Speed of a Wave on a Stretched String

Pluck a guitar's low string and its top string and you hear two very different pitches, though both are the same length and under your same two hands. Tune a string tighter and its pitch rises; wrap a string in heavy wire and its pitch falls. Behind every one of those sounds is a single number — how fast a wave travels along the string — and this article works out exactly what controls it. The answer is startlingly clean. Only two properties of the string matter:

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

the tension $T$ pulling the string taut, and the linear mass density $\mu = m/L$ , the mass per metre of string. Tension is the restoring agency — it is what snaps a displaced bit of string back toward the line and lets the disturbance travel. Mass per metre is the inertia — it is what resists that snapping-back and slows the wave down. Stretch the string harder and the wave speeds up; load it heavier and the wave slows. Nothing else has a vote: not how hard you plucked it (the amplitude), not the pitch of the note (the frequency), not who or what set it moving. This is worth pausing on, because it is the first time in the chapter that a wave speed is pinned to material properties you can measure with a ruler and a set of masses. We will derive $v = \sqrt{T/\mu}$ honestly — by climbing into the frame that moves along with a pulse and applying nothing more than circular motion and a small-angle approximation — and then put it to work on the sonometer, the classroom bench where the whole relation is verified with a wire, two bridges and a hanging weight. Along the way we will meet the single most expensive slip in the topic: getting $\mu$ into the wrong units, which throws the answer off not by a little but by a factor of thirty.

Key takeaways

The wave speed on a stretched string depends on only two things: $v = \sqrt{T/\mu}$ , with $T$ the tension (N) and $\mu = m/L$ the linear mass density (kg m⁻¹).
Tension is the restoring force that carries the wave; $\mu$ is the inertia that slows it. Amplitude, frequency and the source have no effect on $v$ .
For a wire of diameter $d$ and material density $\rho$ , $\mu = \rho\pi d^2/4$ , so at fixed tension $v \propto 1/d$ — a wire twice as thick carries waves at half the speed.
Scaling: to double the speed you must quadruple the tension ( $v \propto \sqrt T$ ). Doubling the tension raises the speed only by $\sqrt2 = 1.41$ , a 41 % gain — not 100 %.
On a sonometer the tension is supplied by a hanging mass: $T = Mg$ , not $M$ . Forgetting the $g$ is a classic error.
Anchors: a wire with $\mu = 2.0\times10^{-3}$ kg m⁻¹ under $T = 80$ N carries waves at exactly $200$ m s⁻¹; a sonometer with $M = 4.0$ kg and $\mu = 1.0$ g m⁻¹ gives $198$ m s⁻¹.
Key idea: a wave on a string travels at $v = \sqrt{T/\mu}$ — set entirely by the tension pulling it back and the mass per metre resisting — so the only ways to change the speed are to change the tension or the string, never to pluck harder or faster.

The core insight

The speed of a wave on a stretched string is $v = \sqrt{T/\mu}$ , fixed by the tension $T$ (the restoring agency) and the linear mass density $\mu = m/L$ (the inertia) and by nothing else — not amplitude, not frequency — so doubling the speed needs four times the tension, and the mass per metre must always be in kg m⁻¹.

Where this lives today

Every stringed instrument is $v = \sqrt{T/\mu}$ made audible. A guitarist tuning a peg is changing $T$ to change the wave speed and so the pitch; the reason the low strings are thick and often wound with wire is to pile on $\mu$ and slow their waves without lengthening the neck. A piano technician spends a day setting the tension of two hundred strings to a fraction of a newton, because the note depends on the square root of that tension. Beyond music, the same formula governs the steel cables of a suspension bridge and the guy-wires of a mast: strike one and the pitch of the ringing tells an engineer the tension it carries — a non-contact tension gauge, used to check bridge cables and even the rigging of racing yachts. It reaches into the human body, too: a vocal-fold model and the surgeon's assessment of a tensioned tendon both lean on the fact that a taut, massive string carries waves at a speed set by exactly two things. And it is the launch pad for the next idea in this chapter — bound a string at both ends, let its waves reflect and add, and the speed $v$ turns into a ladder of resonant frequencies, the physics of every note an instrument can sing.

