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 solve this problem, we need to examine the relationship between the velocity of a transverse wave in a string and the string's physical properties. The velocity \( v \) of a transverse wave on a string is given by the formula:

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

where:

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

Since both strings are made from the same material and are under the same tension, the critical factor that affects the wave's velocity 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 described in terms of its volume as:

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

where:

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

For a circular cross-section, the area \( A \) is given by:

\(A = \pi R^2\)

Therefore, the linear mass density becomes:

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

Since the tension is the same, the ratio of velocities is determined by 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:

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

Calculating the ratio:

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

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

Option: 2

Answer 2

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 \( 2 \).