ARCHETYPE: String Tension & Wave-Speed Numericals

Pluck a guitar's low string and its top string together: the low one sags out a slow, fat wave, the top one a fast, tight buzz. Same length, same player, same instant — wildly different wave speeds. The difference is not in how hard you pluck; it is baked into the wires themselves, into a single number the wire carries whether or not anyone touches it. Every string-and-tension numerical you will ever be set is the story of extracting that number and feeding it to one formula. The formula is $v = \sqrt{T/\mu}$ : wave speed equals the square root of tension over linear mass density. Tension $T$ is the pull that supplies the restoring force; $\mu = m/L$ is the wire's mass per unit length, its inertia. That is the entire physics of the topic. What makes it an ARCHETYPE — a problem-solving article rather than a concept — is that examiners hide the same two ingredients behind a dozen disguises: a sonometer with a hanging mass, a steel wire specified by its diameter, two wires spliced at a junction, a question that asks not for a speed but for the ratio of two speeds. Every one of them is $v = \sqrt{T/\mu}$ with the tension or the linear density dressed up. So this article drills the setup. We fix a five-step master pattern — get $\mu$ honestly in kilograms per metre, get $T$ from the hanging mass as $Mg$ , take the square root, then chain with $v = f\lambda$ if a frequency appears — and we run it forwards through board problems and competitive traps. The single loudest warning on the page is a unit: $\mu$ in grams per metre instead of kilograms per metre is a factor-of-a-thousand landmine that makes a $200\ \mathrm{m/s}$ wave come out faster than steel. One honest scope note before we start: $v = \sqrt{T/\mu}$ is an India-facing outcome — CBSE, ICSE, JEE and NEET — and is not on the A-Level, AP or IB syllabus, which stop at $v = f\lambda$ .

Key takeaways

One formula runs the topic: $v = \sqrt{T/\mu}$ — wave speed from tension $T$ (the restoring pull) and linear mass density $\mu$ (the inertia).
Get $\mu$ in kg/m: $\mu = m/L$ from a mass and a length, or $\mu = \rho\,\pi d^2/4$ from a wire's diameter and material density. Never leave it in g/m.
Get $T$ from the bench: a hanging mass $M$ gives $T = Mg$ — not $T = M$ . State which $g$ the board uses.
Chain with $v = f\lambda$ whenever a frequency is quoted; and at a junction the frequency never changes, so $\lambda$ tracks $v$ .
Speed scales as $\sqrt{T}$ : quadruple the tension to double the speed, not merely double it. Fatten the wire (diameter $\times 2$ , so $\mu \times 4$ ) and the speed halves.
Key idea: get $\mu$ in kg/m and $T$ in newtons honestly, then $v = \sqrt{T/\mu}$ chained with $v = f\lambda$ solves every string numerical.

The core insight

Every stretched-string numerical reduces to two honest extractions — the linear mass density $\mu = m/L$ (or $\rho\,\pi d^2/4$ ) in kilograms per metre, and the tension $T = Mg$ in newtons — fed into $v = \sqrt{T/\mu}$ and, when a frequency is present, chained through $v = f\lambda$ .

Where this lives today

Every stringed instrument on Earth is a live demonstration of $v = \sqrt{T/\mu}$ , and every time a musician tunes one they are solving it. Turning a tuning peg changes $T$ ; because pitch rises with the wave speed, and speed rises as $\sqrt{T}$ , a guitarist tightening a string to raise it by an octave must nearly quadruple its tension. Why then does a guitar not need one impossibly tight string and one impossibly slack one to span its range? Because the maker varies $\mu$ instead: the low strings are wound with extra wire to fatten $\mu$ without stiffening them, so all six sit at comfortable, similar tensions — exactly the $\mu = \rho\,\pi d^2/4$ trade-off from Example 3. A piano does the same across eighty-eight notes, its bass strings copper-wound and its treble strings thin and bright. Beyond music, the same formula sizes the steel cables of suspension bridges, whose engineers must know the speed at which a gust's wave will run along a cable and whether it can build to a resonance — the physics that, left unchecked, once tore apart the Tacoma Narrows bridge. A stretched wire's hidden number, it turns out, is worth getting right.

