The lowest pitch that can be played on a normal flute with all of the keys close
ID: 1703936 • Letter: T
Question
The lowest pitch that can be played on a normal flute with all of the keys closed is middle C. A flute player playing middle C on a particular flute produces a sound with a frequency of 262 s-1. Assume that the speed of sound is 345 m/s.(a) Taking the flute to be a pipe which is open at both ends, estimate the length of the flute.
(b) What are the second- and third-frequencies that can be produced on this flute with all of the
keys closed?
(c) The flute player is playing on the back of a truck which is driving down a parade route at 10
mi/hr. A guy on a tiny motorcycle is traveling in the opposite direction, towards the flute player,
at 15 mi/hr (relative to the road). Supposing that he is actually able to hear the flute over the
sound of his motorcycle, how many compressions per second arrive at his ear from the flute?
Explanation / Answer
The frequency of the flute is f = 262 /s
The speed of the sound is v = 345 m/s
a)
For the open pipe, the resonance occur at
f = n v/2L n= 1,2,3,...........
L is the length of the air coloumn
As all the keys are closed, L is the length of the flute
262 = 1* 345/2*L
L = 0.6584 m
b)
The second resonance will occur at
f2 = 2 v/2L
f2 = 2*345/2*0.6584
f2 = 523.99 = 524 /s (appx)
The third resonance will occur at
f3 = 3 v/2L
f3 = 3*345/2*0.6584
f3 = 785.99 = 786 /s (appx)
c)
The speed os the truck (source) Vs = 10 mi/hr = 4.4704 m/s
the speed of the motorist (observer) Vo = 15 mi.hr = 6.7056 m/s
The observer is approaching the source, therefore the apparent frequency at the observere is
f' = {(v+Vo)/(v-Vs)}f
f' = {(345+6.7056)/(345-4.4704)}*262
f' = 270.59 = 270.6 /s (appx)
Therefore the compressions per second arrive at his ear from the flute is 270.6
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