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An organ tuner is comparing the fundamental pitch from a certain closed-end organ pipe that is 59.9 cm long
to the sound from an electronic signal generator set to a pitch of 151 Hz. The pipe is located in an
environment in which the speed of sound is 346 m/s. How many beats per second will the organ tuner hear?

Respuesta :

skyluke89

Answer:

6.6 beats per second

Explanation:

For a closed-end pipe, the fundamental frequency is given by

[tex]f_1 = \frac{v}{4 L}[/tex]

where

v is the speed of sound

L is the length of the pipe

Here we have

v = 346 m/s

L = 59.9 cm = 0.599 m

So the fundamental frequency is

[tex]f_1 = \frac{346 m/s}{4\cdot 0.599 m}=144.4 Hz[/tex]

The frequency of the signal generator is

[tex]f_2 = 151 Hz[/tex]

So the beat frequency is given by the absolute value of the difference between the two frequencies:

[tex]f_b = |f_1 -f_2|=|144.4 Hz-151 Hz|=6.6 Hz[/tex]

which means that 6.6 beats per second are heard.