Question

An organ tuner is comparing the fundamental pitch from a certain closed-end organ pipe that is 59.9 cm long

Answer 1

Answer:

6.6 beats per second

Explanation:

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

$f_1 = \frac{v}{4 L}$

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

$f_1 = \frac{346 m/s}{4\cdot 0.599 m}=144.4 Hz$

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.