The physical picture

A wave travels on a string because the string is taut. Displace a small piece of it sideways and the tension in the neighbouring pieces pulls it back toward the straight line — that restoring pull is what hands the disturbance on to the next piece, and the next, so the pulse moves along. Take the tension away (let the string go slack) and there is no restoring force at all: a flick just dies where you make it. Tension is the engine of the wave. But the string also has mass, and mass resists being accelerated. The heavier each metre of string, the more sluggishly it responds to the tension's pull, and the slower the wave crawls along. So the wave speed is a tug-of-war between a stiffness that wants to snap the string back fast and an inertia that drags its feet — the same $\sqrt{\text{stiffness}/\text{inertia}}$ structure that governs every wave speed, seen in Wave Speed: v = fλ. Here the stiffness is the tension $T$ and the inertia is the mass per unit length $\mu$ . What is not in the contest is just as important. The amplitude does not appear — a big lazy pulse and a small sharp one race along at the identical speed. The frequency does not appear — the string does not know or care how often the source wiggles. The source itself does not appear. All the wave speed knows about is how tight the string is and how heavy it is per metre. This is why, on a real string, changing the tension (a tuning peg) or the string (a thicker gauge) changes the note, while plucking louder or softer does nothing to the pitch.

Figure

Riding along with a pulse, a small arc of string at the crest moves in a circle of radius R about a centre of curvature below it. The tension T pulls tangentially at each end of the arc; the two pulls lean inward by the small angle theta, and their resultant, T times delta-s over R, points straight at the centre — exactly the centripetal force the arc needs. Balancing them gives v equals the square root of T over mu.

The result, and what goes into it

v = \sqrt{\frac{T}{\mu}}, \qquad \mu = \frac{m}{L} \qquad (1)

where: $v$ is the wave speed (m s⁻¹); $T$ is the tension in the string (N); and $\mu$ is the linear mass density — the mass of the string divided by its length, in kilograms per metre. Read equation (1) as the standard wave-speed competition: tension on top is the restoring stiffness (bigger tension, faster wave), $\mu$ underneath is the inertia (heavier string, slower wave), and the square root is what makes the response gentle — you have to change the inside by a factor of four to change the speed by a factor of two. Everything hangs on getting $\mu$ right, and $\mu$ is where the marks are lost, because it must be in kg m⁻¹. A string quoted as "1.2 grams over 60 centimetres" is $\mu = 1.2\times10^{-3}\ \mathrm{kg} / 0.60\ \mathrm{m} = 2.0\times10^{-3}\ \mathrm{kg\,m^{-1}}$ — convert the grams to kilograms first. And note the tension is a force in newtons; on the sonometer bench it comes from a hanging weight, so $T = Mg$ , never the mass $M$ alone.

What sets the linear density — and the wire-diameter twist

Often you are not handed $\mu$ but must build it from the wire's geometry. A uniform wire of diameter $d$ has a circular cross-section of area $A = \pi d^2/4$ , and its mass per unit length is the material's density times that area:

\mu = \rho\,\frac{\pi d^2}{4} \qquad (2)

where: $\rho$ is the material's ordinary (volume) mass density (kg m⁻³) — steel is $7800$ , met in Wave Speed: v = fλ — and $d$ is the wire diameter (m). Substituting into $v = \sqrt{T/\mu}$ gives a memorable dependence: since $\mu \propto d^2$ , the speed goes as

v = \sqrt{\frac{T}{\rho\pi d^2/4}} = \frac{2}{d}\sqrt{\frac{T}{\rho\pi}} \quad\Rightarrow\quad v \propto \frac{1}{d} \qquad (3)

at fixed tension and material. A wire twice as thick has four times the mass per metre and so carries waves at half the speed. This is exactly why a guitar's low strings are the fat ones: the extra mass per metre slows their waves and drops their pitch, without needing a longer neck.

Scaling — why doubling the tension is not enough

Because $v = \sqrt{T/\mu}$ , the wave speed responds to changes only through a square root, and this trips people constantly. For two situations,

\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}\,\sqrt{\frac{\mu_1}{\mu_2}} \qquad (4)

so doubling the tension multiplies the speed by $\sqrt2 = 1.41$ — a rise of only 41 %, not 100 %. To genuinely double the speed you must make the tension four times larger. To triple it, nine times. On the mass side it is the same: swapping to a wire with four times the $\mu$ (the double-diameter case) halves the speed. For ratio questions, equation (4) is the whole tool — you never need to compute either speed, only the ratios of the quantities that changed.

