May 21st 2011, 10:17 AM
Join Date: Feb 2009
This one’s a little tricky. It hasn’t been mentioned anywhere that the resonances occur at the fundamental frequency.
Let’s write lambda as Y
The end correction is usually small compared to the resonating length. So if we consider it to be resonating at the fundamental at 0.62 (since 0.62>0.6), then we should expect a wavelength lambda close to (4 x 0. 62) or 2.48.
We know that for a tube closed at one end, the allowed resonances correspond to Y/4, 3Y/4 , 5Y/4 i.e. (2n+1) Y/4 ; n = 0,1,2……
The difference between successive allowed lengths thus works out to
(3Y/4) – Y/4 = Y/2 = 0.62 - 0.6 = 0.02 or Y = 2 x 0.02 =0.04
Compare this with the earlier value. Obviously something’s wrong.
What’s actually happening is this. An allowed frequency higher than the fundamental is what causes the resonance, and the wavelength corresponding to this frequency (not the fundamental wavelength) is equal to twice this difference in length,
Thus if L is the effective length and L’ the measured length, we may write
L = L’ + 0.3d.
Since the allowed Ys are given by, (2n+1)Y/4 = L, we may write for the two cases
Y = 4 L / (2n+1) = 4(L’ + 0.3d)/(2n+1).
Note that the two lengths 0.62 and 0.6 correspond to the same wavelength but different numbers of this wavelength. They do not represent resonance at different frequencies. Thus we may write,
4 (0.6+0.3d )/(2n+1) = 4(0.62+0.3d)/[ 2(n+1)+1] ……. (1)
which on simplification gives
(0.6+0.3d )/(2n+1) = (0.62+0.3d)/(2n+3)
Cross multiplication, componenedo dividendo (i think) etc can lead us to
2/(60+30d) = 2/(2n+1).
Equating denominators, we get,
60+30d = 2n+1 or 30d = 2n+1 – 60. Now 30d is a +ve quantity .
The lowest value of n for which this is true is n=30 , which gives us
30d = 1 or d = 1/30 = 0.0333 m or 3.33 cm
The required answer is thus c
As a further check, if we plug n = 30 in eqn (1), we get
L.H.S. = R.H.S = 0.03999