A flutist assembles her flute in a room where the speed of sound is 342m/s . Whe

Question

A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune with a 440Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 343m/s .How many beats per second will she hear if she now plays the note A as the tuning fork is sounded?How far does she need to extend the "tuning joint" of her flute to be in tune with the tuning fork? A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune with a 440Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 343m/s .How many beats per second will she hear if she now plays the note A as the tuning fork is sounded?How far does she need to extend the "tuning joint" of her flute to be in tune with the tuning fork?

Explanation / Answer

let (L) be length of flute used in playing A-note (open ends forming antinodes, L= lamda/2)
f1 = v1/2L = 440 (resonance)
342/2L = 440
2L = 342/440 ---- (1)
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when v2 = 343m/s, resonance with fork will be out-tuned as freq will change
f2 = v2/2L = 343/2L = 343*440/342 > greater than f1
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Beat freq f = f2-f1 = 343*440/342 - 440
f = 1.287(1beats/s) --- (2)
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f2 > f1 because now L length is not able to keep ratio with increased speed of sound, so lets increase L to (L+x)
so that f2' = f1 (again resonates) or beats f=0
343/2(L+x) = 440
2L+2x = 346/440 >>> using (1)
342/440 + 2x = 343/440
2x = 1/440
x = 1/880 = 1.136*10^-3meter
x = 1.136 milli-meters

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