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Velocity of sound in air in a closed air column 
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#1
Aug2510, 04:05 PM

P: 8

Am I missing something and doing this all wrong????
Experiment and data: The level of water in the glass tube was adjusted by raising the supply tank until the sound from the tuning fork was at its loudest. This level corresponds to first resonance position and it was recorded. The mesurement was repeated 2 additional time Room temperature = 20 degree C Frequency= 500Hz 1. 16.5 cm 2. 16.3 cm 3. 16.4 cm The second resonance position was found by lowering the water level in the column until the sound from the tuning fork was at its loudest. 2 additional mesurements weer made at this level Room temperature Frequency= 500Hz 1. 47.5 cm 2. 47.6 cm 3. 47.4 cm Questions: For each length of reasoning air column, calculate the average of the 3 readings Calculate the wavelength of the sound wave in air from the lengths of the resonating air column Calculate the velocity of sound in air from the wavelenth found. Find the average of the 2 values Calculate the velocity of sound in air at room temperature from the known value of velocity of sound at 0 degree C Compare the values of the velocity of sound found in the calculations by finding yhe Percent error Here's what I did so far: first resonance position (would do the same calculations for second resonance position) wavelength = 16.5 cm 16.3 cm = 0.2 cm = 0.002 m L1 = (1/4)(0.002) = 0.0005 m L2 = (5/4)(0.002) = 0.0025 m L3 = (3/4)(0.002) = 0.0015 m f1 = (1)(500) = 500 Hz f2 = (5)(500) = 2500 Hz f3 = (3)(500) = 1500 Hz velocity(v) from wavelength(w) ???? (need an example) velocity from the known value v= 331.4 + 0.6(20) = 343.4 m/s then what? 


#2
Aug2510, 08:32 PM

PF Gold
P: 44

You seem to understand the concept of the closed column, but you are mixing up your data:



#3
Aug2510, 10:45 PM

P: 8

It would really help to see an example? I'm still not getting it



#4
Aug2510, 11:45 PM

PF Gold
P: 44

Velocity of sound in air in a closed air column
An example for which part? Did you try calculating the wavelength? What did you get?



#5
Aug2510, 11:49 PM

PF Gold
P: 44

For example  If I had a closed column that resonated first at 10 cm the wavelength would be 40 cm.



#6
Aug2610, 02:50 PM

P: 8

Is this what you mean:
First resonance position 1. 16.5 x 4 = 66 cm 2. 16.3 x (5/4) = 20.375 cm 3. 16.4 x (3/4) = 12.3 cm 


#7
Aug2610, 03:14 PM

PF Gold
P: 44




#8
Aug2810, 04:45 PM

P: 8

Does this sound right?
First resonance position: I get the average at 32.9 cm wavelength = 66cm  12.3 cm= 53.7 cm = 0.537 m velocity with wavelength = 0.537 m (500Hz) = 268.5 m/s velocity with temperature = 331.4 + 0.6(20C) = 343.4 m/s Second resonance position: 47.6 cm x 4 = 190.4 cm 47.5 cm x 3/4 = 35.625 cm 47.4 cm x 5/4 = 59.25 cm average = 95.09 cm wavelength = 190.4 cm  35.625 cm = 154.775 cm = 1.54775 m velocity with wavelength = 1.54775m (500Hz) = 773.875 m/s velocity with temperature = 331.4 + 0.6(20C) = 343.4 m/s When the level of water is higher, the air column is shorter and the frequency is higher therefore, the wavelength is shorter. The first resonance position was louder than the second resonance position. The only thing now is the formula for the percent error?....as you can see the one i'm using gives me ridiculous answers...can't be right! first resonance position: [(343.4 m/s  268.5 m/s) / 343.4 m/s] x 100 = 22% second resonance positiion: [(343.4 m/s  775 m/s) / 343.4 m/s] x 100 = 126% 


#9
Aug2910, 10:21 PM

PF Gold
P: 44

Also, what mode(s) do you expect to find at the first resonance position? Hint: the tuning fork produced a single frequency. Maybe the hyperphysics demo will make things more clear: http://hyperphysics.phyastr.gsu.edu...es/clocol.html Another hint: the speed of sound should be roughly the same in both parts since you measured them on the same day. 


#10
Aug2910, 11:35 PM

PF Gold
P: 44

To be extremely clear. If I measured 16.3 cm, 16.4 cm and 16.5 cm for some object I would say the average measure was 16.4 cm.



#11
Aug3010, 02:31 PM

P: 8

Ok so I think I have the first resonance position figured out:
average is 16.4 cm wavelength = (4 x 0.164 m) = 0.656 m velocity = wavelength x frequency = 0.656 x 500 = 328 m/s velocity with temperature = 343.4 m/s percent error = [(343.4  328) / 343.4] x 100 = 4.5% The second resonance position should be in the third mode. Not sure I'm doing the right calculations but this is what I came up with: average = 47.5 cm wavelength = 1.5 x 0.475 m = 0.7125 m velocity = 0.7125 x 500 = 356.25 m/s velocity with temperature = 343.4 m/s percent error = [(343.4  356.25) / 343.4] x 100 = 3.7% Shouldn't the frequency in the third mode be 3 x 500 = 1500 Hz ? But if I plugged that in the equation the result is way to high for the velocity. 


#12
Aug3010, 03:25 PM

P: 1,969

In order to excite the higher order without changing the frequency you change the length of the tube. 


#13
Aug3010, 06:37 PM

P: 8

So my calculations are good then?



#14
Aug3110, 01:33 PM

PF Gold
P: 44




#15
Sep210, 12:34 PM

P: 8

If the first resonance position calculations are good what am I doing wrong for the second position?



#16
Sep210, 08:55 PM

P: 1,969

How many quarter waves fit in the tube in this case? 


#17
Sep610, 08:33 PM

P: 8

I have no clue for the second position. Nothing seems to work or make sense to me
wavelength= 3 x 0.475 m = 1.425 m velocity = 1.425 m x 500 Hz = 712.5 m/s that puts the %error off the chart. 


#18
Sep610, 09:25 PM

P: 1,969

If you say 3, as may be assumed from your "3 x 0.475", then write the condition that the length of the tube (0.475 m) contains 3 quarter waves or 3/4 wavelength. Then solve for the wavelength. Remember, the wavelength in the second case is the same as in the first case, within the experimental error. 


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