Question

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?

Answer 1

6.6 beats per second

Step-by-step explanation:

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

`f_1 = (v)/(4 L)`

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

`f_1 = (346 m/s)/(4\cdot 0.599 m)=144.4 Hz`

The frequency of the signal generator is

`f_2 = 151 Hz`

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

`f_b = |f_1 -f_2|=|144.4 Hz-151 Hz|=6.6 Hz`

which means that 6.6 beats per second are heard.