Four closed organ pipes are such that the fundamental frequencies of the three s

Question

Four closed organ pipes are such that the fundamental frequencies of the three shorter pipes are the first three overtones of the longest pipe. The fundamental frequency of the longest pipe is 110 Hz when the temperature is 22 degree C. a. Find the combined length of the four pipes, b. If the longest pipe is sounded at an intensity level of 75 dB and the shortest pipe is sounded at a intensity level of 55 dB, find the ratio of the intensity of the longest pipe to the intensity of the shortest pipe.

Explanation / Answer

a)

speed of sounf,

v = vo + 0.6*Tc

= 331 + 0.6*20

= 343 m/s

fundamental frequency of lognest closed pipe,

f1 = v/(4*L1)

L1 = v/(4*f1)

= 343/(4*110)

= 0.78 m

the possible next frequenies are

first overtone, f3 = 3*f1

= 3*110

= 330 Hz

second overtone, f5 = 5*f1

= 5*100

= 550 hz

third overtone, f7 = 7*f1

= 7*110

= 770 Hz

let L2,L3 and L4 are the lengths of the next higher frequencies.

f3 = v/(4*L2)

L2 = v/(4*f3) = 343/(4*330) = 0.26 m

f5 = v/(4*L3)

L3 = v/(4*f5) = 343/(4*550) = 0.156 m

f7 = v/(4*L4)

L4 = v/(4*f7) = 343/(4*770) = 0.111 m

length of the four pipes = L1 + L2 + L3 + L4

= 0.78 + 0.26 + 0.156 + 0.111

= 1.307 m

b) beta_logest = 75 dB

beta_shortest = 55 dB

we know, beta = 10*log(I/Io)

beta/10 = log(I/Io)

10^(beta/10) = I/10^-12

I = 10^(beta/10 - 12)

so,

I_logest/I_shortest = 10^(beta_longest/10 - 12)/10^(beta_longest/10 - 12)

= 10^(75/10 - 12)/10^(55/10 - 12)

= 100

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