Question

A pipe is 2.37 m long.

(a) Determine the frequencies of the first three harmonics if the pipe is open at both ends. Take 344 m/s as the speed of sound in air.

f1 = 72.6 Correct: Your answer is correct. Hz

f2 = 145.2 Correct: Your answer is correct. Hz

f3 = 217.8 Correct: Your answer is correct. Hz

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

275 Correct: Your answer is correct.

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

f1 = 36.3 Correct: Your answer is correct. Hz

f3 = 108.9 Correct: Your answer is correct. Hz

f5 = 181.5 Correct: Your answer is correct. Hz

_______________________ ___________________________________________

(a) What length of pipe open at both ends has a fundamental frequency of 3.75 102 Hz? Find the first overtone.

lpipe = _________m

fovertone = ________Hz

(b) If the one end of this pipe is now closed, what is the new fundamental frequency? Find the first overtone.

ffundamental = ________Hz

fovertone = __________Hz

(c) 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 =

Answer 1

13.5)Given,

f = 3.75 x 10^2 Hz = 375 Hz

a)We know that

f = n v/2L

L = nv/2f

for n = 1

l(pipe) = 1 x 340/2 x 375 = 0.453 m

f = 2 x 340/2 x 0.453 = 750.55 Hz

Hence, l(pipe) = 0.453 m ; f(overtone) = 750.55 Hz

b)we know that

f = (2n - 1)v/4L

f1 = (2 - 1)340/4 x 0.453 = 187.64 Hz

f2 = (2*2 - 1)340/4 x 0.453 = 562.9

F(fund) = 187.64 Hz ; f(over) = 562.9 Hz

c)interger = 20000/187.64 = 106

range - 187.64 - 106 x 187.64

range = 187.64 - 19889.84 Hz

n = (106 + 2)/2 = 54