two steel guitar strings have the same length. string a has a diameter of 0.50 mm and is under 410.0 n of…

Explanation:

Step1: Recall wave - speed formula

The wave speed on a 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 $\mu=\rho A$, where $\rho$ is the mass per unit volume (constant for the same material, here steel) and $A$ is the cross - sectional area. The cross - sectional area of a circular string is $A=\pi r^{2}=\frac{\pi d^{2}}{4}$, so $\mu=\rho\frac{\pi d^{2}}{4}$. Then $v = \sqrt{\frac{4T}{\rho\pi d^{2}}}$.

Step2: Find the ratio of wave speeds

For string A, $v_A=\sqrt{\frac{4T_A}{\rho\pi d_A^{2}}}$, and for string B, $v_B=\sqrt{\frac{4T_B}{\rho\pi d_B^{2}}}$. The ratio $\frac{v_A}{v_B}=\sqrt{\frac{T_A}{T_B}\times\frac{d_B^{2}}{d_A^{2}}}$. We are given that $T_A = 410\ N$, $T_B = 820\ N$, $d_A=0.5\ mm$, and $d_B = 1.0\ mm$. Substitute the values: $\frac{v_A}{v_B}=\sqrt{\frac{410}{820}\times\frac{(1.0)^{2}}{(0.5)^{2}}}$. First, calculate $\frac{410}{820}=\frac{1}{2}$ and $\frac{(1.0)^{2}}{(0.5)^{2}}=\frac{1}{0.25}=4$. Then $\frac{v_A}{v_B}=\sqrt{\frac{1}{2}\times4}=\sqrt{2}\approx1.4$.

Answer:

a. 1.4