At 20 degrees C, a pipe open at both ends resonates at a frequency of 440 Hertz. At what frequency does the same pipe resonate on a partucularly cold day when the speed of sound is 3% lower than it would be at 20 degrees C.
2006-08-17 16:56:07 · 4 answers · asked by orchard_littlejoe 2 in Science & Mathematics ➔ Physics
First, we have to find the velocity of sound at 20 degrees C.
V=[331.5+0.606(20.0)] m/sec
V=343.6 m/sec
Since we know the frequency and velocity, we can find the wavelength.
f=V/λ
440 Hz=(343.6 m/sec)/λ
λ=0.7809 m
Now we find what the speed of sound is if it's 3% lower.
343.6 m/sec(0.97)=333.3 m/sec
Now we can find the new frequency.
f=(333.3 m/sec)/(0.7809 m)
f=426.8 Hz
2006-08-17 18:01:21 · answer #1 · answered by Anonymous · 1⤊ 0⤋
307
2006-08-17 17:04:26 · answer #2 · answered by fumeluv 2 · 0⤊ 0⤋
400 hertz
cheers
2006-08-17 17:01:45 · answer #3 · answered by Anonymous · 0⤊ 0⤋
426.8 Hz.
2006-08-17 17:37:12 · answer #4 · answered by Dusty 7 · 1⤊ 0⤋