Answer:
The length is [tex]L_d= 0.069 \ m[/tex]
Explanation:
From the question we are told that
The length of the wire [tex]L = 85cm = \frac{85}{100} = 0.85m[/tex]
The mass is [tex]m = 7.25g = \frac{7.25}{1000} = 7.25^10^{-3}kg[/tex]
The tension is [tex]T = 4170N[/tex]
Generally the frequency of oscillation of a stretched wire is mathematically represented as
[tex]f = \frac{n}{2L} \sqrt{\frac{T}{\mu}[/tex]
Where n is the the number of nodes = 3 (i.e the third harmonic)
[tex]\mu[/tex] is the linear mass density of the wire
This linear mass density is mathematically represented as
[tex]\mu = \frac{m}{L}[/tex]
Substituting values
[tex]\mu = \frac{7.2*10^{-3}}{0.85}[/tex]
[tex]= 8.53 *10^{-3} kg/m[/tex]
Substituting values in to the equation for frequency
[tex]f = \frac{3}{2 80.85} * \sqrt{\frac{4170}{8.53*10^{-3}} }[/tex]
[tex]= 1234Hz[/tex]
From the question the we can deduce that the fundamental frequency is equal to the oscillation of a stretched wire
The fundamental frequency is mathematically represented as
[tex]f = \frac{v}{4L_d}[/tex]
Where [tex]L_d[/tex] is the length of the pipe
v is the speed of sound with a value of [tex]v = 343m/s[/tex]
Making [tex]L_d[/tex] the subject of the formula
[tex]L_d = \frac{v}{4f}[/tex]
Substituting values
[tex]L_d = \frac{343}{(4)(1234)}[/tex]
[tex]L_d= 0.069 \ m[/tex]
From the question the we can deduce that the fundamental frequency is equal to the oscillation of a stretched wire
