Speed of a transverse wave on a stretched string

Pluck a guitar string and a kink races along it, bounces off the far end, and comes back — all in a fraction of a second. Tighten the tuning peg and that kink moves faster; swap in a thicker, heavier string and it moves slower. A transverse wave on a stretched string is one of the cleanest wave-speed stories in the whole Waves chapter, because the speed depends on exactly two things you can feel with your hands: how hard the string is pulled, and how heavy it is per metre. The formula behind this feel is short — $v = \sqrt{T/\mu}$ — but it packs in a pattern that keeps reappearing throughout physics: a restoring influence (tension, trying to snap the string back straight) sitting on top of an inertial resistance (linear mass density, the string's reluctance to accelerate), combined under a square root. Get comfortable with this one string formula and you are already halfway to understanding wave speed in a spring, in a rope, and even — once you meet it — the speed of sound. This article rebuilds that formula from Newton's second law applied to a tiny curved element of string, so the $\sqrt{T/\mu}$ is not a fact to memorise but a result you can rederive under exam pressure. Along the way we work two numerical problems (one board-style, one competitive-style), a sonometer setup connecting wave speed to standing waves and frequency, and the traps that most often cost marks — starting with the almost-universal slip of forgetting to convert grams to kilograms.

Key takeaways

The speed of a transverse wave on a stretched string is $v = \sqrt{T/\mu}$ , where $T$ is the tension (newtons) and $\mu$ is the linear mass density (kg/m).
Linear mass density is computed as $\mu = m/L$ — total mass over total length, always in SI units before you touch the formula.
Wave speed depends only on the string's mechanical properties ( $T$ and $\mu$ ); it does not depend on the frequency, amplitude, or waveform of the disturbance sent down the string.
The formula is derived from Newton's second law applied to a small curved element of string, balancing the transverse component of tension against the element's mass times its centripetal acceleration.
Because of the square root, doubling $v$ needs four times the tension, not twice — a favourite twist in competitive-level questions.
Once $v$ is known, it links to the rest of the chapter through $v = f\lambda$ (wavelength for a given frequency) and through standing-wave frequencies $f_n = \dfrac{n}{2L}\sqrt{T/\mu}$ on a fixed string.
Key idea: a transverse wave on a stretched string moves at $v=\sqrt{T/\mu}$ because tension supplies the restoring force and linear mass density supplies the inertia, and this single speed then sets both the wavelength for a given driving frequency and the standing-wave harmonics of the string.

The core insight

The speed of a transverse wave on a stretched string, $v = \sqrt{T/\mu}$ , is fixed entirely by the string's tension (restoring force) and linear mass density (inertia) — never by the wave's own frequency or amplitude.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 Waves chapter (Unit: Oscillations and Waves), immediately after the general relation $v = f\lambda$ for a travelling wave. It is the first concrete case where you compute a wave speed from the medium's mechanical properties rather than being given it directly, and the derivation itself (Newton's second law on a curved string element) is explicitly part of the prescribed syllabus. It is the launch pad for standing waves on strings — sonometer experiments, guitar and sitar strings, and harmonic series problems — all of which reuse $v=\sqrt{T/\mu}$ as an input.

The physical picture

Picture a long, light string stretched horizontally under tension $T$ , lying along the x-axis at equilibrium. A transverse wave is a pattern of sideways (y-direction) displacement that travels along the string without any actual bit of string moving in the x-direction — only up and down. The wave speed $v$ measures how fast that y-displacement pattern advances along x.

