Question

The speed of a transverse wave in a stretched string 'A' is 'v'. Another string 'B' of same length and same radius is subjected to same tension. If the density of the material of the string 'B' is 2\% more than that of 'A', then the speed of the transverse wave in string 'B' is

• A. \(\sqrt{1.04}v\)
• B. \(\sqrt{1.02}v\)
• C. \(\frac{v}{\sqrt{1.04}}\)
• D. \(\frac{v}{\sqrt{1.02}}\)

Hint:

The speed of a transverse wave in a string is inversely proportional to the square root of the linear mass density.

Answer 1

The Correct Option is D

The speed of a transverse wave in a stretched string is given by: $$v = \sqrt{\frac{T}{\mu}}$$ where \(T\) is the tension and \(\mu\) is the linear mass density. The linear mass density is given by: $$\mu = \frac{m}{L} = \frac{\rho V}{L} = \frac{\rho A L}{L} = \rho A$$ where \(\rho\) is the density of the material and \(A\) is the cross-sectional area. Since the length and radius of the strings are the same, the cross-sectional area is the same. Let \(\rho_A\) and \(\rho_B\) be the densities of strings A and B respectively. We are given that \(\rho_B = \rho_A + 0.02\rho_A = 1.02\rho_A\). Let \(v_A\) and \(v_B\) be the speeds of the transverse waves in strings A and B respectively. We have: $$v_A = \sqrt{\frac{T}{\rho_A A}} = v$$ $$v_B = \sqrt{\frac{T}{\rho_B A}} = \sqrt{\frac{T}{1.02\rho_A A}}$$ We can write: $$\frac{v_B}{v_A} = \sqrt{\frac{\rho_A A}{1.02\rho_A A}} = \sqrt{\frac{1}{1.02}}$$ $$v_B = v_A \sqrt{\frac{1}{1.02}} = \frac{v}{\sqrt{1.02}}$$