Respuesta :
Answer:
The length of organ pipe A is [tex]L = 0.3611 \ m[/tex]
The length of organ pipe B is  [tex]L_b = 0.2708 \ m[/tex]
Explanation:
From the question we are told that
  The fundamental frequency is  [tex]f = 475 Hz[/tex]
   The speed of sound is  [tex]v_s = 343 \ m/s[/tex]
The fundamental frequency of the organ pipe A Â is mathematically represented as
    [tex]f= \frac{v_s}{2 L}[/tex]
Where L is the length of  organ pipe
  Now  making L the subject
    [tex]L = \frac{v_s}{2f}[/tex]
substituting values
    [tex]L = \frac{343}{2 *475}[/tex]
    [tex]L = 0.3611 \ m[/tex]
The second harmonic frequency of the  organ pipe A is mathematically represented as
    [tex]f_2 = \frac{v_2}{L}[/tex]
The third harmonic frequency of the  organ pipe B is mathematically represented as   Â
   [tex]f_3 = \frac{3 v_s}{4 L_b }[/tex]
So from the question
    [tex]f_2 = f_3[/tex]
So
  [tex]\frac{v_2}{L} = \frac{3 v_s}{4 L_b }[/tex]
Making  [tex]L_b[/tex] the subject
   [tex]L_b = \frac{3}{4} L[/tex]
substituting values
  [tex]L_b = \frac{3}{4} (0.3611)[/tex]
  [tex]L_b = 0.2708 \ m[/tex]