# Air Temperature

1. Nov 5, 2009

### Physics8944

Hi!!
Q: A column of air length 36cm is closed at both ends, resonates to as sound of third lowest frequceny of 1445Hz. Calculate the air temperature given that the reasonable temperature falls between 10 and 20 C??

Here is what I tried:
Given T= 20 ( 30-10= 20)
Required: vs & T
Analysis: vs= 332m/s+T(0.60m/s C) & T= vs-332/0.60
Solution: vs= 332m/s+(20C)(0.60 C)
= 332m/s+12m/s
= 344m/s
now we the speed, & we can find the temperature:
=344m/s-332m/s/0.60m/s C
= 12/0.60
= 20
So the air temperature is 20 C.
I would really appericate if sum1 could help.
Thanks alot.

2. Nov 5, 2009

### clamtrox

Hi! You have to solve the speed of sound from a measurement first. You can't just just decide that the temperature is 20 C! The air column should give you the speed of sound, which you can then use to calculate the temperature.

3. Nov 5, 2009

### Physics8944

Given: f= 1445 Hz.
Lambda= 0.36m (36 cm is converted)
Required: v and vs
Analysis: v=f(lambda)
Solution: v= (1445Hz)(0.36)
V= 520.2m/s
Now we can find the temperature:
Analysis: T= vs-332m/s
0.60
= 520.2-332/0.6
= -31.13 is the temperature is it correct??

4. Nov 5, 2009

### clamtrox

Does the speed of sound increase or decrease when temperature increases?

5. Nov 5, 2009

### Physics8944

speed increases when the temperature increases

6. Nov 5, 2009

### Physics8944

Speed of sound increases as temperature increases.

7. Nov 5, 2009

8. Nov 5, 2009

### Physics8944

Im still not sure.Confused??

9. Nov 5, 2009

### clamtrox

520 m/s is much larger than the speed of sound in room temperature, therefore you'd expect a much higher temperature as well. -31 Celcius is well below room temperature, so it is clearly wrong. It is not even in your 'reasonable' range of 10-20 C.

First you should calculate the speed of sound again. What is the distance that the sound wave travels in one oscillation? It's not 36 cm.

10. Nov 5, 2009

### Physics8944

Given: f= 1445 Hz.
Length= 0.36m (36 cm is converted)
Required: v and vs
Analysis: v=f(lambda)
Solution:
Lambda=2l/n
=2(.36m)3
=0.24m
v= (1445Hz)(0.24)
V= 346.8m/s
Now we can find the temperature:
Analysis: T= vs-332m/s
0.60
= 346.8-332/0.6
= 24.7
is it correct now?

11. Nov 5, 2009