A. The lowest pedal note on a particular large pipe organ has a fundamental freq

Question

A. The lowest pedal note on a particular large pipe organ has a fundamental frequency of 15.7 Hz. This extreme bass note, about four octaves below middle C, is more felt than heard. What is the length of pipe between the sounding hole and the open end? m

B. Two strings are adjusted to vibrate at exactly 186 Hz. Then the tension in one string is increased slightly. Afterward, three beats per second are heard when the strings vibrate at the same time. What is the new frequency of the string that was tightened? hz

Explanation / Answer

A)

if the pipe is open at both ends
then fundamental frequency is f= v/(2L) = 15.7

length of the pipe is L = v/(2*f) = 342/(2*15.7) = 10.9 m

B) if tension is increased then frequency also increases

So beat frequency is f1-f2 = 3

f1 = 3+f2 = 3+186 = 189 Hz

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