A baritone saxophone acts as if it is a tube that is open at both ends (it is ac

Question

A baritone saxophone acts as if it is a tube that is open at both ends (it is actually a conical tube that is closed at one end, but acoustically these are equivalent). When all the holes are closed, the lowest note it can play is a D-flat two octaves below middle C, which has a frequency of about 68 Hz. (a) What is the overall length of the saxophone? Assume v = 343 m/s for air. (b) What are the frequencies of the next two harmonics of the baritone saxophone? Assume all the holes are closed as before. Show your work. f2= f3= (c) Now suppose the saxophone from part (a) plays its lowest note at the same time in another saxophone plays its own lowest note. However, the second saxophone is out of tune, so it plays a frequency of 71 Hz. What beat frequency will be heard when this happens? v = f lambda. fn = nv/2L, n = 1, 2, 3,... (open-open pipe)

Explanation / Answer

Frequency of instrument = f = 68 Hz

(a) v = 343 m/s

As the instrument is conical tube closed at one end. Then its length will be given by:

L = v / (2 f) = 343/ 2 x 68 = 343 / 136 = 2.52 meters

Hence, its length is L = 2.52 meters

(b)F2 and f3 be the freq of next two harmotics. and will be given by:

f(n) = n v / 2 L

for n = 2

f2 = 2 x 343 / 2 x 2.52 = 136.1 Hz

f3 = 3 x 343 / 2 x 2.52 = 204.1

part (c) freq of second instrument = f' = 71 Hz

f(beat) =f' - f = 71 - 68 = 3 Hz

f2 , f3 andso on, and their corresponding beat freq will be in the multiple of 3 x n. Ex

f2' = 142 hz (using the same formula as part 2)

f2(beat) = f2' - f2 = 142 - 136 = 6 Hz

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