Respuesta :
Answer:
[tex]\dfrac{v_A}{v_B}=1.87[/tex]
Explanation:
We know that the velocity of the wave in the string is given by following relation as :
[tex]v=\sqrt{\dfrac{Tl}{m}}[/tex]
T is the tension in the string
l is the length of the string
m is the mass of the string
We know that density is given by :
[tex]\rho=\dfrac{m}{V}[/tex]
[tex]m=\rho V[/tex] , V is volume, V = Al
Velocity, [tex]v=\sqrt{\dfrac{T}{\rho(\pi d^2/4)}}[/tex]
[tex]v\propto \dfrac{\sqrt{T} }{d}[/tex]..........(1)
The diameter of string A, [tex]d_A=0.491\ mm[/tex]
Tension in string A, [tex]T_A=394\ N[/tex]
The diameter of string B, [tex]d_B=1.32\ mm[/tex]
Tension in string B, [tex]T_B=807\ N[/tex]
From equation (1) :
[tex]\dfrac{v_A}{v_B}=\sqrt{\dfrac{T_A}{T_B}}\times \dfrac{d_B}{d_A}[/tex]
[tex]\dfrac{v_A}{v_B}=\sqrt{\dfrac{T_A}{T_B}}\times \dfrac{d_B}{d_A}[/tex]
[tex]\dfrac{v_A}{v_B}=\sqrt{\dfrac{394}{807}}\times \dfrac{1.32}{0.491}[/tex]
[tex]\dfrac{v_A}{v_B}=1.87[/tex]
So, the ratio of the wave speeds in these two strings is 1.87. Hence, this is the required solution.
The quotient of the two wave speeds is 1.88
How to get the wave speeds?
The wave speed is given by:
[tex]v = \sqrt{\frac{T*l}{m} }[/tex]
Where T is the tension, l is the length of the string, and m is the mass of the string.
The mass is given by:
m = ρ*V
We assume that both strings have the same density, and the volume is given by:
V = 3.14*(d/2)^2*l
Where d is the diameter of the string.
Then we can write:
[tex]v = \sqrt{\frac{T*l}{\rho*3.14*(d/2)^2*l} } = \sqrt{\frac{T}{\rho*3.14*(d/2)^2} }[/tex]
Now, the quotient of the two velocities can be written as:
[tex]\frac{v_a}{v_b} = \frac{\sqrt{\frac{T_a}{\rho*3.14*(d_a/2)^2} } }{\sqrt{\frac{T_b}{\rho*3.14*(d_b/2)^2} } } \\\\\\\frac{v_a}{v_b} = \sqrt{(T_a/T_b)} *\frac{d_b}{d_a} = \sqrt{(394N/807N)} *\frac{1.32mm}{0.491mm} = 1.88[/tex]
So the quotient of the two velocities is 1.88
If you want to learn more about wave speed, you can read:
https://brainly.com/question/14015797
