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}

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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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