Show that the velocity of a travelling wave produced in a string is v=√(T/μ).

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v=\sqrt{T/\mu}

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Concept understanding — Wave Speed on String

Wave Speed on a String – From Intuition to Formula

Imagine you and a friend hold a long, taut rope between you. If you give your end a quick flick upward, a bump travels along the rope toward your friend. That bump is a wave, and the speed at which it moves is the wave speed.

Now ask yourself: what determines how fast that bump travels? Two things stand out from everyday experience:

Tension – If you pull the rope tighter, the bump zips along faster. A loose rope makes the wave crawl.
Mass – If the rope is heavy (like a thick clothesline), the wave moves slower than on a light, thin string under the same tension.

So wave speed increases with tension and decreases with the "heaviness" of the string. That's the core intuition.

The Precise Statement

For a wave traveling along a stretched string, the wave speed $v$ is given by:

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

where:

$T$ is the tension in the string (in newtons, N)
$\mu$ is the linear mass density – the mass per unit length of the string (in kg/m)

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

This formula is exact for an ideal string (perfectly flexible, no stiffness, no damping). It comes from solving the wave equation for a string, but you can understand it physically.

Why the Square Root? A Quick Physical Argument

Think of a small segment of the string. The tension provides the restoring force that tries to straighten the string when it's bent. A higher tension means a stronger restoring force, so the wave accelerates faster – hence higher speed.

The mass per unit length $\mu$ is the inertia of the string. A heavier string resists acceleration more, so the wave slows down.

The square root appears because the relationship between force, mass, and acceleration isn't linear when you derive it properly. But the key takeaway is:

Important

Wave speed on a string depends only on the string's tension and its linear density – not on the frequency or amplitude of the wave.

This is a surprising and important result. Whether you send a slow, gentle ripple or a fast, sharp pulse, both travel at the same speed on the same string.

A Simple Example

A steel guitar string has $\mu = 0.002\ \text{kg/m}$ and is under tension $T = 100\ \text{N}$ . What is the wave speed?

v = \sqrt{\frac{100}{0.002}} = \sqrt{50000} \approx 224\ \text{m/s}

That's about half the speed of sound in air – fast enough that the wave reaches the other end almost instantly.

Common Mistakes to Avoid

Watch out

Do not confuse wave speed with the speed of the string's particles. The string itself moves up and down (transverse motion), but the wave travels horizontally. These are different speeds.
Wave speed does NOT depend on frequency. Changing how fast you flick your hand changes the frequency, but the wave still travels at $v = \sqrt{T/\mu}$ .
Tension is not the same as force applied at the end. If the string is under tension $T$ everywhere (ideal case), that's the value you use – not the force you apply to create the wave.

Where This Formula Comes From (A Glimpse)

If you're curious, the derivation uses Newton's second law on a tiny curved segment of the string. For small displacements, the net vertical force from tension equals $\mu \Delta x$ times the acceleration. This leads to the wave equation:

\frac{\partial^2 y}{\partial t^2} = \frac{T}{\mu} \frac{\partial^2 y}{\partial x^2}

Comparing with the standard wave equation $\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2}$ gives $v^2 = T/\mu$ , hence $v = \sqrt{T/\mu}$ .

Note

For exams, you only need to remember and apply the formula $v = \sqrt{T/\mu}$ . The derivation is for understanding, not memorization – unless your syllabus explicitly asks for it.

Quick Summary

Quantity	Symbol	Effect on wave speed
Tension	$T$	Higher tension → faster wave
Linear density	$\mu$	Heavier string → slower wave
Frequency	$f$	No effect
Amplitude	$A$	No effect

Final takeaway: Wave speed on a string is determined entirely by the string's material and how tightly it's stretched. It's a property of the medium, not the wave itself.

Many students find this page while searching "Wave Speed on String formula physics" or "Wave Speed on String important questions and answers"; the concept sits firmly within the Class 11 Physics NCERT/CBSE syllabus. It's also a frequent building block for numericals in JEE Main, NEET and state engineering/medical entrance exams, so treating it as a one-time memorisation task rather than an understood idea tends to backfire later.

