ARCHETYPE: Sonometer & String-Harmonic Problems

Open any Indian physics paper — CBSE, NEET, a JEE-Main set — and somewhere in the waves section sits a stretched wire. A mass hangs off a pulley, two bridges pin down a length, a tuning fork is struck, and you are asked for a frequency, a tension, a mass per metre, or how many beats two wires make together. It is the sonometer, the single most-drilled apparatus in the school sound syllabus, and the striking thing is that every one of these questions — dozens of variants across a decade of papers — is the same problem wearing different clothes. That one problem is a chain of four ideas you already met when a string's normal modes were worked out: the wave speed on a string $v = \sqrt{T/\mu}$ , the harmonic formula $f_n = n\,v/(2L)$ , the tension from a hanging weight $T = Mg$ , and the linear mass density $\mu = m/L$ . Bolt them together and you get the master equation of this entire archetype,

f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}},

and the whole skill is knowing which of its symbols a given question hands you and which it hides. This article treats the method itself as the concept. We build the master recipe, the ratio shortcut that saves a minute on every "double the tension" question, the two-wire beat variant, and the traps examiners reuse every single year — the grams-versus-kilograms slip, the tension-that-is-a-weight, and the fatal habit of computing the wave speed when a clean ratio would have done. Learn the machine once and an entire question bank turns into fill-in-the-blanks.

Key takeaways

Every sonometer question is one master equation: $f_n = \dfrac{n}{2L}\sqrt{\dfrac{T}{\mu}}$ , with $T = Mg$ and $\mu = m/L$ (or $\mu = \rho\pi d^2/4$ for a wire of diameter $d$ ).
Master recipe: get $\mu$ honestly in kg per metre, then $T = Mg$ , then $v = \sqrt{T/\mu}$ , then $f_n = n\,v/(2L)$ .
For a ratio question, never compute $v$ . Use $\dfrac{f'}{f} = \sqrt{\dfrac{T'}{T}}\,\sqrt{\dfrac{\mu}{\mu'}}\,\dfrac{L}{L'}$ and cancel everything unchanged.
Because $f \propto \sqrt{T}$ , a $4\%$ tension change gives only a $\approx 2\%$ frequency change — the square root halves it.
Two-wire beats: two sonometer wires at slightly different tension beat at $f_{\text{beat}} = |f_A - f_B|$ .
Key idea: one equation $f_n = (n/2L)\sqrt{T/\mu}$ solves every sonometer problem — the art is spotting which variable is unknown, and using ratios (not wave speeds) whenever two states are compared.

The core insight

The sonometer archetype is the single relation $f_n = (n/2L)\sqrt{T/\mu}$ with $T = Mg$ and $\mu = m/L$ ; direct questions substitute once for the unknown, ratio questions divide two states so the constants cancel, and the recurring traps are unit conversions and the square-root dependence on tension.

Where this lives today

The sonometer is a teaching instrument, but the equation it drills is everywhere a taut wire makes a controlled pitch. A piano technician setting the tension of two hundred strings, a guitar luthier choosing string gauges so that six very different wires all sit at a playable tension, a violinist retuning between movements — all are solving $f = (1/2L)\sqrt{T/\mu}$ by ear and by feel. Engineers use the same relation in reverse: pluck a bridge's stay cable or a crane's guy wire, measure the note it sounds, and $T = \mu(2Lf)^2$ gives the tension without a load cell — a standard field method for checking cable tensions on suspension bridges. The MEMS resonators inside a phone's clock and the vibrating wires in some pressure sensors read out a frequency that maps, through this same formula, to a force. The bench in the figure is small; the equation on it holds up bridges.

The setup

The apparatus is deliberately simple, because in an archetype the apparatus is the physics. A metal wire runs horizontally over two sharp bridges resting on a hollow wooden box (the box amplifies the sound). One end of the wire is fixed; the other passes over a pulley at the edge of the box and carries a hanging mass $M$ . The wire between the two bridges — length $L$ — is the part that vibrates, and sliding a bridge changes $L$ without touching anything else. Everything you can vary has a clean handle: the hanging mass sets the tension, the bridge spacing sets the length, and swapping the wire sets the mass per metre.