Figure

A sonometer bench: a wire stretched over two knife-edge bridges on a hollow wooden box, running over a pulley at one end to a hanging mass M. The tension in the wire is not M but its weight, T equals M times g. Sliding the bridges changes the vibrating length; changing the hanging mass changes the tension and so the wave speed.

The sonometer: reading the tension off a hanging mass

The bench that turns $v = \sqrt{T/\mu}$ into a measurement is the sonometer (or monochord): a wire stretched along a hollow wooden box, resting on two movable knife-edge bridges that fix its vibrating length, and led over a pulley at one end to a pan of masses. The masses provide the tension — but through their weight, so $T = Mg$ . Hang $M = 4.0$ kg and, taking $g = 9.8\ \mathrm{m\,s^{-2}}$ (the CBSE/JEE/NEET value), the tension is $T = 4.0\times9.8 = 39.2\ \mathrm{N}$ , not $4.0$ . Writing $T = M$ is a guaranteed way to lose the marks and get an answer $\sqrt{9.8} = 3.1$ times too small. With the tension known and $\mu$ measured by weighing a length of the wire, the wave speed follows, and — one chapter ahead — the frequencies at which the bounded string resonates. Sliding the bridges to change $L$ , or loading the pan to change $T$ , lets a student trace out how the pitch depends on each, reproducing the three laws Mersenne found by ear in 1636. This article stops at the wave speed; the resonant frequencies of the bounded string are the next chapter's business.

Two derivations

Derivation ADeriving v = √(T/μ) by riding along with the pulse

This is the derivation the whole article is built around, and it uses nothing heavier than circular motion and a small-angle approximation — no calculus, no wave equation. The goal is to obtain $v = \sqrt{T/\mu}$ from Newton's second law alone, by a trick that turns a moving pulse into a static problem: climb into the reference frame that travels along with the pulse at speed $v$ . In that frame the pulse sits still and the string streams through it, and a small arc at the crest is simply a bit of matter moving in a circle — which we know how to handle.

Step 1. Move into the pulse's frame. Now the pulse is frozen and the whole string flows backward through it at speed $v$ . Focus on a small arc of string right at the crest, short enough to be treated as part of a circle of radius $R$ (the radius of curvature there), subtending a small angle $2\theta$ at the centre of curvature $O$ below. Its length is $\Delta s = R\,(2\theta)$ and its mass is $\Delta m = \mu\,\Delta s$ .

Step 2. Find the force on the arc. The only force is the string's tension $T$ , pulling tangentially at each end of the arc. By symmetry the two horizontal components cancel; the two components pointing inward toward $O$ add. Each inward component is $T\sin\theta$ , so the net inward force is $F = 2T\sin\theta$ . For a small angle, $\sin\theta \approx \theta$ , and since $2\theta = \Delta s/R$ , this is

F = 2T\theta = T\,\frac{\Delta s}{R}.

Step 3. Apply Newton's second law for circular motion. In the pulse frame the arc travels along a circular path of radius $R$ at speed $v$ , so it requires a centripetal (inward) force of $\Delta m\,\dfrac{v^2}{R} = \mu\,\Delta s\,\dfrac{v^2}{R}$ .

Step 4. The tension supplies exactly that centripetal force, so set them equal:

T\,\frac{\Delta s}{R} = \mu\,\Delta s\,\frac{v^2}{R}.

The arc length $\Delta s$ and the radius $R$ cancel from both sides — a signal that the result depends on neither the size of the piece we chose nor the particular curvature of the pulse. What remains is $T = \mu v^2$ , that is

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

The single approximation was $\sin\theta \approx \theta$ , exact in the limit of a small-amplitude wave — which is why $v = \sqrt{T/\mu}$ holds so long as the string is not violently displaced. Verified: T = μv² gives v = 200 m s⁻¹ for μ = 2.0×10⁻³ kg m⁻¹, T = 80 N, in verify_hs2-speed-on-a-string.py.

Go deeper · Derivation B · 1 minLinear density from a wire's diameter — why v ∝ 1/d

Derivation BLinear density from a wire's diameter — why v ∝ 1/d

In real problems the string is a wire specified by its diameter and its material, not by a mass and a length, so you must build $\mu$ before you can find the speed. The goal here is to express $\mu$ through the wire's geometry and read off how the wave speed depends on the wire's thickness — a favourite of competitive papers, which love to ask what happens to the speed when a wire is swapped for a thicker one of the same metal. The only physics is that mass is density times volume.