v = \sqrt{\frac{T}{\mu}}

where: $T$ is the tension in the string, in newtons (N) — the pulling force that keeps the string straight and taut; and $\mu$ is the linear mass density, in kilograms per metre (kg/m) — the mass of string per unit length. The physical reading: $T$ is a restoring quantity — a displaced bit of string feels a net sideways pull from its stretched, tilted neighbours, always directed back toward the straight line, and a bigger $T$ means a stronger pull-back, hence a faster wave. Meanwhile $\mu$ is an inertial quantity — the more mass packed into each metre of string, the more sluggishly that mass responds to the restoring pull, so $v$ falls as $\mu$ rises. The square root means $v$ is not simply proportional to either quantity; it responds more gently than a straight-line relationship would suggest. Linear mass density itself is computed from bulk quantities you can measure with a scale and a ruler:

\mu = \frac{m}{L}

where: $m$ is the total mass of the string (kilograms) and $L$ is its total length (metres). Always convert mass to kilograms and length to metres before dividing — $\mu$ must come out in kg/m for the wave-speed formula to give an answer in m/s.

Figure

A stretched string with a transverse pulse travelling along it, tension arrows pulling outward at both ends, and a small curved element of the string shown with the tension force at each end tilted at a small angle to the horizontal so their vertical components add to a net restoring force

Once $v$ is known, it plugs straight into the general travelling-wave relation $v = f\lambda$ : for a string held at a given tension and driven at a chosen frequency $f$ , the wavelength $\lambda$ that appears on the string is fixed by $\lambda = v/f$ . The same $v$ also governs the string's natural standing-wave frequencies, $f_n = \dfrac{n}{2L}\sqrt{T/\mu}$ for a string of length $L$ fixed at both ends — the reason a sonometer or a guitar can be tuned by adjusting tension.

Derivation AWave speed on a stretched string from Newton's second law

The goal is to derive $v=\sqrt{T/\mu}$ from first principles — Newton's second law applied to one small curved bit of string — using only the small-angle approximation and algebra, so that the formula is a provable result rather than a fact to memorise. We model a short segment of string near the crest of a pulse as an arc of a circle of radius $R$ , moving with the pulse at speed $v$ ; in the wave's own reference frame the pulse pattern is frozen and this bit of string is momentarily undergoing circular motion, which lets us use the familiar centripetal-force relation instead of calculus.

Step 1. Isolate a small element of string of length $\Delta l$ , subtending a small angle $2\theta$ at the centre of curvature, so that $\Delta l = R(2\theta)$ . Its mass is $\Delta m = \mu\, \Delta l$ , using the linear mass density $\mu$ .

Step 2. Identify the forces on this element: tension $T$ pulling tangentially at each end. Because the element is curved, the two tension vectors are not quite opposite — each makes a small angle $\theta$ with the horizontal chord, so each has a vertical (radially inward) component $T\sin\theta$ . For small $\theta$ , $\sin\theta \approx \theta$ , so each end contributes an inward force of about $T\theta$ , and the two ends together give a net inward (centripetal) force of $2T\theta$ .

Step 3. Apply Newton's second law for circular motion. In the wave's frame, this element moves on a circular arc of radius $R$ at speed $v$ , so it needs a centripetal force $\Delta m\, v^2/R$ directed toward the centre. Setting the net tension force equal to the required centripetal force: $2T\theta = \dfrac{\Delta m\, v^2}{R}$ .

Step 4. Substitute $\Delta m = \mu\,\Delta l = \mu\,(2R\theta)$ from Step 1. The equation becomes $2T\theta = \dfrac{\mu\,(2R\theta)\,v^2}{R} = 2\mu\theta\, v^2$ . The factor $2\theta$ cancels from both sides (it is common and non-zero), leaving simply $T = \mu v^2$ .

Step 5. Solve for $v$ . Rearranging $T = \mu v^2$ gives $v^2 = T/\mu$ , and taking the positive square root (speed is a positive quantity) gives $v = \sqrt{T/\mu}$ — the result, obtained without ever needing to know the shape or size of the pulse, only that it moves at a constant speed $v$ while every bit of string briefly traces a circular arc. Check by dimensions: $[T]$ is newtons $= \text{kg}\,\cdot\,\text{m/s}^2$ , and $[\mu]$ is kg/m, so $[T/\mu] = \dfrac{\text{kg}\,\cdot\,\text{m/s}^2}{\text{kg/m}} = \text{m}^2/\text{s}^2$ , whose square root is m/s — correctly a speed, confirming the algebra is dimensionally consistent.