Step 1. Consider a small elemental length $dl$ of a stretched string, of mass $dm=\mu\,dl$ where $\mu$ is the linear mass density. As a transverse pulse passes, this element momentarily traces a circular arc of radius $R$ , subtending an angle $\theta=dl/R$ at the arc's centre O.

Step 2. Viewed from a frame moving with the pulse at speed $v$ , the element requires a centripetal force $F_{cp}=(dm)v^2/R=\mu(dl)v^2/R$ directed toward O.

Step 3. This force is supplied by the string's tension T acting tangentially at both ends of the element. The horizontal components of T at the two ends cancel by symmetry; the vertical components (each $\approx T\sin(\theta/2)\approx T\theta/2$ for small $\theta$ ) add up to give a net radial force $F_r=2\times T\theta/2=T\theta=T\,dl/R$ .

Step 4. By Newton's second law, this net radial (restoring) force must equal the required centripetal force: $T\,dl/R=\mu(dl)v^2/R$ .

Step 5. Cancelling the common factor $dl/R$ from both sides gives $T=\mu v^2$ , i.e. $v^2=T/\mu$ , so $v=\sqrt{T/\mu}$ .

✓Final answer

v=\sqrt{T/\mu}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The speed of a wave is determined by the product of its$
(a) Frequency and amplitude
(b) Wavelength and frequency
(c) Amplitude and period
(d) Wavelength and amplitude
›Reveal solutionSolution

The basic wave equation states that wave speed equals the product of its wavelength and its frequency: v = lambda f.

For any periodic wave, in the time of one full period T, the wave advances forward by exactly one wavelength lambda. So:

Speed v = distance / time = lambda / T

Since frequency f = 1/T, this becomes:

v = lambda x f

This is the fundamental relation connecting a wave's speed, wavelength, and frequency -- not amplitude or period alone, which describe the wave's size and the interval of one oscillation respectively but do not by themselves determine how fast the wave pattern propagates.

✓Final answer
$(b) Wavelength and frequency.$
CBSE 2026Set ANN1 markMCQ
$Q. The frequency, wavelength and velocity of a wave are related by the equation$
(a) v = f/λ
(b) f = v/λ
(c) v = λ/f
(d) 1/v = fλ
›Reveal solutionSolution

The wave relation is v = f lambda, which rearranges to f = v/lambda - option (b).

The fundamental relation between the speed v, frequency f and wavelength lambda of a wave is

v = f lambda,

since a wave advances one wavelength in each time period T = 1/f. Rearranging gives

f = v/lambda.

Checking the options, only (b) f = v/lambda is consistent with v = f lambda; the others do not satisfy the relation dimensionally or algebraically.

✓Final answer
$(b) f = v/lambda.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Speed of transverse vibration in stretched string is (A) \sqrt(T/m) (B) T/m (C) Tm (D) m/T$
›Reveal solutionSolution

The speed of a transverse wave on a stretched string is $v=\sqrt{T/m}$ .

For a string under tension $T$ with linear mass density (mass per unit length) $m$ , applying Newton's second law to a small string element in a transverse disturbance gives the wave equation with wave speed:

$v = \sqrt{\dfrac{T}{m}}$

This shows the speed increases with greater tension and decreases with a heavier (denser) string.

✓Final answer
$(A) \sqrt(T/m).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. A 10 m long steel wire has mass 5 g. If the wire is under a tension of 80 N, the speed of transverse waves on the wire is$
(a) 100 ms^-1
(b) 200 ms^-1
(c) 400 ms^-1
(d) 500 ms^-1
›Reveal solutionSolution

Wave speed on a stretched string is v = sqrt(T/mu); computing the linear mass density from the given length and mass, then substituting, gives v = 400 m/s.

Length of wire, L = 10 m. Mass, m = 5 g = 5 x 10^-3 kg.