The setup

The geometry is always the same wire under always the same two influences, so in this archetype the setup is a bookkeeping discipline. A uniform string of mass $m$ and length $L$ is pulled taut by a tension $T$ . The wave speed it will carry is

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

where: $v$ — the speed of a transverse wave along the string (metres per second); $T$ — the tension, the restoring pull, in newtons; and $\mu$ — the linear mass density, the mass per unit length, in kilograms per metre. Read it physically: tension is the spring that snaps a displaced bit of string back, so more tension means a faster wave; $\mu$ is the inertia that resists being snapped back, so a heavier string carries a slower wave. Amplitude and frequency are nowhere in it — the string's own speed is set before any wave is even launched. (The derivation of (1) itself is the previous article's, Speed of a Wave on a Stretched String; here we use it.) Two ingredients feed the formula, and each has an honest and a careless way to get it. The linear density can come from a mass and a length, or, for a wire specified by its thickness, from its diameter and material:

\mu = \dfrac{m}{L} \qquad\text{or}\qquad \mu = \rho\,\dfrac{\pi d^2}{4} \qquad (2)

where: $\rho$ — the material's ordinary (volume) density in $\mathrm{kg\,m^{-3}}$ ; $d$ — the wire's diameter in metres. The second form is just "density times cross-sectional area": a wire is a long thin cylinder, and its mass per length is $\rho$ times $\pi d^2/4$ . And the tension, on the standard bench where a mass hangs over a pulley, is a weight:

T = Mg \qquad (3)

where: $M$ — the hanging mass in kilograms; $g$ — the acceleration due to gravity, which CBSE, ICSE, JEE and NEET take as $9.8\ \mathrm{m\,s^{-2}}$ unless the paper says otherwise. The commonest one-mark loss on the whole topic is writing $T = M$ and forgetting the $g$ .

Figure

The sonometer bench — a hanging mass sets the tension T equals M g, which drives the wave speed

The master results are (1), (2) and (3), and the whole skill is choosing which to reach for and keeping every unit in SI. Two more facts, stated now, save time later. First, the scaling: because $v \propto \sqrt{T}$ , a ratio question never needs a single value of $v$ — you write $v_2/v_1 = \sqrt{T_2/T_1}\cdot\sqrt{\mu_1/\mu_2}$ and read the answer off. Second, the junction: when a wave passes from one wire into another, the joint is driven at one rate, so the frequency is common to both — and since $v = f\lambda$ , whatever happens to $v$ happens to $\lambda$ in step, while $f$ sits still.

The master solution pattern

Five steps solve every problem on this page, whether it is a three-mark bench question or a junction trap. Fix the order and you will never be lost.

Step 1. Get $\mu$ in kilograms per metre — honestly. If given a mass and a length, $\mu = m/L$ (put $m$ in kilograms first: $1.2\ \mathrm{g}$ is $1.2\times10^{-3}\ \mathrm{kg}$ ). If given a diameter and a material, $\mu = \rho\,\pi d^2/4$ . Write the unit next to the number and confirm it reads $\mathrm{kg/m}$ .

Step 2. Get the tension in newtons. On a hanging-mass bench, $T = Mg$ ; state the $g$ your board uses. If the tension is given directly, use it — but check it is a force in newtons, not a mass in kilograms.

Step 3. Take the square root: $v = \sqrt{T/\mu}$ . Do the division inside first, then the root. A wave on a school wire lands in the range of a couple of hundred metres per second; if you get $6\ \mathrm{km/s}$ or $6\ \mathrm{m/s}$ , a unit has slipped by a factor of a thousand.

Step 4. Chain with $v = f\lambda$ if — and only if — a frequency or wavelength is in play. Given $f$ , the wavelength is $\lambda = v/f$ ; given $\lambda$ , the frequency is $f = v/\lambda$ .

Step 5. If the question asks for a ratio or a "what factor" change, do not compute $v$ at all. Use $v_2/v_1 = \sqrt{T_2/T_1}\cdot\sqrt{\mu_1/\mu_2}$ , and at a junction remember $f$ is unchanged so $\lambda_2/\lambda_1 = v_2/v_1$ .