Worked examples

Board level - speed on a stretched laboratory string

A string of length 2.5 m has a mass of 25 g and is stretched with a tension of 200 N. Find the speed of a transverse wave on this string.

Step 1. Convert all quantities to SI units first: $L = 2.5$ m, $m = 25\ \text{g} = 0.025$ kg, $T = 200$ N.

Step 2. Compute the linear mass density: $\mu = \dfrac{m}{L} = \dfrac{0.025}{2.5} = 0.01\ \text{kg/m}$ .

Step 3. Substitute into the wave-speed formula: $v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{200}{0.01}} = \sqrt{20000}$ .

Step 4. Evaluate the square root: $\sqrt{20000} \approx 141.4\ \text{m/s}$ .

A transverse wave on this string travels at about $141.4\ \text{m/s}$ .

Competitive level - sonometer wire speed, wavelength, and a tension change

A sonometer wire of linear mass density $4\times10^{-3}$ kg/m is stretched by a tension of 64 N. A wave of frequency 200 Hz travels along it. Find (a) the wave speed, (b) the wavelength of the wave on the wire, and (c) the factor by which the tension must be changed to double the wave speed while keeping $\mu$ fixed.

Step 1. Find the wave speed directly from $v=\sqrt{T/\mu}$ : $v = \sqrt{\dfrac{64}{4\times10^{-3}}} = \sqrt{16000} \approx 126.5\ \text{m/s}$ .

Step 2. Find the wavelength using the general travelling-wave relation $v=f\lambda$ , rearranged as $\lambda = v/f$ : $\lambda = \dfrac{126.5}{200} \approx 0.633\ \text{m} = 63.3\ \text{cm}$ .

Step 3. For part (c), note that $v \propto \sqrt{T}$ when $\mu$ is fixed, since $v = \sqrt{T/\mu} = \sqrt{T}\cdot\sqrt{1/\mu}$ and $\sqrt{1/\mu}$ is constant here. To double $v$ , we need $\sqrt{T_{\text{new}}}$ to be twice $\sqrt{T_{\text{old}}}$ , which means $T_{\text{new}} = 4\,T_{\text{old}}$ — squaring both sides of the proportionality.

Step 4. Apply this to the numbers: $T_{\text{old}} = 64$ N, so $T_{\text{new}} = 4 \times 64 = 256$ N is needed to reach $v_{\text{new}} = 2 \times 126.5 \approx 253\ \text{m/s}$ .

The wave speed is $126.5\ \text{m/s}$ , the wavelength is $63.3\ \text{cm}$ , and doubling the speed at fixed $\mu$ requires quadrupling the tension, to $256\ \text{N}$ . Timing note: In a competitive exam, do not compute $v_{\text{new}}$ by first finding a new $T$ and then reapplying the full square-root formula — that wastes time. Recognise directly that $v \propto \sqrt{T}$ at fixed $\mu$ , so any ratio question about $v$ and $T$ collapses to squaring or square-rooting the given ratio; this shortcut alone should save 30-40 seconds per question in this pattern.