Figure

A sonometer wire stretched over two bridges by a hanging mass, with the tension equal to the weight and the fundamental loop drawn between the bridges, annotated with the master frequency formula

Three facts feed the master equation, and each is a place a question can hide a step:

f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}, \qquad T = Mg, \qquad \mu = \frac{m}{L_{\text{wire}}} = \frac{\rho\,\pi d^2}{4} \qquad (1)

where: $n$ — the harmonic number ( $1$ for the fundamental); $L$ — the vibrating length between bridges; $T$ — the tension, equal to the hanging weight $Mg$ ; $\mu$ — the linear mass density in kg per metre, either the wire's total mass over its length or, for a wire of density $\rho$ and diameter $d$ , the value $\rho\pi d^2/4$ . That last form matters: a question that gives you a diameter and a material density expects you to build $\mu$ from them. For a steel wire ( $\rho = 7800\ \mathrm{kg\,m^{-3}}$ ) of diameter $0.40\ \mathrm{mm}$ , $\mu = 7800 \times \pi \times (4.0\times10^{-4})^2/4 = 9.8\times10^{-4}\ \mathrm{kg\,m^{-1}}$ , about $0.98\ \mathrm{g\,m^{-1}}$ — the kind of thin wire a real sonometer uses.

The master solution pattern

Every direct question follows the same four moves, and doing them in this order keeps the units honest:

Step 1. Get $\mu$ in kg per metre. If the wire is quoted in grams per metre, multiply by $10^{-3}$ ; if you are given a diameter, use $\mu = \rho\pi d^2/4$ . Step 2. Get the tension, $T = Mg$ — a weight in newtons, never a mass in kilograms. State your $g$ . Step 3. Wave speed, $v = \sqrt{T/\mu}$ . Step 4. Frequency, $f_n = n\,v/(2L)$ , or, if a fork drives the wire at resonance, read the speed straight off with $v = 2Lf_n/n$ . For any question that compares two states — "the tension is doubled", "the length is halved", "raise the pitch a fifth" — throw the recipe away and divide instead. Writing the master equation for both states and dividing, every unchanged constant cancels:

\frac{f'}{f} = \sqrt{\frac{T'}{T}}\;\sqrt{\frac{\mu}{\mu'}}\;\frac{L}{L'} \qquad (2)

where: each factor is the ratio of one variable between the two states, and any variable you did not change contributes a factor of $1$ . This is the single most time-saving idea in the archetype. If only the tension changes, $f'/f = \sqrt{T'/T}$ and nothing else — you never compute a wave speed, never touch $\mu$ or $L$ , and the answer is one square root. The exam's favourite graph, $f$ against $\sqrt{T}$ , is exactly equation (2) with $\mu$ and $L$ fixed: a straight line through the origin.

Figure

A straight-line graph of fundamental frequency against the square root of tension, passing through the origin, marked at the tension of 64 newtons where the frequency is 63.25 hertz

Derivation AThe master recipe worked end to end

The goal is to run the four-step recipe once, in full, on a clean board-level case, so the order of the moves becomes automatic — because the order is what keeps the units from betraying you. We take a wire of length $1.0\ \mathrm{m}$ and mass $4.0\ \mathrm{g}$ under $64\ \mathrm{N}$ of tension and find its wave speed, fundamental and third harmonic. The strategy is: density first, then tension, then speed, then frequency — never out of sequence.

Step 1. Linear mass density in SI. The wire is $4.0\ \mathrm{g}$ over $1.0\ \mathrm{m}$ , so $\mu = m/L_{\text{wire}} = 4.0\times10^{-3}\ \mathrm{kg}/1.0\ \mathrm{m} = 4.0\times10^{-3}\ \mathrm{kg\,m^{-1}}$ . The grams became kilograms here, once, deliberately.

Step 2. Tension. It is given directly as $T = 64\ \mathrm{N}$ (no hanging mass to convert this time), so no $Mg$ step is needed.