Derivation AThe master pattern on a bench — mass, length, tension, speed, wavelength

This derivation runs the five-step pattern end to end on one clean bench problem, so the sequence becomes automatic: a wire of known mass and length, a known tension, a known driving frequency, and we want its wave speed and the wavelength on it. The discipline being trained is not algebra — the algebra is one square root — but unit hygiene, because every mark lost on this topic is lost by carrying grams into a kilograms formula or forgetting the $g$ in $T = Mg$ . Follow the units as closely as the numbers.

Step 1. Linear density, in kilograms per metre. A wire of mass $m = 1.2\ \mathrm{g} = 1.2\times10^{-3}\ \mathrm{kg}$ and length $L = 0.60\ \mathrm{m}$ has $\mu = m/L = 1.2\times10^{-3}/0.60 = 2.0\times10^{-3}\ \mathrm{kg/m}$ . Note the conversion of grams to kilograms happened first, before the division.

Step 2. Tension, in newtons. Here it is given directly as $T = 80\ \mathrm{N}$ — a force, already in newtons, so nothing to convert. (Had a hanging mass been quoted instead, this step would be $T = Mg$ .)

Step 3. Wave speed. $v = \sqrt{T/\mu} = \sqrt{80/(2.0\times10^{-3})} = \sqrt{40000} = 200\ \mathrm{m/s}$ — a clean value because the setter chose the numbers, and comfortably in the few-hundred range a real wire gives.

Step 4. Chain with $v = f\lambda$ . Driven at $f = 250\ \mathrm{Hz}$ , the wavelength on the wire is $\lambda = v/f = 200/250 = 0.80\ \mathrm{m}$ . The frequency came from the source; the speed came from the wire; the wavelength is whatever the two agree on. Verified: $\mu = 2.0\times10^{-3}$ kg/m, $v = 200$ m/s, $\lambda = 0.80$ m are recomputed in verify_hs2-string-wave-problems.py.

Go deeper · Derivation B · 1 minThe variant — scalings and the wire-geometry route, where you never compute v

Derivation BThe variant — scalings and the wire-geometry route, where you never compute v

The second skill an archetype must build is the ratio move: recognising that "what factor" and "by how much" questions are answered without ever finding a single speed, straight from $v \propto \sqrt{T}$ and $v \propto 1/\sqrt{\mu}$ . This derivation collects the scalings — what quadrupling the tension does, what fattening the wire does — and derives the wire-geometry form $\mu = \rho\,\pi d^2/4$ that turns a diameter into a linear density. The reward is speed under exam pressure: a ratio you can write in one line beats a pair of square roots you can slip a unit in.

Step 1. From $v = \sqrt{T/\mu}$ , form the ratio of two states of the same wire: $v_2/v_1 = \sqrt{T_2/T_1}$ . To double the speed you need $\sqrt{T_2/T_1} = 2$ , that is $T_2 = 4T_1$ — quadruple the tension, not double it. This is the single most-tested scaling on the topic.

Step 2. Now vary the wire instead of the tension. A wire of diameter $d$ and material density $\rho$ is a cylinder, so its mass per length is density times cross-sectional area: $\mu = \rho\,\pi d^2/4$ . Linear density therefore grows as the square of the diameter.

Step 3. Combine. At fixed tension and material, $v \propto 1/\sqrt{\mu} \propto 1/d$ . Double the diameter and $\mu$ quadruples, so the speed halves. A thick bass string is slow for exactly this reason — its extra girth is extra inertia, growing as $d^2$ .

Step 4. The junction, as a ratio. Two wires under the same tension are joined; the second is four times as dense, $\mu_2 = 4\mu_1$ . Then $v_2/v_1 = \sqrt{\mu_1/\mu_2} = \sqrt{1/4} = 1/2$ : the wave halves its speed crossing into the heavier wire. Because the joint is driven at one frequency, $f$ is unchanged, so $\lambda_2/\lambda_1 = v_2/v_1 = 1/2$ — the wavelength halves too. Verified: the scalings ( $T\times4 \Rightarrow v\times2$ ; $d\times2 \Rightarrow v\div2$ ; $\mu_2 = 4\mu_1 \Rightarrow v_2 = v_1/2$ ) are recomputed in verify_hs2-string-wave-problems.py.

