Q.Two wires of the same material and the same cross section are stretched on a sonometer in succession. Length of one wire is 60 cm and that of the other is 30 cm. An unknown load is applied to the first wire and second wire is loaded with 1.5 kg. If both the wires vibrate with the same fundamental frequencies, calculate the unknown load.
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.
n₁ = n₂ with same m gives T₁/l₁² = T₂/l₂²; T₁ = T₂(l₁/l₂)² = 1.5×(60/30)² = 6 kg-wt.
Both wires are of the same material and cross-section, so their mass per unit length m is the same. Equal fundamental frequencies with n = (1/2l)√(T/m) require
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 …