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 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
[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]
