- #1
kell
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Homework Statement
An observer listens to a vibrating string using a cardboard tube of length 1.4m placed close to his ear (one end closed). The string is excited so as to vibrate at its fundamental frequency, and the tension is increased slowly. The intensity of the sound heard by the observer increases at 130N and 360N (I assume these are two resonant frequencies for the tube then). The linear density of the string is 1.6g/m and its length is 40cm. Deduce the speed of sound in this air.
Homework Equations
v=sqrt(tension/linear density)
f(string)=nv/2L
f(tube)=nv/4L
That's all I've been using.
The Attempt at a Solution
My reasoning was: Find the two frequencies of the string, find the difference, and this is the fundamental of the tube; put it through the f(tube) equation to get v.
v(string)= sqrt(130N/1.6E-3kg/m)= 285.04 m/s
Thus f(string)= nv(2L) where n=1 because it's at its fundamental, f=285.04m/s/0.8m= 356.3Hz
v(string)2= (same method) = 474.34m/s
Thus f(string)2= (same method) = 592.9 Hz
The difference in the two is 236.6 Hz; I took this to be the fundamental frequency and things start to feel queasy:
v=λf= (4L)*f = (5.6)(236.6)... This comes out at 1324.96 m/s. This is, what, 4 times too big.
I tried this too:
356.3= (n*v)/5.6 592.9=(n+1)*v/5.6
v=1995/n 3320=(n+1)*v = (n+1)(1995/n)
3320n=(n+1)(1995)=1995n+1995
1325n=1995
n=... 1.5. Nonsense.
So I'm a bit stuck!