# Velocity of sound in air in a closed air column

1. Aug 25, 2010

### enirak

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. Aug 25, 2010

### presbyope

You seem to understand the concept of the closed column, but you are mixing up your data:
• The first resonance position corresponds to 1/4 wavelength. Take the average of the three values.
• Calculate the wavelength again using the data for the second position
• Calculate the velocity using the frequency and wavelength
• Compare with the value predicted by the formula

Last edited: Aug 25, 2010
3. Aug 25, 2010

### enirak

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

4. Aug 25, 2010

### presbyope

An example for which part? Did you try calculating the wavelength? What did you get?

5. Aug 25, 2010

### presbyope

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

6. Aug 26, 2010

### enirak

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. Aug 26, 2010

### presbyope

This is what I mean. The instructions said to average the values though. If I measured 9 cm, 13 cm and 11 cm the average would be 11 cm.

Think about this. You are saying that the column vibrates in three modes at the first position. Did it sound the same when you vibrated the second position, or did you hear a higher pitch sound at the second position?

8. Aug 28, 2010

### enirak

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%

Last edited: Aug 28, 2010
9. Aug 29, 2010

### presbyope

What quantity are you averaging? You can reduce measurement error by averaging the actual measurements you make. Is that what you are doing here?

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.phy-astr.gsu.edu/hbase/Waves/clocol.html

The first resonance position was louder? Good. What do you conclude from that? Did you hear a low tone or a high tone in the second position (when the air column was longer)? What mode(s) did you expect to find? Did it match with what you heard?

Another hint: the speed of sound should be roughly the same in both parts since you measured them on the same day.

The percent error formula is correct.

10. Aug 29, 2010

### presbyope

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. Aug 30, 2010

### enirak

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. Aug 30, 2010

### nasu

No. The frequency is determined by the tuning fork and should be the same.
In order to excite the higher order without changing the frequency you change the length of the tube.

13. Aug 30, 2010

### enirak

So my calculations are good then?

14. Aug 31, 2010

### presbyope

Your calculations for the first part are good.

15. Sep 2, 2010

### enirak

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

16. Sep 2, 2010

### nasu

The wavelength is not 1.5 * Length of tube.
How many quarter waves fit in the tube in this case?

17. Sep 6, 2010

### enirak

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. Sep 6, 2010

### nasu

I am asking again, how many quarter waves (or quarters of wavelengths)?
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.