Step 3. Wave speed. $v = \sqrt{T/\mu} = \sqrt{64/(4.0\times10^{-3})} = \sqrt{16000} = 126.5\ \mathrm{m\,s^{-1}}$ . Sense-check the size: a hundred-odd metres per second is right for a taut thin wire.

Step 4. Frequencies. Fundamental $f_1 = v/(2L) = 126.5/(2\times1.0) = 63.25\ \mathrm{Hz}$ . The third harmonic is $f_3 = 3f_1 = 189.7\ \mathrm{Hz}$ (a stretched string carries every harmonic, so $f_3$ is simply three times $f_1$ ). Verified: v = √(64 / 0.004) = 126.5 m/s, f₁ = 126.5/2 = 63.25 Hz, f₃ = 3·63.25 = 189.7 Hz — recomputed in verify_hs2-string-harmonic-problems.py.

Go deeper · Derivation B · 1 minThe ratio method — solving a comparison without a wave speed

Derivation BThe ratio method — solving a comparison without a wave speed

Here we derive and use equation (2) to show why, whenever two states of the same wire are compared, computing a wave speed is wasted effort. The goal is to make the ratio move second nature on the two classic comparisons — a tension change and a target pitch — using the locked numbers. The strategy is to write the master equation twice and divide, letting the unchanged constants cancel before any arithmetic begins.

Step 1. Write $f_n = (n/2L)\sqrt{T/\mu}$ for state 1 and state 2 and divide. Everything held fixed cancels, leaving $f'/f = \sqrt{T'/T}\,\sqrt{\mu/\mu'}\,(L/L')$ . For the same wire at the same length and harmonic, only the tension factor survives: $f'/f = \sqrt{T'/T}$ .

Step 2. Apply it to a $4\%$ tension drop (the two-wire beat setup). With $T' = 0.96\,T$ , $f'/f = \sqrt{0.96} = 0.980$ , so the frequency falls by only $2.0\%$ , not $4\%$ — the square root halves the fractional change. On a $256\ \mathrm{Hz}$ wire this gives $f' = 256 \times 0.980 = 250.8\ \mathrm{Hz}$ .

Step 3. Apply it to a target pitch. To raise the pitch by a musical fifth means $f'/f = 1.5$ . Since $f'/f = \sqrt{T'/T}$ , square both sides: $T'/T = 1.5^2 = 2.25$ . You must multiply the tension by $2.25$ — a $125\%$ increase — to raise a note a fifth, which is exactly why over-tuning snaps strings.

Step 4. Note what never happened: no wave speed, no $\mu$ , no $L$ , no fundamental computed. Two lines each. The ratio method is not just faster — it removes the very steps where unit slips live. Verified: √0.96 = 0.980 so 256 → 250.8 Hz (a 2.0% drop from a 4% tension change), and 1.5² = 2.25 for a musical fifth — recomputed in verify_hs2-string-harmonic-problems.py.

Limiting cases and scalings

change	effect on $f$	why
tension $\times 4$	$f \times 2$	$f \propto \sqrt{T}$ , and $\sqrt{4} = 2$ (one octave)
tension $\times 2$	$f \times \sqrt{2} = 1.414$	doubling tension is only a musical fifth, not an octave
tension $-4\%$	$f -2.0\%$	the square root halves the fractional change
length halved	$f \times 2$	$f \propto 1/L$ , no square root — a clean octave
$\mu \times 4$ (four times heavier)	$f \times \tfrac12$	$f \propto 1/\sqrt{\mu}$ — an octave lower
pitch up a fifth ( $f \times 1.5$ )	tension $\times 2.25$	invert $f \propto \sqrt{T}$ : $T \propto f^2$

Common trapTwo silent unit slips: μ in grams, and T as a mass

The sonometer punishes exactly two shortcuts. First, $\mu$ must be in kilograms per metre. A wire quoted as " $4\ \mathrm{g\,m^{-1}}$ " is $4\times10^{-3}\ \mathrm{kg\,m^{-1}}$ ; drop the conversion and the factor of a thousand becomes $\sqrt{1000} = 31.6$ in your wave speed, throwing the frequency out by a factor of thirty in one direction or the other. Second, the tension is a weight, $T = Mg$ , not the mass $M$ . A $9\ \mathrm{kg}$ load is $88.2\ \mathrm{N}$ at $g = 9.8$ , not $9\ \mathrm{N}$ ; using the mass directly makes your tension nearly ten times too small. Both errors are invisible in the algebra and only show up in a wildly wrong number, so build the habit of writing every quantity with its unit and checking $\mu$ reads kg per metre and $T$ reads newtons before you take the square root.

