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Exercises · 14.3

Q.A steel wire has a length of 12.0 m12.0\ \text{m} and a mass of 2.10 kg2.10\ \text{kg}. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 ∘C=343 m s−120\ ^\circ\text{C} = 343\ \text{m s}^{-1}?

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The wave speed on a string depends only on tension and linear mass density. We find the linear density from the given mass and length, then solve for the tension that makes the wave speed equal to 343 m/s. The required tension is about 2.06×104 N2.06 \times 10^4\ \text{N}.

The speed of a transverse wave on a stretched string is a beautiful example of how a simple mechanical property — tension — controls wave propagation. The formula is clean and intuitive: a tighter string (more tension) makes waves travel faster; a heavier string (more mass per unit length) slows them down. Here, we’re told the string’s total mass and length, so we can find its linear mass density. Then we set the wave speed equal to the given speed of sound and solve for the tension.

Let’s go step by step.

  1. Find the linear mass density μ\mu of the wire. Linear mass density is mass per unit length:

μ=masslength=2.10 kg12.0 m=0.175 kg/m.\mu = \frac{\text{mass}}{\text{length}} = \frac{2.10\ \text{kg}}{12.0\ \text{m}} = 0.175\ \text{kg/m}.

  1. Recall the wave speed formula for a transverse wave on a string under tension TT:

v=Tμ.v = \sqrt{\frac{T}{\mu}}.

This comes from Newton’s second law applied to a small segment of the string — the restoring force is proportional to tension, and the inertia is proportional to μ\mu.

  1. Set the wave speed equal to the given speed v=343 m/sv = 343\ \text{m/s} and solve for TT:

343=T0.175.343 = \sqrt{\frac{T}{0.175}}.

  1. Square both sides to remove the square root:

3432=T0.175.343^2 = \frac{T}{0.175}.

  1. Multiply through by 0.1750.175 to isolate TT:

T=0.175×3432.T = 0.175 \times 343^2.

  1. Calculate:

3432=117649,343^2 = 117649,

T=0.175×117649=20588.575 N.T = 0.175 \times 117649 = 20588.575\ \text{N}.

Rounding to three significant figures (since the given data has three significant figures: 12.0 m, 2.10 kg, 343 m/s), we get:

T≈2.06×104 N.T \approx 2.06 \times 10^4\ \text{N}.

Watch out

A common mistake is to forget that the wave speed formula uses linear mass density, not total mass. Always divide mass by length first. Also, don’t confuse this with the speed of sound in the wire material itself — that’s a different concept (bulk modulus vs. tension).

Tip

Notice that the tension here is enormous — over 20,000 N. That’s because steel wire is heavy (high μ\mu), so to make waves travel as fast as sound in air, you need a huge pull. In practice, such a wire would be near its breaking point.

✓Final answer

The required tension is 2.06×104 N\boxed{2.06 \times 10^4\ \text{N}}.

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