Common trapGrams per metre in the speed formula — the factor-of-a-thousand landmine

The linear density $\mu$ in $v = \sqrt{T/\mu}$ must be in kilograms per metre. Feed it grams per metre and you are wrong by a factor of $1000$ inside a square root, so the speed comes out wrong by $\sqrt{1000} = 31.6$ — and the sign of that error is worth internalising, because the giveaway is different each way. If your $\mu$ is a thousand times too small (the classic over-conversion, dividing an already-SI value by $1000$ "to get kilograms"), $v$ comes out $31.6$ times too big: a genuine $200\ \mathrm{m/s}$ wave balloons to about $6.3\ \mathrm{km/s}$ — faster than sound travels in solid steel (about $5\ \mathrm{km/s}$ ), which is physically impossible for a school wire and should stop you cold. If instead $\mu$ is a thousand times too large (plugging the raw gram-number $2.0$ straight in for $2.0\times10^{-3}$ ), $v$ comes out $31.6$ times too small: that $200\ \mathrm{m/s}$ wave shrinks to about $6.3\ \mathrm{m/s}$ , slower than a jogger. Either way the fix is the same reflex: before the square root, look at $\mu$ and read its unit aloud. If it does not say "kilograms per metre," you are not ready to take the root. Convert grams to kilograms once, at the moment you first write $\mu$ down, and never again.

Worked examples

1 · Board level — mass, length, tension, then a wavelength (CBSE/ICSE pattern, 3 marks)

A uniform wire of length $0.60\ \mathrm{m}$ and mass $1.2\ \mathrm{g}$ is stretched to a tension of $80\ \mathrm{N}$ . Find (a) the speed of a transverse wave on it, and (b) the wavelength when the wire is driven at $250\ \mathrm{Hz}$ .

Step 1. Linear density in kg/m: convert the mass first, $m = 1.2\ \mathrm{g} = 1.2\times10^{-3}\ \mathrm{kg}$ , then $\mu = m/L = 1.2\times10^{-3}/0.60 = \boxed{2.0\times10^{-3}\ \mathrm{kg/m}}$ .

Step 2. Tension is given as a force, $T = 80\ \mathrm{N}$ — no conversion.

Step 3. Speed: $v = \sqrt{T/\mu} = \sqrt{80/(2.0\times10^{-3})} = \sqrt{40000} = \boxed{200\ \mathrm{m/s}}$ .

Step 4. Chain with $v = f\lambda$ : $\lambda = v/f = 200/250 = \boxed{0.80\ \mathrm{m}}$ . Sanity: $200\ \mathrm{m/s}$ is a believable wire speed, and $0.80\ \mathrm{m}$ is a sensible wavelength on a $0.60\ \mathrm{m}$ wire being driven hard.

2 · Board level — the sonometer, where the tension is a hanging weight (CBSE/NEET pattern, 3 marks)

A sonometer wire of linear mass density $1.0\ \mathrm{g/m}$ carries a wave when a mass of $4.0\ \mathrm{kg}$ is hung from it over the end pulley. Taking $g = 9.8\ \mathrm{m\,s^{-2}}$ , find the tension and the wave speed.

Step 1. Linear density in kg/m: $\mu = 1.0\ \mathrm{g/m} = 1.0\times10^{-3}\ \mathrm{kg/m}$ . This single conversion is the whole trap of the question.

Step 2. Tension from the hanging mass: $T = Mg = 4.0 \times 9.8 = \boxed{39.2\ \mathrm{N}}$ — not $4.0\ \mathrm{N}$ , and certainly not $4.0$ .

Step 3. Speed: $v = \sqrt{T/\mu} = \sqrt{39.2/(1.0\times10^{-3})} = \sqrt{39200} = \boxed{198.0\ \mathrm{m/s}}$ . Had the paper set $g = 10$ , the tension would be $40\ \mathrm{N}$ and $v = 200\ \mathrm{m/s}$ — quote whichever $g$ the question states, and never mix the two.

