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

To address this issue, we must analyze the relationship between the speed of a transverse wave on a string and its physical characteristics. The speed \( v \) of a transverse wave on a string is determined by the equation:

\(v = \sqrt{\frac{T}{\mu}}\)

where:

• \(T\) represents the tension in the string.
• \(\mu\) is the linear mass density of the string.

Given that both strings are composed of the same material and are subjected to identical tension, the primary determinant of the wave's speed is the linear mass density \(\mu\). Linear mass density is defined as:

\(\mu = \frac{m}{L}\),

where \(m\) is the mass of the string and \(L\) is its length. For a string with a circular cross-section, \( \mu \) can also be expressed in terms of its volume as:

\(\mu = \rho \cdot A\)

where:

• \(\rho\) is the density of the material
• \(A\) is the cross-sectional area

For a circular cross-section, the area \( A \) is calculated as:

\(A = \pi R^2\)

Consequently, the linear mass density is:

\(\mu = \rho \pi R^2\)

With constant tension, the ratio of velocities depends on the ratio of the square roots of the inverse of their linear densities:

\(\frac{v_2}{v_1} = \sqrt{\frac{\mu_1}{\mu_2}}\)

For the first string with radius \( R \), the linear density is:

\(\mu_1 = \rho \pi R^2\)

For the second string with radius \( \frac{R}{2} \), the linear density is:

\(\mu_2 = \rho \pi \left(\frac{R}{2}\right)^2 = \rho \pi \frac{R^2}{4}\)

The ratio calculation yields:

\(\frac{v_2}{v_1} = \sqrt{\frac{\rho \pi R^2}{\rho \pi \frac{R^2}{4}}} = \sqrt{4} = 2\)

Therefore, the ratio \(\frac{v_2}{v_1}\) is \(2\). The correct answer is:

Option: 2