A tube that is closed at one end produces tones of 75 Hz, 225 Hz, and 375 Hz. If the speed of sound in air is 340 m/s, what is the length of the pipe?
In a closed-end tube, only odd harmonics are produced. The fundamental frequency or the first harmonic is given by:
f1 = v/2L,
where f1 is the fundamental frequency, v is the speed of sound, and L is the length of the pipe.
The frequencies produced by the tube are 75 Hz, 225 Hz, and 375 Hz, which correspond to the 3rd, 7th, and 11th harmonics, respectively.
So, according to the pattern of odd harmonics, we have the following equations:
f3 = 3f1 = 75 Hz
f7 = 7f1 = 225 Hz
f11 = 11f1 = 375 Hz
Rearranging the equation for the fundamental frequency, we get:
f1 = v/2L
Substituting the value of f3 = 75 Hz:
75 = v/2L
Simplifying the equation, we have:
L = v/2f3
L = 340 m/s / (2 * 75 Hz)
L = 2.27 m or 2.3 meters
Therefore, the length of the pipe is approximately 2.3 meters.