Worked examples

1 · Board level — wave speed, fundamental and third harmonic (CBSE and ICSE pattern)

A wire of length $1.0\ \mathrm{m}$ and mass $4.0\ \mathrm{g}$ is stretched by a tension of $64\ \mathrm{N}$ . Find the wave speed, the fundamental frequency, and the frequency of the third harmonic.

Step 1. Density in SI: $\mu = 4.0\times10^{-3}\ \mathrm{kg}/1.0\ \mathrm{m} = 4.0\times10^{-3}\ \mathrm{kg\,m^{-1}}$ .

Step 2. Wave speed: $v = \sqrt{T/\mu} = \sqrt{64/(4.0\times10^{-3})} = \sqrt{16000} = \boxed{126.5\ \mathrm{m\,s^{-1}}}$ .

Step 3. Fundamental: $f_1 = v/(2L) = 126.5/2 = \boxed{63.25\ \mathrm{Hz}}$ .

Step 4. Third harmonic (all harmonics present on a string): $f_3 = 3f_1 = \boxed{189.7\ \mathrm{Hz}}$ . Sense-check: the harmonics are evenly spaced by $63.25\ \mathrm{Hz}$ , as a fixed-fixed string requires.

2 · Board level — two wires beating (JEE-Main and NEET pattern)

Two identical sonometer wires are tuned to the same fundamental of $256\ \mathrm{Hz}$ . The tension in one is then reduced by $4.0\%$ . How many beats per second are heard when both are plucked at their fundamentals?

Step 1. Recognise it as a ratio problem — same wire, same length, only the tension changes — so use $f'/f = \sqrt{T'/T}$ and do not compute a wave speed.

Step 2. The slackened wire: $T' = 0.96\,T$ , so $f' = 256\sqrt{0.96} = 256 \times 0.980 = \boxed{250.8\ \mathrm{Hz}}$ . Notice the $4\%$ tension cut produced only a $2.0\%$ pitch drop — the square root halved it.

Step 3. Beat frequency between the two wires: $f_{\text{beat}} = |f_A - f_B| = |256 - 250.8| = \boxed{5.17\ \mathrm{Hz}}$ — a touch over five beats a second.

Step 4. The teaching point to carry away: a small tension mismatch makes an audible beat, and the beat is about half the percentage tension error times the frequency — here $\tfrac12 \times 4\% \times 256 \approx 5\ \mathrm{Hz}$ . This is why a sonometer beats so sensitively.

3 · Competitive level — tension from a driving fork (JEE-Main pattern)

A tuning fork of frequency $320\ \mathrm{Hz}$ sets a sonometer wire vibrating in its fundamental when the vibrating length is $0.50\ \mathrm{m}$ . The wire's linear mass density is $1.0\times10^{-3}\ \mathrm{kg\,m^{-1}}$ . Find the wave speed and the tension in the wire.

Step 1. Read the speed straight off the resonance. At the fundamental the fork's frequency is the wire's $f_1$ , so $v = 2Lf_1 = 2 \times 0.50 \times 320 = \boxed{320\ \mathrm{m\,s^{-1}}}$ . No tension or density needed for the speed — the resonance gives it directly.

Step 2. Now invert $v = \sqrt{T/\mu}$ for the tension: $T = \mu v^2$ .

Step 3. Substitute: $T = 1.0\times10^{-3} \times (320)^2 = 1.0\times10^{-3} \times 102400 = \boxed{102.4\ \mathrm{N}}$ . A sensible sonometer tension (about a $10\ \mathrm{kg}$ load).

