A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If

NCERT Exemplar · Q20

Q. A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?

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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) …

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The frequency of a vibrating string depends inversely on its radius when length, material, and tension are fixed. Tripling the radius reduces the frequency to one-third of its original value.

The key to this problem lies in understanding how the fundamental frequency of a stretched string depends on its physical properties. For a sitar wire (or any string fixed at both ends), the fundamental frequency is given by:

f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}

where $L$ is the length, $T$ is the tension, and $\mu$ is the mass per unit length (linear mass density). The wire is replaced by another of the same length and material, with the same tension, but with three times the radius. Since the material is the same, the density $\rho$ is unchanged. The linear mass density $\mu$ is mass per unit length, which for a cylindrical wire is $\mu = \rho \cdot \text{(cross-sectional area)} = \rho \pi r^2$ .

So $\mu$ is proportional to $r^2$ . If the radius becomes $3$ times larger, $\mu$ becomes $9$ times larger. Now look at the frequency formula: $f \propto 1/\sqrt{\mu}$ . Therefore, if $\mu$ increases by a factor of $9$ , the frequency decreases by a factor of $\sqrt{9} = 3$ .

Let’s walk through it step by step.

Write the formula for fundamental frequency. For a string fixed at both ends, the fundamental frequency is

f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}

Here $L$ and $T$ are constant in this problem.

Express $\mu$ in terms of radius. Since the wire is cylindrical and made of the same material (density $\rho$ ),

\mu = \rho \times \text{area} = \rho \pi r^2

So $\mu \propto r^2$ .

Relate frequency to radius. Substitute $\mu \propto r^2$ into the frequency expression:

f \propto \frac{1}{\sqrt{r^2}} = \frac{1}{r}

That is, frequency is inversely proportional to the radius. …

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Q.A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?

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