Question

Transverse waves of the same frequency are generated in two steel wires A and B. The diameter of A is twice that of B, and the tension in A is half that in B. The ratio of the velocities of the waves in A and B is:

• A. \(1:2\)
• B. \(1:\sqrt{2}\)
• C. \(1:2\sqrt{2}\)
• D. \(3:2\sqrt{2}\)

Hint:

For wave velocity in a stretched wire: - \( v = \sqrt{T / \mu} \), where \( T \) is tension and \( \mu \) is linear mass density. - If the diameter changes, mass per unit length scales with \( d^2 \). - If the tension changes, velocity scales with \( \sqrt{T} \).

Answer 1

The Correct Option is C

Step 1: Understanding Wave Velocity in a Wire The velocity of a transverse wave in a stretched string is given by: \[ v = \sqrt{\frac{T}{\mu}} \] where: \( T \) is the tension in the wire, \( \mu \) is the mass per unit length of the wire. 
Step 2: Expressing Mass Per Unit Length For a cylindrical wire of density \( \rho \) and diameter \( d \): \[ \mu = \frac{{mass}}{{length}} = \rho \times \frac{\pi d^2}{4} \] Since the diameter of A is twice that of B: \[ d_A = 2d_B \] \[ \mu_A = \rho \times \frac{\pi (2d_B)^2}{4} = 4 \rho \times \frac{\pi d_B^2}{4} = 4\mu_B \] 
Step 3: Expressing Velocity Ratio Given that the tension in A is half that of B: \[ T_A = \frac{T_B}{2} \] Using the velocity formula: \[ v_A = \sqrt{\frac{T_A}{\mu_A}} = \sqrt{\frac{\frac{T_B}{2}}{4\mu_B}} = \sqrt{\frac{T_B}{8\mu_B}} = \frac{1}{\sqrt{8}} v_B \] \[ = \frac{1}{2\sqrt{2}} v_B \] Thus, the ratio of velocities is: \[ v_A : v_B = 1 : 2\sqrt{2} \]