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A pipe is 2.40 m long. (a) Determine the frequencies of the first three harmonic

ID: 1514183 • Letter: A

Question

A pipe is 2.40 m long.

(a) Determine the frequencies of the first three harmonics if the pipe is open at both ends. Take 349 m/s as the speed of sound in air. f1 = Hz f2 = Hz f3 = Hz

(b) How many harmonic frequencies of this pipe lie in the audible range, from 20 Hz to 20000 Hz?

(c) What are the three lowest possible frequencies if the pipe is closed at on end and open at the other? f1 = Hz f3 = Hz f5 = Hz

(d) What length of pipe open at both ends has a fundamental frequency of 3.63 102 Hz? Find the first overtone. lpipe = m fovertone = Hz (

e) If the one end of this pipe is now closed, what is the new fundamental frequency? Find the first overtone. ffundamental = Hz fovertone = Hz

(f) If the pipe is open at one end only, how many harmonics are possible in the normal hearing range from 20 to 20000 Hz? n =

Explanation / Answer

L = 2.4 m , v =349 m/s

for open pipe fundamental frequency f = v/2L

(a) f1 = 349/(2*2.4) = 72.71 Hz

f2 = 2f1 =2*72.71 = 145.42 Hz

f3 = 3f1 = 3*72.71 = 218.13 Hz

(b) fn = n*f1

20000 = n*72.71

n = 275 harmonics are possible

(c) fro closed pipe fundamental frequncy f = v/4L

f1 = 349/(4*2.4) = 36.36 Hz

f3 = 3f1 = 3*36.36 = 109.1 Hz

f5 = 5f1 =5*36.36= 181.8 Hz

d) f = v/2L

3.63*10^2 = 349/2L

L = 0.481 m

first overtone f2 = 2*f1 = 2*363 = 726 Hz

e) f1 = v/4L = 349/(4*0.481) =181.4 Hz

f3 = 3*f1 = 3*181.4 = 544.2 Hz

f) 20000= (n+(1/2))363

for close pipe n = 55 harmonics are possible

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