3 · Competitive level — a wire specified by its diameter (JEE-Main/NEET pattern)

A steel wire of diameter $0.50\ \mathrm{mm}$ (steel density $\rho = 7800\ \mathrm{kg\,m^{-3}}$ ) is stretched to a tension of $60\ \mathrm{N}$ . (a) Find its linear mass density and the wave speed. (b) What tension would be needed to double the wave speed?

Step 1. Linear density from the geometry, in SI. Diameter $d = 0.50\ \mathrm{mm} = 5.0\times10^{-4}\ \mathrm{m}$ , so $\mu = \rho\,\pi d^2/4 = 7800 \times \pi \times (5.0\times10^{-4})^2/4 = \boxed{1.532\times10^{-3}\ \mathrm{kg/m}}$ .

Step 2. Speed: $v = \sqrt{T/\mu} = \sqrt{60/(1.532\times10^{-3})} = \sqrt{39177} = \boxed{197.9\ \mathrm{m/s}}$ .

Step 3. Part (b), the scaling — do not recompute anything. Since $v \propto \sqrt{T}$ , doubling $v$ needs $T \to 4T = 4 \times 60 = \boxed{240\ \mathrm{N}}$ . Quadruple the tension for double the speed; the tempting "double the tension" only gives a factor $\sqrt{2} = 1.41$ . Timing note: under two minutes. The only real work is Step 1's $\rho\,\pi d^2/4$ — keep the diameter in metres and square it before multiplying. Part (b) is a five-second scaling if you resist the urge to find the new speed; candidates who recompute $v$ at $240\ \mathrm{N}$ are wasting a minute to confirm what $v \propto \sqrt{T}$ already told them.

4 · Competitive level — the junction of two wires (JEE-Advanced pattern)

Two wires, the second four times as dense as the first ( $\mu_2 = 4\mu_1$ ), are joined and kept under the same tension $T = 100\ \mathrm{N}$ ; $\mu_1 = 1.0\times10^{-3}\ \mathrm{kg/m}$ . A wave of frequency $100\ \mathrm{Hz}$ travels along the first wire and crosses into the second. Find the wave speed and wavelength in each wire.

Step 1. Speeds from $v = \sqrt{T/\mu}$ . In wire 1: $v_1 = \sqrt{100/(1.0\times10^{-3})} = \sqrt{100000} = \boxed{316.2\ \mathrm{m/s}}$ . In wire 2, with $\mu_2 = 4.0\times10^{-3}\ \mathrm{kg/m}$ : $v_2 = \sqrt{100/(4.0\times10^{-3})} = \sqrt{25000} = \boxed{158.1\ \mathrm{m/s}}$ , which is exactly $v_1/2$ as the ratio $\sqrt{\mu_1/\mu_2} = 1/2$ promised.

Step 2. The frequency is set by the source and is the same in both wires — the joint cannot oscillate at two rates at once. So $f = 100\ \mathrm{Hz}$ throughout.

Step 3. Wavelengths from $v = f\lambda$ . Wire 1: $\lambda_1 = v_1/f = 316.2/100 = \boxed{3.162\ \mathrm{m}}$ . Wire 2: $\lambda_2 = v_2/f = 158.1/100 = \boxed{1.581\ \mathrm{m}}$ — halved, tracking the halved speed while $f$ stands still.

Step 4. Read the physics off the numbers: crossing into the denser wire, the wave slows and its wavelength shrinks in the same proportion, but its frequency — the pitch you would hear — is untouched. (What reflects at that junction, and why it flips, is the next article's story, Reflection of Waves & Phase Change at Boundaries.) Timing note: about two minutes. The shortcut is to compute $v_1$ once and halve it for $v_2$ rather than taking a second square root, and to write $\lambda_2 = \lambda_1/2$ directly. The examiner's bait is a candidate who "recomputes" a new frequency in wire 2 — there is none; $f$ never changes at a junction.