Step 4. The shape to memorise: whenever a fork drives the wire at resonance, the fastest first move is $v = 2Lf$ , and only then $T = \mu v^2$ . Reaching for $\sqrt{T/\mu}$ before you have the speed is the slow road. Timing note: about 90 seconds once you see $v = 2Lf$ . The trap is not spotting that the fork frequency is the wire's fundamental, and grinding through $\sqrt{T/\mu}$ with an unknown $T$ .

4 · Competitive level — the tension to raise a pitch a fifth (JEE-Advanced pattern)

A sonometer wire sounds a certain note. Keeping its length and material unchanged, by what factor must its tension be increased to raise the pitch by a musical fifth, i.e. to multiply the frequency by $1.5$ ? Comment on why over-tightening breaks strings.

Step 1. Same wire, same length — a pure ratio problem. With $\mu$ and $L$ fixed, $f'/f = \sqrt{T'/T}$ .

Step 2. Set the target: $f'/f = 1.5$ . Square both sides to isolate the tension ratio: $T'/T = (f'/f)^2 = (1.5)^2 = \boxed{2.25}$ .

Step 3. Interpret: you must multiply the tension by $2.25$ — a $125\%$ increase — for a rise of just a fifth. Because $T \propto f^2$ , pitch is a steep function of tension near the top of a string's range.

Step 4. Why strings snap: a full octave ( $f'/f = 2$ ) needs $T'/T = 4$ , quadrupling the tension. A wire tuned near its breaking stress cannot survive that, which is why the highest strings on a piano or guitar are the ones that break, and why you tune up to a note gently from below rather than overshooting. Timing note: about 60 seconds — it is one squaring. The trap is multiplying the tension by $1.5$ instead of by $1.5^2$ ; the tension goes as the frequency squared, not linearly.

Exam traps

Grams to kilograms, every time. $\mu$ in $\sqrt{T/\mu}$ must be kg per metre. A g-per-metre value needs $\times 10^{-3}$ ; a wire given by diameter needs $\mu = \rho\pi d^2/4$ with $d$ in metres. The slip is a factor of $\sqrt{1000} = 31.6$ in the speed — the commonest wrong answer in the whole topic.

Tension is a weight. $T = Mg$ . A hanging $9\ \mathrm{kg}$ is $88.2\ \mathrm{N}$ at $g = 9.8$ , not $9\ \mathrm{N}$ . Feeding the mass in place of the weight makes the tension about ten times too small and the frequency about three times too low.

Use the linear density, not the total mass. In $v = \sqrt{T/\mu}$ the symbol is mass per unit length. A question that gives "a $2.0\ \mathrm{g}$ wire $0.80\ \mathrm{m}$ long" wants $\mu = 2.0\times10^{-3}/0.80\ \mathrm{kg\,m^{-1}}$ , not $2.0\times10^{-3}\ \mathrm{kg}$ .

$f \propto \sqrt{T}$ , not $\propto T$ . Doubling the tension raises the pitch by $\sqrt{2}$ , quadrupling gives the octave, and to raise a note a fifth you multiply the tension by $1.5^2 = 2.25$ . Any "double the tension, double the pitch" reasoning is wrong.

In a ratio problem, do not compute $v$ . If two states share a wire, divide the master equation and cancel — equation (2). Computing wave speeds twice is slower and doubles your chances of a unit slip. The whole point of the shortcut is to skip the steps where errors live.

Which $g$ , and which board. State $g$ ( $9.8$ for CBSE and NEET, $9.81$ for A-Level) because $T = Mg$ depends on it. And note the scope: the sonometer numerical is an India-facing staple (CBSE, ICSE, JEE-Main, JEE-Advanced, NEET). It is not an A-Level, IB or AP requirement — those syllabuses do the stationary-wave physics but not $v = \sqrt{T/\mu}$ or the hanging-mass drill.