Step 1. A uniform wire of diameter $d$ has a circular cross-section of area $A = \pi d^2/4$ . A length $L$ of it has volume $V = AL = \pi d^2 L/4$ and, at material density $\rho$ , mass $m = \rho V = \rho\pi d^2 L/4$ .

Step 2. Divide by the length to get the linear mass density: $\mu = m/L = \rho\pi d^2/4$ — equation (2). Everything about the wire's build is now in one place: its material ( $\rho$ ) and its thickness ( $d$ ), with the mass per metre growing as the square of the diameter.

Step 3. Substitute into $v = \sqrt{T/\mu}$ : $v = \sqrt{\dfrac{T}{\rho\pi d^2/4}} = \dfrac{2}{d}\sqrt{\dfrac{T}{\rho\pi}}$ . At fixed tension and material, everything under the root except $d$ is constant, so $v \propto 1/d$ .

Step 4. Read the payoff and check a number. Doubling the diameter quarters nothing and halves the speed: $\mu$ goes up by $2^2 = 4$ , and $v$ down by $\sqrt4 = 2$ . Concretely, a steel wire ( $\rho = 7800$ kg m⁻³) of diameter $d = 0.50\ \mathrm{mm} = 5.0\times10^{-4}$ m has $\mu = 7800\times\pi\times(5.0\times10^{-4})^2/4 = 1.532\times10^{-3}\ \mathrm{kg\,m^{-1}}$ — a plausible thousandth of a kilogram per metre, which is the sanity check that you did the $\pi d^2/4$ correctly. Verified: μ = ρπd²/4 = 1.532×10⁻³ kg m⁻¹ for d = 0.50 mm steel, and v halving when d doubles, in verify_hs2-speed-on-a-string.py.

Limiting cases

Case	Result	Meaning
tension $T$ up	$v$ up as $\sqrt T$	double $T$ → $v \times 1.41$ ; to double $v$ you need $4T$
linear density $\mu$ up	$v$ down as $1/\sqrt\mu$	a heavier string carries slower waves
diameter $d$ doubled (same metal)	$\mu \times 4$ , $v \div 2$	$\mu \propto d^2$ , so $v \propto 1/d$
amplitude changed	$v$ unchanged	a louder pluck does not travel faster
frequency changed	$v$ unchanged	the medium fixes $v$ ; then $\lambda = v/f$
tension $\to 0$ (slack string)	$v \to 0$	no restoring force, no wave
$\mu$ left in g m⁻¹	$v$ off by $\sqrt{1000} \approx 31.6\times$	the units disaster below

Common trapThe g m⁻¹ disaster — a factor of thirty, either way

The costliest mistake in this whole topic is feeding the linear density into $v = \sqrt{T/\mu}$ in the wrong units. The formula demands $\mu$ in kilograms per metre while $T$ is in newtons; a value in grams per metre is a thousand times different, and because $v$ depends on $\sqrt{1/\mu}$ , getting it wrong throws the answer off by $\sqrt{1000} = 31.6$ — a factor of thirty, not a rounding error. And it can go either way. A genuine $200\ \mathrm{m\,s^{-1}}$ wave, worked with $\mu$ a thousand times too small (over-converting, dividing an already-SI value by 1000), balloons to $6.3\ \mathrm{km\,s^{-1}}$ — faster than sound in steel, and physically absurd. Worked with $\mu$ a thousand times too large (leaving grams per metre in the formula as though it were kg m⁻¹), it collapses to $6.3\ \mathrm{m\,s^{-1}}$ — slower than a jog. Both are nonsense you can catch in one glance if you carry a benchmark in your head: waves on a musical string run at a few hundred metres a second. Always convert $\mu$ to kg m⁻¹ before substituting, and sanity-check the size of your answer.

Worked examples

Board level — a wire, and a sonometer (CBSE/ICSE pattern)

(a) A string of length 0.60 m has mass 1.2 g and is held under a tension of 80 N. Find its linear mass density and the speed of a transverse wave on it. (b) A sonometer wire of linear density 1.0 g m⁻¹ is stretched by hanging a mass of 4.0 kg over its pulley ( $g = 9.8\ \mathrm{m\,s^{-2}}$ ). Find the tension and the wave speed.

Step 1. Part (a). Convert the mass to kilograms first: $m = 1.2\ \mathrm{g} = 1.2\times10^{-3}$ kg. Then $\mu = m/L = 1.2\times10^{-3}/0.60 = \boxed{2.0\times10^{-3}\ \mathrm{kg\,m^{-1}}}$ .

