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)
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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.
The speed of a wave, denoted as \( v \), in a rope is determined by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Here, \( v \) represents the wave speed, \( T \) is the tension applied to the rope, and \( \mu \) is the rope's linear mass density. The linear mass density is calculated as:
\[ \mu = \frac{m}{L} = \frac{\rho A}{L} \]
In this equation, \( m \) is the rope's mass, \( L \) is its length, \( \rho \) is the density of the material composing the rope, and \( A \) is the rope's cross-sectional area. For a cylindrical rope, the cross-sectional area is given by \( A = \pi R^2 \).
Consequently, the linear mass density \( \mu \) can be expressed as:
\[ \mu = \frac{\rho \pi R^2}{L} \]
Substituting this into the wave speed formula yields:
\[ v = \sqrt{\frac{T}{\frac{\rho \pi R^2}{L}}} = \sqrt{\frac{T L}{\rho \pi R^2}} \]
When the tension is constant across two ropes, the ratio of their wave speeds is dependent on their radii. If the second rope has a radius of \( \frac{R}{2} \) compared to the first, the ratio of wave speeds (\( v_2 \) to \( v_1 \)) is:
\[ \frac{v_2}{v_1} = \frac{\sqrt{\frac{T}{\mu_2}}}{\sqrt{\frac{T}{\mu_1}}} = \sqrt{\frac{\mu_1}{\mu_2}} = \sqrt{\frac{R^2}{\left( \frac{R}{2} \right)^2}} = \sqrt{4} = 2 \]
Therefore, the wave speed in the second rope is twice that of the first rope.