Origins · 1 minOsborne Reynolds and how this was discovered

Osborne Reynolds

1842–1912 · Manchester

Osborne Reynolds was the Manchester engineer whose name rides on the Reynolds number, and his signature tool — dimensional analysis, grouping the variables of a problem into the combinations that actually matter — is exactly what cracks every sonometer ratio question. The master relation $f = (n/2L)\sqrt{T/\mu}$ is a dimensional statement: it tells you that doubling the tension can only raise the frequency by $\sqrt{2}$ , because that is the sole way $T$ can enter a frequency built from $T$ , $\mu$ and $L$ . Reynolds also worked directly in acoustics — his memoir On the Refraction of Sound by the Atmosphere (1874–76) explained why sound bends over warm ground and carries far on a still evening. He did not invent the sonometer bench; what he gave this topic is the method for reasoning about its ratios without grinding through every number.

History

1636

— Marin Mersenne states the three laws of the vibrating string: pitch varies as one over the length, as the square root of the tension, and inversely as the square root of the mass per length.

1713

— Brook Taylor gives the first theoretical derivation of a string's fundamental frequency from its tension and density.

1747

— Jean d'Alembert writes down the wave equation of the string, whose standing-wave solutions are the harmonics the sonometer measures.

1877

— Lord Rayleigh's Theory of Sound systematises the vibrating string and establishes the sonometer as a precision instrument for comparing frequencies.

1883

— Osborne Reynolds's study of flow in pipes introduces the dimensionless-grouping method that underlies every sonometer ratio problem.

Going further

Texts. NCERT Physics Class 11, Ch. 15 ("Waves"), the vibrating string and sonometer sections · H. C. Verma, Concepts of Physics Vol. 1, "Waves on a String", worked sonometer problems · D. C. Pandey, Waves and Thermodynamics, the sonometer problem set · previous years' JEE-Main and NEET waves questions.

Provenance. Standard results. Recompute: $\mu = 4.0\times10^{-3}\ \mathrm{kg\,m^{-1}}$ , $v = \sqrt{64/0.004} = 126.5\ \mathrm{m\,s^{-1}}$ , $f_1 = 63.25$ , $f_3 = 189.7\ \mathrm{Hz}$ ; two-wire beat $f_B = 256\sqrt{0.96} = 250.8\ \mathrm{Hz}$ , beat $|256-250.8| = 5.17\ \mathrm{Hz}$ , tension-to-frequency $\sqrt{0.96} = 0.980$ ( $2.0\%$ drop from $4\%$ ); fork-driven $v = 2\times0.5\times320 = 320\ \mathrm{m\,s^{-1}}$ , $T = 10^{-3}\times320^2 = 102.4\ \mathrm{N}$ ; fifth $T'/T = 1.5^2 = 2.25$ ; octave $T'/T = 4$ ; steel wire $\mu = 7800\pi(4\times10^{-4})^2/4 = 9.8\times10^{-4}\ \mathrm{kg\,m^{-1}}$ ; graph slope $1/(2L\sqrt\mu) = 7.906$ , marked point $(\sqrt{T}=8, f=63.25)$ .

Prerequisites. Normal Modes and Harmonics on a String — the physics of $f_n = n\,v/(2L)$ and $v = \sqrt{T/\mu}$ that this archetype drills. Leads to. Normal Modes and Standing Waves on a String and Discrete Loaded String and the Continuum Limit (university mechanics) — where $v = \sqrt{T/\mu}$ is derived from first principles.

Go deeper (UG). How the continuous string's wave speed emerges from a chain of beads with springs between them is the subject of Discrete Loaded String and the Continuum Limit in the university mechanics track. Enrichment only — the archetype needs none of it.

When should I use the ratio method instead of the full recipe?

Use the ratio method (equation 2) whenever a question compares two states of the same setup — "the tension is doubled", "the length is halved", "raise the pitch a fifth", "two wires differ by 4 % in tension". In all of these, most of the variables are unchanged and will cancel, so writing the master equation twice and dividing turns a four-step calculation into one square root. Use the full recipe (density, tension, speed, frequency) only when you need an actual number for a single state — a specific frequency, wave speed or tension. A good tell: if the question never asks for a wave speed and only wants "the new frequency" or "the factor", the ratio method is the intended route and computing $v$ is wasted work.

Why does a 4 percent change in tension give only about a 2 percent change in frequency?