Linear mass density, mu = m/L = (5 x 10^-3 kg) / (10 m) = 5 x 10^-4 kg/m.

Tension, T = 80 N.

Speed of transverse waves on a stretched wire/string:

v = sqrt(T / mu) = sqrt(80 / (5 x 10^-4)) = sqrt(160000) = 400 m/s.

✓Final answer
$(c) 400 ms^-1.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. With increase in tension in a string, frequency of transverse vibration (A) increases (B) decreases (C) remains constant (D) first increases then decreases$
›Reveal solutionSolution

Frequency of a vibrating string increases with tension.

For a string of length $L$ and mass per unit length $\mu$ under tension $T$ , the fundamental frequency of transverse vibration is $f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ . Since $f\propto\sqrt T$ , increasing the tension increases the frequency — this is exactly why tightening a guitar or violin string raises its pitch.

✓Final answer
$(A) increases.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Match the column: Speed v of a transverse wave in a string — match with the correct expression.$
(a) sqrt(2gR)
(b) sqrt(T/m)
(c) GMm/r^2
(d) I*omega
(e) 2pisqrt(l/g)
(f) sqrt(gR)
(g) m*R^2
›Reveal solutionSolution

The speed of a transverse wave on a stretched string is v = sqrt(T/m) (T = tension, m = linear mass density), matching option (b).

For a string under tension T with mass per unit length m (also written as mu), a transverse disturbance travels along it with speed:

v = sqrt(T/m)

This shows that a more tightly stretched string (higher T) carries waves faster, while a heavier string (higher m per unit length) carries waves more slowly -- exactly why tightening a guitar string raises the pitch, and thicker strings sound lower. Among the given options, only (b), sqrt(T/m), matches this form.

✓Final answer
$(b) sqrt(T/m).$
CBSE 2020Set ANNUAL1 mark
$Q. Write down the formula for the speed of transverse waves in a stretched string.$
›Reveal solutionSolution
The speed of a transverse wave on a string is $v=\sqrt{T/\mu}$ — it increases with tension and decreases with a heavier string.

Step 1 — Physical basis.

A transverse wave on a stretched string is a restoring-force phenomenon: tension $T$ provides the restoring force that pulls a displaced element of the string back, while the string's inertia (its mass per unit length $\mu$ ) resists that acceleration. A wave equation derived from Newton's second law applied to a small string element under tension gives:

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

Step 2 — Meaning of the symbols.

$T$ = tension in the string (in newtons)

$\mu = m/L$ = mass per unit length of the string (in kg/m)

Step 3 — Interpretation.

A tighter string (larger $T$ ) carries waves faster; a heavier string (larger $\mu$ ) carries waves more slowly — exactly matching everyday observation on musical instrument strings.

✓Final answer
$v = \sqrt{T/\mu}$
CBSE 2017Set ANNUAL1 mark
$Q. Write the relation between angular frequency (w), angular wave number (k) and wave velocity (v).$
›Reveal solutionSolution

The wave velocity equals the angular frequency divided by the angular wave number: v = ω/k.

For a progressive wave y = A sin(kx - ωt), a point of constant phase (kx - ωt = constant) moves with the wave. Differentiating, k(dx/dt) - ω = 0, so dx/dt = ω/k. This dx/dt is exactly the wave (phase) velocity v. Also, since ω = 2πf (f = frequency) and k = 2π/λ (λ = wavelength), ω/k = (2πf)/(2π/λ) = fλ, which is the familiar relation v = fλ. So both forms — v = ω/k and v = fλ — express the same fact: the wave travels a distance of one wavelength in one time period.

✓Final answer
$v = ω / k (wave velocity equals angular frequency divided by angular wave number).$

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Q.Show that the velocity of a travelling wave produced in a string is v=T/μv=\sqrt{T/\mu}v=T/μ​.

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Q.Show that the velocity of a travelling wave produced in a string is $v=\sqrt{T/\mu}$ .

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