Calculating Air Temperature from Frequency of Sound

[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.

 

[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.

 

[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??

 

[clamtrox]

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

 

[Physics8944]

speed increases when the temperature increases

 

[Physics8944]

Speed of sound increases as temperature increases.

 

[clamtrox]

That's right. Then you can answer yourself if your answer is right or not :)

 

[Physics8944]

Im still not sure.Confused??

 

[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.

 

[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?

 

[Physics8944]

Help please!Is the above Answer correct?