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Organ pipe A, with both ends open, has a fundamental frequency of 360 Hz. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. How long are (a) pipe A and (b) pipe B? (Take the speed of sound to be 343 m/s.)

Respuesta :

Answer:

The length of the pipe A is [tex]L_A[/tex] = 0.4763 m &  the length of the pipe B is [tex]L_B =[/tex] 0.357 m

Explanation:

Fundamental  frequency  = 360 Hz

Velocity = 343 [tex]\frac{m}{s}[/tex]

(a). Length of the pipe is given by

[tex]L = \frac{V}{2 f}[/tex]

Put all the values in above equation we get

[tex]L = \frac{343}{2 (360)}[/tex]

[tex]L_A[/tex] = 0.4763 m

(b). Given that

The third harmonic of organ pipe B = the second harmonic of pipe A

[tex]\frac{n_B}{4L_B} = \frac{n_A}{2L_A}[/tex]

Thus

[tex]L_B = \frac{2 n_{B} L_A }{4n_A}[/tex]

Put all the values in above formula we get

[tex]L_B = \frac{2 (3)(0.4763) }{4(2)}[/tex]

[tex]L_B =[/tex] 0.357 m

Therefore the length of the pipe A is [tex]L_A[/tex] = 0.4763 m &  the length of the pipe B is [tex]L_B =[/tex] 0.357 m

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