > 0, the tangent at x points up-and-to-the-right, so the tension — which pulls away from the segment toward -x — has a downward vertical component). Subtracting:

Why: the tension on each end acts along the tangent, pulling the segment outward. At the right end, "outward" has a component along +y if the string is sloping up; at the left end, "outward" is toward -x, and the tangent there is pointing into +x, so the y-component of the pulling force is downward. Subtracting gives the net upward force.

Step 3. Rewrite the right-hand side as a derivative times \Delta x.

The bracket is exactly the difference of \partial y/\partial x at two nearby points. For small \Delta x, that difference is

So the net vertical force on the segment is

Why: this is the definition of the second partial derivative — the rate of change of the slope. The curvature \partial^2 y/\partial x^2 tells you how rapidly the slope is changing along the string; multiplied by \Delta x, it is the first-order change in the slope across the segment.

Step 4. Apply Newton's second law in the vertical direction.

The mass of the segment is \mu\, \Delta x; its vertical acceleration is \partial^2 y/\partial t^2 (the second time-derivative of the vertical position — since each piece of string at position x moves only vertically). Newton's second law says F_y = m a_y:

Step 5. Cancel \Delta x.

Both sides have a factor of \Delta x; dividing through,

Rearrange:

Why: the \Delta x cancellation is what makes the result a local (differential) statement, independent of the size of the segment. The equation now applies at every point of the string.

Step 6. Identify this as the wave equation.

The general wave equation is

and every solution of it has the form y = f(x - vt) + g(x + vt) — a right-moving and a left-moving disturbance, each travelling at speed v. Comparing with equation (2):

Why: pattern-match equation (2) with the general wave equation. The coefficient of \partial^2 y/\partial x^2 must be v^2. Taking the positive root — because v is the magnitude of the propagation speed — gives the formula we set out to derive.

This is the result. Three paragraphs of careful bookkeeping about small angles and tensions, and you reach the cleanest possible statement: on a stretched string, the wave speed depends only on tension and mass per length, and in exactly the combination \sqrt{T/\mu}.

A dimensional sanity check

Let us check the units. T has units of newtons = kg·m/s². \mu has units of kg/m. Their ratio:

Taking the square root,

which is a speed. The dimensions work.

Why the amplitude doesn't appear

Nothing in the derivation referred to the amplitude A or the frequency f of any particular wave. The formula v = \sqrt{T/\mu} is a property of the medium — the string — not of any particular wave running along it. A large-amplitude wave and a tiny-amplitude wave travel at the same speed (as long as both are small enough for the small-angle approximation to hold). A high-frequency wave and a low-frequency wave also travel at the same speed. That is what it means for a string to be a non-dispersive medium.

The small-angle approximation is load-bearing, though. If the wave is so large that the slope \partial y/\partial x is not small — so large that \sin\theta \ne \tan\theta — the tension itself changes (because a more stretched piece of string pulls harder, by Hooke's law applied to the string's own elasticity), and the wave speed starts depending on the amplitude. That is when the medium becomes nonlinear. For everything in this article, we stay well within the linear regime.

Explore it: drag tension and density

Drag the red point below to change the ratio T/\mu (in units where the speed range is easy to read). Watch the wave speed respond.

How tension and density set the pitch

The formula v = \sqrt{T/\mu} is a statement about wave speed, but musicians and instrument designers mostly think in terms of pitch. On a string of length L fixed at both ends (a sitar string pinned to the bridge and the nut), the lowest-frequency standing wave — the fundamental — has wavelength \lambda_1 = 2L (you will see this in full in standing waves). The corresponding fundamental frequency is

Three physical knobs control pitch:

• Length L. Shorten the string (by pressing a fret, or by pulling the sliding gourd of a sarod): f_1 \propto 1/L, so halving the length doubles the pitch.
• Tension T. Tighten the string: f_1 \propto \sqrt{T}, so quadrupling the tension doubles the pitch. This is how a sitar player fine-tunes with the peg.
• Mass per length \mu. Change the string to a thicker or heavier one: f_1 \propto 1/\sqrt{\mu}. Quadrupling the mass halves the pitch. This is why the low strings of a sarangi are thick gut and the high strings of a sitar are thin steel.

The square roots are the reason musicians tune with large, smooth peg rotations for small pitch adjustments — a 5% change in tension produces only a 2.5% change in pitch. A linear tuner would be far more twitchy.

Worked examples

Energy and power carried by a sinusoidal wave

A travelling wave carries energy from one place to another. On a string, each small element receives kinetic energy (it is moving) and stores elastic potential energy (it is stretched slightly by the displacement). As the wave passes, that energy flows along the string. The question this section answers: how much energy flows past a fixed point, per second? That is the power the wave transmits.

Take a sinusoidal wave

with amplitude A, wavenumber k = 2\pi/\lambda, and angular frequency \omega = 2\pi f.

