A string of mass 2.5 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the tran

NCERT Exemplar · Q9

Q. A string of mass 2.5 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in

(a) one second

(b) 0.5 second

(d) data given is insufficient.

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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 disturbance travels as a transverse wave whose speed depends only on tension and linear mass density. Using $v = \sqrt{T/\mu}$ , the time to travel 20.0 m is 0.5 second.

The key idea is that a transverse jerk on a string creates a wave pulse that travels along the string at a speed determined by the string's tension and its mass per unit length. This is a fundamental result from wave mechanics on a string — the speed does not depend on how hard you jerk, only on the properties of the string itself.

Let’s work through it step by step.

Find the linear mass density $\mu$ The string has total mass $m = 2.5\ \text{kg}$ and length $L = 20.0\ \text{m}$ . Linear mass density is mass per unit length:

\mu = \frac{m}{L} = \frac{2.5}{20.0} = 0.125\ \text{kg/m}

Apply the wave speed formula for a string For a stretched string under tension $T$ , the speed of a transverse wave is:

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

This formula comes from balancing the restoring force (tension) against the inertia of the string. It’s one of the most important results in wave physics — the wave speed is higher when tension is larger, and lower when the string is heavier.

Plug in the values:

v = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40\ \text{m/s}

Calculate the time to travel the length The disturbance must travel the full length $L = 20.0\ \text{m}$ at speed $v = 40\ \text{m/s}$ . …

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Q.A string of mass 2.5 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in

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