Question:

In the diagram shown below, both the strings AB and CD are made of the same material and have the same cross-section. The pulleys are light and frictionless. If the speed of the wave in string AB is \( v_1 \) and in CD is \( v_2 \), then the ratio \( \frac{v_1}{v_2} \) is: 
 

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For wave speed in stretched strings: - \( v = \sqrt{T / \mu} \), - If the material and cross-section are the same, the speed ratio depends only on tension. - In pulley-based problems, check the force distribution to determine tension relationships.
Updated On: Mar 24, 2025
  • \(1\)
  • \(\sqrt{2}\)
  • \(2\)
  • \(\frac{1}{\sqrt{2}}\)
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The Correct Option is D

Solution and Explanation

Step 1: Understanding Wave Speed in a Stretched String The speed of a transverse wave in a stretched string is given by: \[ v = \sqrt{\frac{T}{\mu}} \] where: \( T \) is the tension in the string, \( \mu \) is the mass per unit length. Since both strings are made of the same material and have the same cross-section, their linear mass densities (\(\mu\)) are equal. 
Step 2: Analyzing Tension in the Strings - Let the tension in string AB be \( T_1 \). - The tension in string CD, which supports an additional load due to the pulley arrangement, is given by: \[ T_2 = 2T_1 \] 
Step 3: Finding the Velocity Ratio Using the wave speed formula: \[ v_1 = \sqrt{\frac{T_1}{\mu}}, \quad v_2 = \sqrt{\frac{T_2}{\mu}} \] \[ v_2 = \sqrt{\frac{2T_1}{\mu}} = \sqrt{2} v_1 \] Thus, the ratio is: \[ \frac{v_1}{v_2} = \frac{v_1}{\sqrt{2} v_1} = \frac{1}{\sqrt{2}} \]

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