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Exercises · 14.1

Q.A string of mass 2.50 kg2.50\ \text{kg} is under a tension of 200 N200\ \text{N}. The length of the stretched string is 20.0 m20.0\ \text{m}. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Yanam BieapTextbookSubjective· 3mImportance★★★★★est
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✓ Free question

The disturbance travels as a transverse wave on the string. Its speed depends only on tension and linear mass density, not on amplitude. The time taken is 0.5 s0.5\ \text{s}.

The key idea here is that a transverse jerk (a pulse) propagates along a stretched string as a wave. The speed of such a wave is determined by two properties of the string: how tightly it is stretched (tension) and how heavy it is per unit length (linear mass density). Once we know the speed, the time to travel a given distance is simply distance divided by speed.

Let’s work through it step by step.

  1. Find the linear mass density μ\mu The string has a total mass m=2.50 kgm = 2.50\ \text{kg} and a total length L=20.0 mL = 20.0\ \text{m}. Linear mass density is mass per unit length:

μ=mL=2.5020.0=0.125 kg/m\mu = \frac{m}{L} = \frac{2.50}{20.0} = 0.125\ \text{kg/m}

  1. Recall the wave speed formula for a string For a transverse wave on a string under tension TT, the wave speed vv is given by:

v=Tμv = \sqrt{\frac{T}{\mu}}

This formula comes from combining Newton’s second law with the restoring force due to tension. Intuitively: higher tension pulls the string back faster (higher speed), while heavier string resists motion more (lower speed).

v=Tμv = \sqrt{\frac{T}{\mu}}

  1. Plug in the values Tension T=200 NT = 200\ \text{N}, μ=0.125 kg/m\mu = 0.125\ \text{kg/m}:

v=2000.125=1600=40 m/sv = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = 40\ \text{m/s}

  1. Calculate the time to travel the length The pulse must travel the entire length L=20.0 mL = 20.0\ \text{m} at speed v=40 m/sv = 40\ \text{m/s}:

t=Lv=20.040=0.5 st = \frac{L}{v} = \frac{20.0}{40} = 0.5\ \text{s}

Watch out

A common mistake is to forget that the mass given is the total mass of the string, not the mass per unit length. Always divide by the length first to get μ\mu.

Tip

Notice that the time does not depend on how hard you jerk the string — the wave speed is fixed by tension and density. A bigger jerk just makes a bigger pulse, but it still travels at the same speed.

✓Final answer

The disturbance takes 0.5 s\boxed{0.5\ \text{s}} to reach the other end.

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