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. Â Â
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 = Ď*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
