Question

Problem 3: One of the harmonic frequencies of tube A with two open ends is 325 Hz. The next-highest harmonic frequency is 390 Hz. (a) What is the fundamental frequency of the tube A?! (b) What harmonic frequency is next highest after the harmonic frequency 195 Hz? (c) What is the number (the integer multiplied by the fundamental frequency to get the requested frequency) of this next-highest harmonic? One of the harmonic frequencies of tube B with only one open end is 1080 Hz. The next highest harmonic frequency is 1320 Hz. (d) What is the fundamental frequency of the tube B? (e) What harmonic frequency is next highest after the harmonic frequency 600 Hz? (f) What is the number (the integer multiplied by the fundamental frequency to get the requested frequency) of this next highest harmonic?

Answer 1

a) For two open ends fundamental frequency, F = nu, where v- speed of sound al l= length of tube Given Fn = 325 Hz n = integer Fnti = 390 H₂ - 21 nu = 325-(1) & (n+1) v = 390 (2) 21 Dividing (2) by (1) we get 325 =n 390 ntl =) n= 5 Putting n=5 in eq (1) we get y = 65 Hz .. For n=1 F = 65Hz 6) Given, Fn = 195Hz =) 195= ny = nx 65 Now = =) n = 3 next highest frequency will be for n= 4 FA 4X 65=260 Hz

c) Fourth Harmonic (n=4) d) Given Fn = 1080 hz Fura = 1320hz Fundamental frequency for one open end F = nu , Only odd harmonics of the fundamental are natural frequencies for a one open end. 40 Now, my 1080 - 0 l (n+20v = 1320 - © 41 4 l Dividing both equation n = 1080 nta 1320 =) n = 9 Putting n=9 in ① we get a = 1080 = V = 120 Hz 40 41 =) For n=1 F = 120 Hz, Fundamental frequency

I el Given Fn = 600 Hz =) 60o = nxu = nx 120 41 =) n = 5 i the next highest frequency will be for n+2=7 = F = 7x120= 840 Hz F) Seventh harmonic for n=7