Step 2. Speed: $v = \sqrt{T/\mu} = \sqrt{80/(2.0\times10^{-3})} = \sqrt{40000} = \boxed{200\ \mathrm{m\,s^{-1}}}$ — a clean, exact number, and comfortably in the "few hundred m s⁻¹" range a musical string should give.

Step 3. Part (b). The tension is the weight of the hanging mass: $T = Mg = 4.0\times9.8 = \boxed{39.2\ \mathrm{N}}$ — not $4.0$ . Convert the linear density: $\mu = 1.0\ \mathrm{g\,m^{-1}} = 1.0\times10^{-3}\ \mathrm{kg\,m^{-1}}$ .

Step 4. Speed: $v = \sqrt{T/\mu} = \sqrt{39.2/(1.0\times10^{-3})} = \sqrt{39200} = \boxed{198\ \mathrm{m\,s^{-1}}}$ (to three figures, $198.0$ ). Had you used $T = 4.0$ N instead of $39.2$ N, you would have got $63\ \mathrm{m\,s^{-1}}$ , $\sqrt{9.8} = 3.1$ times too small — the tell-tale of a dropped $g$ .

Competitive level — a steel wire and a thicker twin (JEE-Main/NEET pattern)

A steel wire of diameter 0.50 mm and density 7800 kg m⁻³ is stretched to a tension of 60 N. (a) Find its linear mass density. (b) Find the wave speed. (c) The wire is replaced by one of the same steel but twice the diameter, at the same tension. Find the new wave speed without recomputing from scratch.

Step 1. Part (a). Build $\mu$ from the geometry, keeping SI throughout: $d = 0.50\ \mathrm{mm} = 5.0\times10^{-4}$ m, so $\mu = \rho\pi d^2/4 = 7800\times\pi\times(5.0\times10^{-4})^2/4 = \boxed{1.532\times10^{-3}\ \mathrm{kg\,m^{-1}}}$ .

Step 2. Part (b). $v = \sqrt{T/\mu} = \sqrt{60/(1.532\times10^{-3})} = \sqrt{39177} = \boxed{197.9\ \mathrm{m\,s^{-1}}}$ — again a few hundred m s⁻¹, so the units are right.

Step 3. Part (c). Do not recompute — scale. Twice the diameter means $\mu' = 2^2\mu = 4\mu$ at the same tension, and $v \propto 1/\sqrt\mu$ , so $v' = v/\sqrt4 = v/2 = 197.9/2 = \boxed{99.0\ \mathrm{m\,s^{-1}}}$ . The thicker wire carries waves at half the speed — the $v \propto 1/d$ rule in action. Timing note: about 2 minutes. Part (c) is the whole point: reach for the ratio $v \propto 1/d$ rather than grinding $\sqrt{T/\mu}$ a second time. The trap is to forget that diameter enters $\mu$ as $d^2$ , so that doubling $d$ changes the speed by a factor of 2, not 4 or $\sqrt2$ .

Exam traps

$\mu$ in the wrong units. The single biggest error: $\mu$ must be in kg m⁻¹, not g m⁻¹, or the answer is off by $\sqrt{1000} = 31.6\times$ . Convert grams to kilograms before you touch the square root, and sanity-check against "a few hundred m s⁻¹".

$T = M$ instead of $T = Mg$ . On a sonometer or any hanging-mass bench the tension is the weight of the load, $Mg$ . Dropping the $g$ makes the answer $\sqrt g \approx 3.1$ times too small. State which $g$ you use — $9.8\ \mathrm{m\,s^{-2}}$ for CBSE/JEE/NEET.

Thinking a louder pluck travels faster. Amplitude does not enter $v = \sqrt{T/\mu}$ . A hard pluck and a gentle one send waves along the string at the same speed; only the tension and the string decide $v$ .

Using the total mass $m$ instead of $\mu$ . The formula is $v = \sqrt{T/\mu}$ with $\mu = m/L$ , not $v = \sqrt{T/m}$ . Forgetting to divide the mass by the length inflates $\mu$ and shrinks the speed.

Doubling the tension to double the speed. Because $v \propto \sqrt T$ , doubling $T$ gives only $\sqrt2 = 1.41$ times the speed (a 41 % rise). To double the speed you need four times the tension.

