Question

Two ropes of the same material of radius $ R $ and $ \frac{R}{2} $, what will be the ratio of wave speed in the second rope to the first? (They both are with the same tension)

Hint:

The wave speed in a rope is dependent on the tension and the linear mass density. The wave speed is proportional to the inverse of the square root of the rope's radius.

Answer 1

Correct Answer: 2

The wave speed \( v \) in a rope is given by the formula: \[ v = \sqrt{\frac{T}{\mu}} \] where: - \( v \) is the wave speed, - \( T \) is the tension in the rope, - \( \mu \) is the linear mass density of the rope, which is given by: \[ \mu = \frac{m}{L} = \frac{\rho A}{L} \] where \( m \) is the mass of the rope, \( L \) is its length, \( \rho \) is the density of the rope material, and \( A \) is the cross-sectional area of the rope.
For a cylindrical rope, \( A = \pi R^2 \). Thus, the linear mass density \( \mu \) is: \[ \mu = \frac{\rho \pi R^2}{L} \] Now, the wave speed becomes: \[ v = \sqrt{\frac{T}{\frac{\rho \pi R^2}{L}}} = \sqrt{\frac{T L}{\rho \pi R^2}} \] Since the tension is the same for both ropes, the ratio of wave speeds depends on the radii of the ropes.
If the radius of the second rope is \( \frac{R}{2} \), then the ratio of wave speeds in the second rope to the first rope is: \[ \frac{v_2}{v_1} = \frac{\sqrt{\frac{T}{\mu_2}}}{\sqrt{\frac{T}{\mu_1}}} = \sqrt{\frac{\mu_1}{\mu_2}} = \sqrt{\frac{R^2}{\left( \frac{R}{2} \right)^2}} = \sqrt{4} = 2 \] Thus, the ratio of wave speed in the second rope to the first rope is \( 2 \).