Imagine a guitar string. Pluck it, and you hear a note. Press your finger down at a different fret — the vibrating part of the string gets shorter — and the note gets higher. Tighten the tuning peg, and the note rises again. Swap the string for a thicker one, and the pitch drops. That's the entire physics of a sonometer, stripped down to its bones.
A sonometer is simply a laboratory version of that guitar string. It's a long, hollow wooden box with a thin wire stretched tightly over two fixed bridges. You can change three things about the wire: how long the vibrating segment is (by moving a third, movable bridge), how tight the wire is (by hanging weights on one end), and what kind of wire you use (different materials or thicknesses). A small, light paper rider placed on the wire helps you see when the wire is vibrating strongly — it dances or falls off at resonance.
Note
The hollow box isn't decorative. It acts as a sounding board, amplifying the faint sound of the wire so you can hear the note clearly.
The Core Relationship: Frequency and Its Dependence
The sonometer exists to verify one central formula. For a stretched string, the fundamental frequency f (the lowest note it can produce) is:
f=2L1μT
where:
L is the vibrating length of the string (in metres)
T is the tension in the string (in newtons)
μ is the linear mass density — mass per unit length of the string (in kg/m)
This isn't a random equation. It comes from the wave equation for a string fixed at both ends. The wave speed on a string is v=T/μ, and the fundamental standing wave has a wavelength λ=2L. Since v=fλ, you get f=v/(2L)=(1/2L)T/μ.
f=2L1μT
What the Sonometer Actually Shows You
You can test each variable one at a time, keeping the others constant.
Length: Move the movable bridge to change L. Pluck the wire and find the tuning fork that matches its pitch. You'll discover that f∝1/L — halve the length, double the frequency. That's why guitar frets get closer together as you go up the neck.
Tension: Hang different weights on the end of the wire. More weight means more tension. You'll find f∝T. To double the frequency, you need four times the tension.
Linear density: Use wires of different thicknesses or materials. A thicker wire has larger μ, so f∝1/μ. Heavy strings on a piano are thick and produce low notes; thin strings produce high notes.
Watch out
A common mistake: thinking frequency is proportional to tension itself. It's proportional to the square root of tension. Doubling tension only raises frequency by a factor of about 1.414, not 2.
The laws of transverse vibrations of a stretched string relate the fundamental frequency to the string's length, tension, and mass per unit length, combining into $f = \frac{1}{ …
The three laws of transverse vibrations describe how a stretched string's fundamental frequency depends on its length, tension, and mass per unit length, combining into f=2L1T/μ.
For a stretched string vibrating transversely, three experimental laws govern its fundamental frequency f:
First law (Law of length): For a given tension T and mass per unit length (linear density) μ, the fundamental frequency is inversely proportional to the vibrating length L of the string:
f∝L1(T,μ constant)
Second law (Law of tension): For a given length L and linear density μ, the fundamental frequency is directly proportional to the square root of the tension:
f∝T(L,μ constant)
Third law (Law of mass/linear density): For a given length L and tension T, the fundamental frequency is inversely proportional to the square root of the mass per unit length:
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2025Set ANNUAL1 markMCQ
Q.Frequency of a sonometer with increase of length of its wire
(A) increases
(B) decreases
(C) remains unchanged
(D) sometimes increases and sometimes decreases
›Reveal solutionSolution
Increasing the length of a sonometer wire decreases its frequency of vibration.
The fundamental frequency of a stretched wire vibrating between two fixed points a distance L apart is:
Q.A sonometer wire vibrates with frequency n1 in air under a suitable load of specific gravity 'σ'. When the load is immersed in water, the frequency of vibration of the wire n2 will be ______.
(A) n1σσ+1
(B) n1σσ−1
(C) n1σ+1σ
(D) n1σ−1σ
›Reveal solutionSolution
Frequency of a stretched wire depends on the tension; find how the tension (net weight) changes when the load is immersed in water (buoyancy reduces the net downward pull).
The frequency of vibration of a stretched sonometer wire under tension T is
n=2L1mT⇒n∝T
(all else — length L, linear mass density m — being unchanged).
In air, the tension in the wire equals the weight of the suspended load:
T1=Vsdg
where V is the volume of the load, sd (i.e. σρw, with ρw the density of water) is the density of the load material expressed via specific gravity σ, so T1=Vσρwg.
In water, the load additionally experiences an upward buoyant force equal to the weight of water it displaces, Vρwg. The net tension becomes …