Because frequency depends on the square root of tension, $f \propto \sqrt{T}$ . When a quantity depends on the square root of another, a small fractional change is halved on the way through: a $4\%$ change in $T$ becomes roughly a $2\%$ change in $f$ (precisely, $\sqrt{0.96} = 0.980$ , a $2.0\%$ drop). This "square root halves the percentage" rule is worth internalising, because it appears all over the topic — it is why doubling the tension ( $100\%$ up) gives only a $41\%$ rise in pitch ( $\sqrt 2 = 1.414$ ), and why tuning by tension is a delicate business near the top of a string's range.

How do I get μ when the question gives me the wire's diameter and material?

Use $\mu = \rho\pi d^2/4$ , the mass per unit length of a cylinder. Here $\rho$ is the material's density (steel is about $7800\ \mathrm{kg\,m^{-3}}$ ), $d$ is the diameter in metres, and $\pi d^2/4$ is the wire's cross-sectional area, so $\rho \times \text{area}$ is mass per length. For a $0.40\ \mathrm{mm}$ steel wire, $\mu = 7800 \times \pi \times (4.0\times10^{-4})^2/4 = 9.8\times10^{-4}\ \mathrm{kg\,m^{-1}}$ . The two traps are leaving $d$ in millimetres (it must be metres, so $0.40\ \mathrm{mm} = 4.0\times10^{-4}\ \mathrm{m}$ ) and using the radius where the formula wants the diameter — if you prefer the radius, the formula is $\mu = \rho\pi r^2$ .

If a fork and a wire beat, how do I know the wire's exact frequency?

A beat count alone gives you two candidates. If a $320\ \mathrm{Hz}$ fork beats $n$ times a second with a wire, the wire is at $320 + n$ or $320 - n$ , and the beat rate cannot tell you which. You resolve it by changing the wire and watching the beats: slightly increase the tension (which raises the wire's frequency) and see whether the beats speed up (the wire was above the fork) or slow down (it was below). That "change one source, watch the beats move" logic is the entire subject of the beat-problem archetype in this chapter. For a pure sonometer question where the wire is driven by the fork at resonance, there is no ambiguity — the wire simply takes the fork's frequency, and you work backwards to tension or density.

Does the harmonic number n change any of the method?

Only the last step. Everything up to the wave speed — density, tension, $v = \sqrt{T/\mu}$ — is identical for every harmonic, because $v$ is a property of the wire and its tension, not of how it happens to be vibrating. The harmonic enters only in $f_n = n\,v/(2L)$ : the fundamental is $n = 1$ , and because a stretched string carries all harmonics, the higher ones are just whole-number multiples, $f_3 = 3f_1$ and so on. So if a question asks for the third harmonic, find $f_1$ by the usual recipe and multiply by three. This is different from a closed pipe, which skips its even harmonics — a string has no gaps.

What is the hollow wooden box under the wire for?

The wire itself moves very little air — it is thin, so on its own it is almost inaudible. The hollow box it is mounted on acts as a sounding board: the wire's vibration is transmitted through the bridges to the box's large wooden surfaces, which push on a much greater volume of air and radiate the sound efficiently, so you can actually hear the note and judge the beats. The box does not change the frequency — that is fixed entirely by $L$ , $T$ and $\mu$ through the master equation — it only makes the sound loud enough to work with. It is the same trick as the body of a guitar or violin: a light string sets the pitch, a resonating box makes it audible. In calculations the box never appears; it is there purely so the physics can be heard.

Is the sonometer on my syllabus?

It depends on your board. The sonometer numerical — hanging masses, bridge spacings, $v = \sqrt{T/\mu}$ and its ratios — is a staple of the Indian boards: CBSE, ICSE, JEE-Main, JEE-Advanced and NEET all expect it, and it is a prescribed practical on several of them. It is not part of A-Level (CIE 9702), IB or AP Physics 2: those syllabuses teach the stationary-wave physics on strings (nodes, antinodes, $f_n = n\,v/2L$ ) but do not use the $\sqrt{T/\mu}$ tension formula or drill the hanging-mass apparatus. If you are on one of those boards, read this for the physical insight, but you will not be examined on the numerical machinery.