Figure

A wave crossing from a thin wire into a thick one — speed and wavelength halve, frequency stays fixed

Exam traps

The limiting behaviours sit in this table too — the scaling extremes are exactly where "by what factor" distractors come from.

the trap	why it is wrong	the fix
$\mu$ in g/m	a factor of $1000$ inside a square root is $31.6$ in the speed	convert to kg/m once, when you first write $\mu$ ; read the unit aloud before the root
$T = M$ (dropping $g$ )	tension is a force; a hanging mass supplies $Mg$ newtons	on any hanging-mass bench, $T = Mg$ — and state the $g$
"double the tension doubles the speed"	$v \propto \sqrt{T}$ , so $T\times2$ gives only $v\times1.41$	to double $v$ , quadruple $T$ ; to treble $v$ , use $9T$
using the wire's total mass as $\mu$	$\mu$ is mass per length, $m/L$ , not $m$	always divide the mass by the length; check the unit is kg/m
changing $f$ at a junction	the joint is driven at one rate; $f$ is common to both wires	hold $f$ fixed; let $v$ and $\lambda$ change together, $\lambda = v/f$
the limits	$T \to 0$ gives $v \to 0$ (a slack string carries nothing); $\mu \to 0$ sends $v \to \infty$ (a massless string would be infinitely fast)	a wave faster than steel or slower than a walk means a unit slipped by $1000$

Origins · 1 minFélix Savart and how this was discovered

Félix Savart

1791–1841 · France

Félix Savart trained as a physician but gave his life to the physics of vibrating strings and plates. Fascinated by the violin, he studied how a bowed string sets its wood singing and even built a trapezoidal violin to test his ideas about resonance; his memoir on stringed and bowed instruments, communicated to the Paris Académie, brought the vibrating string under laboratory scrutiny. To measure frequency directly he devised the toothed "Savart wheel," whose teeth strike a card a countable number of times a second, turning a pitch into a number — the ancestor of every sonometer bench. His name also rides on the Biot–Savart law of magnetism, but that is a different Savart paper from a different year; on this page he is the acoustician, the man who made the frequency of a string something you could weigh and count.

History

1636

— Marin Mersenne, in Harmonie universelle, states the laws of the vibrating string: its pitch rises as it is shortened, as the tension is raised, and as the string is made lighter — all found by ear and balance-scale.

1713

— Brook Taylor calculates the fundamental frequency of a stretched string from its tension, length and weight, the first theory to sit beneath Mersenne's measurements.

1747

— Jean le Rond d'Alembert shows the stretched string obeys the wave equation, fixing the travelling-wave speed that √(T/μ) sets.

1819

— Félix Savart brings the physics of the violin and the bowed string into the laboratory, and later measures frequency directly with his toothed wheel.

1877

— Lord Rayleigh's The Theory of Sound gathers string speed, tension and linear density into one systematic account.

Going further

Texts. NCERT Physics Class 11, Ch. 15 "Waves", §15.3 · H. C. Verma, Concepts of Physics Vol. 1, "Waves on a String" — the sonometer and junction problems live here · D. C. Pandey, Waves and Thermodynamics, string-speed problem sets.

Provenance. Every number is recomputed in `verify_hs2-string-wave-problems.py`: the bench $\mu = 2.0\times10^{-3}$ kg/m, $v = 200$ m/s, $\lambda = 0.80$ m; the sonometer $T = 39.2$ N (and $40$ N at $g = 10$ ), $v = 198.0$ m/s (and $200$ m/s); the steel wire $\mu = 1.532\times10^{-3}$ kg/m, $v = 197.9$ m/s, and the doubling tension $240$ N; the junction $v_1 = 316.2$ m/s, $v_2 = 158.1$ m/s, $\lambda_1 = 3.162$ m, $\lambda_2 = 1.581$ m; and the unit-slip tells, $200\times\sqrt{1000} = 6.3$ km/s and $200/\sqrt{1000} = 6.3$ m/s.

Prerequisites. Speed of a Wave on a Stretched String — where $v = \sqrt{T/\mu}$ is derived rather than used · Wave Speed: v = fλ — the chain rule of this whole topic. Leads to. Standing Waves & Stationary-Wave Formation — where this same $v$ and $\lambda$ build the harmonics of a fixed string. Go deeper. The university account, in which a real string of discrete beads becomes a continuous one and $\sqrt{T/\mu}$ emerges in the limit, is Discrete Loaded String and the Continuum Limit — enrichment, not a dependency.

