Question

Two steel guitar strings have the same length. String A has a diameter of 0.513 mm and is under 403 N of tension. String B has a diameter of 1.29 mm and is under a tension of 800 N. Calculate the ratio of the wave speeds, vA/vB, in these two strings.

Answer 1

Answer:

$\frac{v_{A}}{v_{B}} = 1.785$

Explanation:

$T_{A}$ = Tension force in string A = 403 N

$T_{B}$ = Tension force in string B = 800 N

$d_{A}$ = diameter of string A = 0.513 mm

$d_{B}$ = diameter of string B = 1.29 mm

$v_{A}$ = wave speed of string A

$v_{B}$ = wave speed of string B

Ratio of the wave speeds is given as

$\frac{v_{A}}{v_{B}} = \sqrt{\frac{T_{A}}{T_{B}}} \left ( \frac{d_{B}}{d_{A}} \right )$

$\frac{v_{A}}{v_{B}} = \sqrt{\frac{403}{800}} \left ( \frac{1.29}{0.513} \right )$

$\frac{v_{A}}{v_{B}} = 1.785$