Board scope. This derivation is core only for the Indian boards (CBSE, ICSE, JEE-Main, JEE-Advanced, NEET). A-Level (9702), IGCSE, AP Physics 2 and IB candidates: $v = \sqrt{T/\mu}$ is not a required outcome for you — A-Level meets the stretched string only qualitatively, through Melde's experiment and stationary waves, and IGCSE, AP and IB need just the wave model and $v = f\lambda$ . Treat this page as enrichment, not syllabus.

Origins · 1 minBrook Taylor and how this was discovered

Brook Taylor

1685–1731 · England

Brook Taylor is remembered by every calculus student for the Taylor series, published in his 1715 Methodus Incrementorum Directa et Inversa — but two years earlier, in a 1713 paper "De motu nervi tensi" ("On the motion of a stretched string"), he did something no one had managed before: he calculated the frequency at which a taut string vibrates, from its tension, its length and its weight. Marin Mersenne had found the same relationships by ear in 1636, but Taylor derived them, treating the string as a curve pulled back toward its rest line by tension — the first genuine piece of theory behind the sound of a plucked string, and the direct ancestor of the $v = \sqrt{T/\mu}$ on this page. A fellow and later secretary of the Royal Society, he worked at the frontier where the new calculus met real vibrating matter, and it is fitting that the string, humblest of instruments, gave mathematical physics one of its first exact results.

History

1636

— Marin Mersenne, in Harmonie universelle, states the three laws of the vibrating string empirically: pitch falls with length, rises with the square root of tension, and falls with the square root of mass per unit length.

1713

— Brook Taylor, in "De motu nervi tensi", makes the first theoretical calculation of a stretched string's vibration frequency from its tension, length and weight.

1715

— Taylor publishes Methodus Incrementorum Directa et Inversa, developing the underlying mathematics of increments.

1747

— Jean le Rond d'Alembert derives the wave equation for the string, whose travelling-wave speed is exactly

\sqrt{T/\mu}

1859

— Franz Melde's tuning-fork experiment shows directly how a string's standing-wave pattern depends on its tension and its mass per unit length.

Going further

Texts. NCERT Physics Class 11, Ch. 15 "Waves", §15.5 (speed of a transverse wave on a stretched string) · H. C. Verma, Concepts of Physics Vol. 1, "Wave Motion and Waves on a String" — the pulse-frame derivation and sonometer problems · Halliday, Resnick & Walker, "Waves — I", wave speed on a stretched string.

Provenance. Standard results. Recompute in `verify_hs2-speed-on-a-string.py`: $\mu = 1.2\times10^{-3}/0.60 = 2.0\times10^{-3}$ kg m⁻¹ and $v = \sqrt{80/0.002} = 200$ m s⁻¹; the sonometer $T = 4.0\times9.8 = 39.2$ N, $\mu = 1.0\times10^{-3}$ , $v = \sqrt{39200} = 198.0$ m s⁻¹, and the dropped- $g$ value $\sqrt{4.0/1.0\times10^{-3}} = \sqrt{4000} = 63.2$ m s⁻¹; the steel wire $\mu = 7800\pi(5.0\times10^{-4})^2/4 = 1.532\times10^{-3}$ kg m⁻¹, $v = \sqrt{60/\mu} = 197.9$ m s⁻¹ and the double-diameter $99.0$ m s⁻¹; the scalings $\sqrt2 = 1.41$ (double $T$ ) and the units-blunder factor $\sqrt{1000} = 31.6$ ( $200 \to 6.3\ \mathrm{km\,s^{-1}}$ or $6.3\ \mathrm{m\,s^{-1}}$ ); the quiz value $T = 5.0\times9.8 = 49$ N.

Prerequisites. Wave Speed: v = fλ — the $\sqrt{\text{stiffness}/\text{inertia}}$ family this article's formula belongs to, and the dimensional argument that already told us the form. Leads to. The string-tension numericals archetype, which drills $v = \sqrt{T/\mu}$ chained with $v = f\lambda$ · Standing Waves & Stationary-Wave Formation, where a bounded string turns this one speed into a whole ladder of resonant notes.

Go deeper (UG). Where $v = \sqrt{T/\mu}$ comes from when the string is imagined as a row of tiny masses joined by springs, and how the continuous formula emerges as the masses are packed infinitely close, is developed in Discrete Loaded String and the Continuum Limit in the university mechanics track. It is enrichment; the pulse-frame derivation here is complete on its own.

