Question

Two steel guitar strings have the

Answer 1

Answer:

$\dfrac{v_A}{v_B}=1.47$

Explanation:

given,

diameter of string A = 0.5 mm

tension in string A = 440 N

diameter of string B = 1 mm

tension in string B = 820 N

wave speed =

$v = \sqrt{\dfrac{T}{M}}$

M is the linear density that is mass per unit length =mass / length

 mass = density x volume

$M = \dfrac{\pi r^2d}{l}$

$v = \sqrt{\dfrac{T}{\pi r^2d}}$

from the above equation

$v \alpha\ \sqrt{\dfrac{T}{ r^2}}$

now,

$\dfrac{v_A}{v_B}=\sqrt{\dfrac{T_A}{ T_B}\times \dfrac{r_B^2}{r_A^2}}$

$\dfrac{v_A}{v_B}=\sqrt{\dfrac{440}{820}\times \dfrac{1^2}{0.5^2}}$

$\dfrac{v_A}{v_B}=1.47$

ratio of the wave speed Va/Vb is 1.47