Why must μ be in kilograms per metre and not grams per metre?

Because $v = \sqrt{T/\mu}$ mixes $\mu$ with a tension in newtons, and newtons are built from kilograms, metres and seconds — all SI. If $\mu$ is in grams per metre it is a thousand times larger as a number than its SI value, and that thousand sits inside a square root, so it distorts the speed by $\sqrt{1000} = 31.6$ . Convert grams to kilograms the instant you write $\mu$ down: $1.0\ \mathrm{g/m} = 1.0\times10^{-3}\ \mathrm{kg/m}$ . Then read the unit back to yourself before taking the root. This one habit prevents the most common wrong answer on the entire topic.

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

Because tension is a force and $M$ is a mass — different things with different units. The hanging block is in equilibrium: gravity pulls it down with a force $Mg$ (its weight), and the wire's tension pulls it up with an equal force. So the tension equals the weight, $T = Mg$ , not the mass $M$ . A $4.0\ \mathrm{kg}$ block gives $T = 39.2\ \mathrm{N}$ at $g = 9.8$ , a number roughly ten times the mass. Writing $T = 4.0$ silently treats a mass as a force and throws away a factor of $g$ — an automatic mark loss, and a distractor the examiner will have placed in the options.

Does plucking harder make the wave travel faster?

No. Plucking harder increases the amplitude — how far the string swings — and therefore the energy the wave carries and the loudness of the note, but it does not touch the wave speed. The speed is fixed by the tension and the linear density through $v = \sqrt{T/\mu}$ , both of which are properties of the taut wire, set before you pluck it. This is why a guitar string sounds the same pitch whether you pluck it gently or hard: the speed, and hence the frequency of its note, is unchanged. Amplitude changes volume, not speed.

Why does the frequency stay the same when a wave crosses into a new wire?

Because the junction between the two wires is a single point, and it can only move up and down at one rate — the rate the incoming wave drives it. That rate is the frequency, so both wires share it. What differs is the speed, set by each wire's own $\mu$ , and therefore the wavelength, since $\lambda = v/f$ . Crossing into a denser wire the wave slows and its wavelength shrinks, but its frequency — the pitch — is carried through untouched. Any solution that "recomputes" a new frequency in the second wire has misunderstood what a junction does.

When do I use the ratio method instead of computing the speed?

Whenever the question asks "by what factor" or gives you a change rather than raw numbers. If a problem says "the tension is quadrupled, what happens to the speed," or "wire 2 is four times as dense, find the ratio of speeds," you should never find a single value of $v$ . Write $v_2/v_1 = \sqrt{T_2/T_1}\cdot\sqrt{\mu_1/\mu_2}$ and substitute the factors directly. It is faster, and it dodges the unit slips that full calculations invite. Reserve the full $v = \sqrt{T/\mu}$ for when an actual speed in metres per second is wanted.

A wire is given by its diameter, not its mass. How do I get μ?

Treat the wire as the thin cylinder it is: its linear mass density is the material's density times the cross-sectional area, $\mu = \rho\,\pi d^2/4$ . Put the diameter in metres first ( $0.50\ \mathrm{mm} = 5.0\times10^{-4}\ \mathrm{m}$ ), square it, multiply by $\pi/4$ for the area, then by $\rho$ for the mass per length. Because the area grows as $d^2$ , a wire twice as thick has four times the $\mu$ and therefore, at the same tension, half the wave speed — which is exactly why bass strings are made fat.

Is v = √(T/μ) on my board's syllabus?

Only on some. The formula and its numericals are core for CBSE, ICSE, JEE-Main, JEE-Advanced and NEET — the India-facing boards this article is written for. A-Level (Cambridge 9702), AP Physics 2 and IB all teach waves and use $v = f\lambda$ , but they do not require $v = \sqrt{T/\mu}$ ; a stretched-string speed from tension and linear density is simply not one of their assessed outcomes. If you sit one of those boards, read this for the physical picture — tension speeds the wave, mass slows it — but you will not be asked to compute a string speed from $T$ and $\mu$ on your paper.