Why does the amplitude not affect the wave speed on a string?

Because the speed comes from a balance between the tension pulling each bit of string back and the mass of that bit resisting — and neither of those depends on how far the string was displaced. A bigger pluck moves the string further, so each bit travels faster up and down (its particle speed grows), but the restoring pull per unit displacement and the mass per metre are unchanged, so the wave travels along at the same $\sqrt{T/\mu}$ . This is really the small-angle approximation in the derivation doing its job: as long as the displacement is modest, the geometry that produced $v = \sqrt{T/\mu}$ does not know the amplitude. Only at violently large amplitudes, where the string stretches noticeably and the tension itself changes, does the speed start to depend on how hard you plucked — a nonlinear effect outside this course.

On a sonometer, why is the tension Mg and not M?

Because tension is a force, measured in newtons, and the hanging mass provides that force through its weight, not its mass. A mass $M$ pulled down by gravity exerts a downward force of $Mg$ , and the wire, running over the pulley, feels exactly that as its tension. Using $M$ alone confuses a mass (kilograms) with a force (newtons) — dimensionally wrong, and numerically it makes your speed $\sqrt g \approx 3.1$ times too small. So always write $T = Mg$ , and state your $g$ (9.8 m s⁻² for the Indian boards). The pulley's only job is to turn the vertical weight into a horizontal pull without changing its size; a good frictionless pulley passes the full $Mg$ into the wire.

What exactly is linear mass density, and how is it different from ordinary density?

Ordinary (volume) density $\rho$ is mass per unit volume, in kg m⁻³ — a property of the material (steel is 7800 kg m⁻³ whatever its shape). Linear mass density $\mu$ is mass per unit length, in kg m⁻¹ — a property of a particular string, depending on both what it is made of and how thick it is. The two are linked by the cross-section: $\mu = \rho A = \rho\pi d^2/4$ for a wire of diameter $d$ . So two wires of the same steel but different thickness share one $\rho$ but have different $\mu$ . In $v = \sqrt{T/\mu}$ it is the linear density that matters, because a wave runs along the string and "feels" the mass sitting in each metre of its path, not the mass packed into each cubic metre.

Why is the answer so sensitive to the units of μ — is a factor of 30 really right?

Yes, and the culprit is the square root. A gram is a thousandth of a kilogram, so $\mu$ in g m⁻¹ is a thousand times the number it should be in kg m⁻¹ (or a thousandth, if you mis-convert the other way). The speed depends on $\sqrt{1/\mu}$ , and $\sqrt{1000} = 31.6$ , so a factor-of-1000 slip in $\mu$ becomes a factor-of-31.6 slip in $v$ . That is why a genuine 200 m s⁻¹ wave can come out as 6.3 m s⁻¹ or 6.3 km s⁻¹ depending on which way you err. The defence is simple: convert $\mu$ to kg m⁻¹ before substituting, and keep a benchmark in mind — musical strings carry waves at a couple of hundred metres per second, so any answer wildly off that should make you recheck the units.

Does this formula work for longitudinal waves too, or only transverse ones?

$v = \sqrt{T/\mu}$ is specifically the speed of a transverse wave on a string — the up-and-down (or side-to-side) waves you make by flicking a rope, where the restoring force is the string's tension. A string can also carry a longitudinal wave (a compression running along its length), but that one is restored by the string's stretching stiffness, not its tension, and travels at a quite different, usually much higher, speed set by the material's Young's modulus, $\sqrt{Y/\rho}$ . When a problem talks about waves on a string, a plucked guitar, or a sonometer, it means the transverse wave, and $v = \sqrt{T/\mu}$ is the right tool. The longitudinal case belongs with sound in a solid rod.

The derivation jumps into a frame moving with the pulse. Is that a legitimate trick?

Completely legitimate, and it is one of the most useful moves in physics. Newton's laws hold in any frame moving at constant velocity (an inertial frame), so as long as the pulse travels at a steady speed $v$ , the frame gliding along with it is inertial and we may apply $F = ma$ inside it. The payoff is that a hard, time-dependent problem — a pulse sweeping along a string — becomes a simple, static one: a fixed hump of string with material flowing through it, each arc executing steady circular motion. We compute the centripetal force that circular motion demands and note the tension supplies it, and out drops $v = \sqrt{T/\mu}$ . The moving frame does not change the physics; it just chooses the vantage point from which the physics is easiest to see.