Common traps

Forgetting to convert grams to kilograms (or centimetres to metres) before computing $\mu=m/L$ . This is the single most common numerical error in this topic and throws the final answer off by a large, easy-to-miss factor (often 1000, from the gram-to-kilogram slip alone).
Plugging total mass $m$ or total length $L$ directly into $v=\sqrt{T/\mu}$ instead of computing $\mu=m/L$ first. The formula needs the ratio $\mu$ , not either raw quantity on its own — always compute $\mu$ as a clearly labelled separate step.
Assuming wave speed depends on the frequency or amplitude of the disturbance. It does not: $v=\sqrt{T/\mu}$ is fixed entirely by the string's own mechanical properties; frequency only decides the wavelength through $v=f\lambda$ , and amplitude does not enter the speed formula at all.
Treating $v$ as directly proportional to $T$ (or inversely proportional to $\mu$ ). Because of the square root, doubling $T$ multiplies $v$ by only $\sqrt2\approx1.41$ , not by 2; doubling $\mu$ divides $v$ by $\sqrt2$ , not by 2. Ratio questions that ignore the square root are a frequent source of lost marks.
Reversing which quantity increases speed. Higher tension $T$ gives a faster wave; higher linear mass density $\mu$ (a heavier or thicker string) gives a slower wave. Mixing these up is a common conceptual slip under exam pressure.

You should be able to…

State the formula $v=\sqrt{T/\mu}$ , identify each symbol, and give its correct SI unit.
Compute linear mass density $\mu=m/L$ from a given mass and length, with correct unit conversion to kilograms and metres.
Derive $v=\sqrt{T/\mu}$ from Newton's second law applied to a small curved element of string, using the small-angle approximation.
Calculate wave speed on a string given tension and linear mass density, and combine it with $v=f\lambda$ to find wavelength or frequency.
Reason correctly about square-root scaling: state the tension multiplier needed for a given speed multiplier, and vice versa.
Explain physically why higher tension increases speed while higher linear mass density decreases it.

Ask the tutor: Want to see how this same small-element, small-angle derivation reappears (with different symbols) when you meet the speed of sound in air later in this chapter — ask the AI tutor to walk you through the parallel.

Why does frequency not appear anywhere in v equals square root of T over mu?

Wave speed on a string is a property of the medium — how tightly it is stretched and how heavy it is per metre — not a property of any particular wave riding on it. Think of it like sound in air: air's temperature and density fix the speed of sound, and every sound wave, from a low hum to a high whistle, travels through that same air at that same speed. Frequency only decides how closely spaced the wave crests are (the wavelength), through $v=f\lambda$ , once the medium has already fixed $v$ .

If I pluck a string harder, giving it a bigger amplitude, does the pulse travel faster?

No — amplitude does not appear in $v=\sqrt{T/\mu}$ at all, so a gentle pluck and a hard pluck send pulses down the same string at the identical speed $v$ . What changes with a harder pluck is how much energy the wave carries and how large the displacement is, not how fast the disturbance pattern moves along the string. This is a genuinely counter-intuitive result the first time you meet it, but it follows directly from the derivation, where amplitude never enters the force balance.

Why do thicker guitar strings sound lower even though the formula says higher mu means slower waves, and slower should mean higher pitch by my intuition?

Your intuition about the direction is backwards in a way worth fixing carefully. Higher $\mu$ (thicker string) gives a lower wave speed $v$ . For a string of fixed length $L$ fixed at both ends, the fundamental frequency is $f_1 = \dfrac{1}{2L}\sqrt{T/\mu}$ , which is directly proportional to $v$ . So a lower $v$ from a thicker string directly gives a lower $f_1$ — a deeper, lower-pitched note. That is exactly why bass strings on a guitar are thick and wound with extra winding wire: more mass per unit length, lower wave speed, lower pitch, at a comparable tension to the thinner treble strings.

Is the full circular-arc derivation something I actually need to reproduce in a board exam, or just the final formula?

CBSE board papers can ask for the derivation of $v=\sqrt{T/\mu}$ using exactly this small-element, small-angle approach, so it is worth being able to reproduce the five steps cleanly: isolate the element, identify the tension components, apply Newton's second law for circular motion, substitute for $\Delta m$ , and cancel to isolate $v$ . Even if a particular question only asks you to "state and use" the formula, knowing the derivation makes the dimensional check and the physical reasoning about tension versus mass density much more secure under exam pressure.