Question

Two strings with circular cross section and made of same material are stretched to have same amount of tension. A transverse wave is then made to pass through the strings. The velocity of the wave in the first string having the radius of cross section $ R $ is $ v_1 $, and that in the other string having radius of cross section $ R/2 $ is $ v_2 $. Then, $ \frac{v_2}{v_1} $ is:

• A. \( 4 \)
• B. \( \sqrt{2} \)
• C. \( 8 \)
• D. \( 2 \)

Hint:

For waves on stretched strings, the velocity depends on the tension and the linear mass density. When comparing strings with different cross-sectional areas, use the formula \( v = \sqrt{T / \mu} \) and account for changes in area to find the velocity ratio.

Answer 1

The Correct Option is D

The velocity of a wave in a stretched string is given by the formula: \[ v = \sqrt{\frac{T}{\mu}} \] Where \( T \) is the tension in the string and \( \mu \) is the linear mass density of the string, which is given by: \[ \mu = \frac{m}{L} = \frac{\rho A}{L} \] where: - \( \rho \) is the density of the material of the string, - \( A \) is the cross-sectional area of the string, - \( L \) is the length of the string. Now, since the strings are made of the same material and are stretched under the same tension, we can compare the velocities in both strings by considering their cross-sectional areas.
For a string with a circular cross section, the area \( A \) is given by: \[ A = \pi r^2 \] For the first string, the radius is \( R \), so the area is: \[ A_1 = \pi R^2 \] For the second string, the radius is \( \frac{R}{2} \), so the area is: \[ A_2 = \pi \left( \frac{R}{2} \right)^2 = \frac{\pi R^2}{4} \] Thus, the linear mass density for the two strings will be: \[ \mu_1 = \frac{\rho A_1}{L} = \frac{\rho \pi R^2}{L}, \quad \mu_2 = \frac{\rho A_2}{L} = \frac{\rho \pi \left( \frac{R}{2} \right)^2}{L} = \frac{\rho \pi R^2}{4L} \] The ratio of the velocities is given by: \[ \frac{v_2}{v_1} = \frac{\sqrt{\frac{T}{\mu_2}}}{\sqrt{\frac{T}{\mu_1}}} = \sqrt{\frac{\mu_1}{\mu_2}} = \sqrt{\frac{\frac{\rho \pi R^2}{L}}{\frac{\rho \pi R^2}{4L}}} = \sqrt{4} = 2 \] Thus, the ratio \( \frac{v_2}{v_1} \) is \( 2 \).