Kinetic energy per unit length

The transverse velocity of a piece of string at position x is

A small element of string of length dx has mass \mu\, dx, so its kinetic energy is

The kinetic energy per unit length is therefore

Potential energy per unit length

When the string stretches transversely, its length increases slightly. For a small element originally of length dx and currently at slope \partial y/\partial x, the new length is

The extension of the element is \tfrac{1}{2}(\partial y/\partial x)^2 dx, and the work done against the tension to stretch it is T \cdot (extension):

Why: \partial y/\partial x = Ak\cos(kx - \omega t). Squaring gives A^2 k^2 \cos^2(\cdot). The factor \tfrac12 comes from the expansion of \sqrt{1+u} for small u.

The potential energy per unit length is

The kinetic and potential energies are equal

Compare equations (4) and (5):

But v = \omega/k = \sqrt{T/\mu}, so \omega^2/k^2 = T/\mu. Substitute:

So dK/dx = dU/dx pointwise. At every point and every instant, the kinetic energy and the elastic potential energy are equal. Their sum is

Why: for a travelling sinusoidal wave, kinetic and potential energies oscillate in phase — when one is big, so is the other. This is different from SHM, where they are 90° out of phase. The reason is that each point of the medium is being driven by the wave in a particular way, and the combined energy density tracks the square of the local velocity (and, equivalently, the square of the local slope).

Average power

The total energy per unit length, averaged over a cycle, uses \langle \cos^2(\cdot)\rangle = \tfrac12:

This energy density moves along the string at speed v. The power — energy per unit time flowing past a fixed point — is therefore energy density times speed:

Why: think of the energy as a density (joules per metre of string), being conveyed along the string at speed v (metres per second). Multiplying gives joules per second, which is watts — power.

Physical scaling

Two scaling relations in equation (7) are worth burning in:

• P \propto A^2. Double the amplitude, quadruple the power. This is why the "intensity" of a wave (energy per unit area, or per unit length for strings) scales as the square of the amplitude — a fact that shows up in every branch of wave physics, from sound loudness (the decibel) to the probability density |\psi|^2 of a quantum wavefunction.
• P \propto \omega^2 (equivalently P \propto f^2). Double the frequency, quadruple the power. High-frequency waves of the same amplitude carry more power than low-frequency ones — because each particle of the medium completes twice as many oscillations per second, and each oscillation involves kinetic energy \propto \omega^2.

Together: intensity \propto A^2 f^2. Both factors matter. A softly-played high note and a loudly-played low note can carry the same power.

A worked power estimate

Take the sitar string from Example 1: \mu = 3.0 \times 10^{-4} kg/m, v \approx 466 m/s. Suppose it is oscillating at its fundamental frequency f_1 = 291 Hz with amplitude A = 2 mm = 0.002 m (reasonable for a plucked string). Then \omega = 2\pi f = 2\pi \times 291 \approx 1830 rad/s, and

Nearly one watt of power is flowing along the string at peak amplitude. Most of that energy gradually leaks into the sitar's wooden gourd and then into the air as sound (and some goes to friction at the bridge). This is why plucked notes decay over a few seconds — the wave energy is being drained faster than the player replenishes it.

Common confusions

• "A heavier string means faster waves." Wrong direction. Heavier string (\mu larger) means slower waves, because the string has more inertia per length. Check: v = \sqrt{T/\mu}, so v shrinks as \mu grows.

• "Tension changes pitch because it changes the string length." No. Tension changes pitch because it changes the wave speed. The length L is set by the frets (or the nut-bridge geometry) and is barely affected by tuning. f_1 = v/(2L) depends on both v and L separately; tuning pegs adjust v via T.

• "Wave speed depends on amplitude — a big wave moves faster." Not for a linear medium (which includes all small-amplitude waves on a string). v = \sqrt{T/\mu} contains neither A nor f. The formula is a property of the medium alone.

• "The power P = \frac12 \mu \omega^2 A^2 v is the average or the instantaneous?" It is the average over one cycle. The instantaneous power oscillates between 0 and \mu \omega^2 A^2 v; averaging over \cos^2 gives the factor of \tfrac12.

• "Kinetic and potential energies are 90° out of phase, like in SHM." For a travelling wave, they are in phase — both proportional to \cos^2(kx - \omega t). For a standing wave (superposition of two opposite-going travelling waves), they are indeed out of phase, oscillating past each other the way they do in a pendulum.

• "A wave with zero amplitude still has energy." No — zero amplitude is no wave at all. Energy density goes to zero as A\to 0. (This is one place where the A^2 scaling is important: it means the energy comes entirely from motion, not from some residual "wave energy" independent of amplitude.)

Where this leads next

• Principle of Superposition — when two waves on the same string meet, they add linearly, giving interference.
• Reflection and Transmission of Waves — what happens at a fixed or free end, and at a junction between two strings of different \mu.
• Standing Waves and Normal Modes — when a wave reflects back and forth between two ends, the superposition gives the discrete resonances that set musical pitch.
• Sound Waves — Nature and Propagation — the same v = \sqrt{\text{restoring/inertia}} applied to compression waves in air.
• The Wave Equation — the universal partial differential equation derived here, and the sinusoidal solutions that satisfy it in any linear medium.