How Can the Speed of Sound Help Determine Room Temperature?

[SelenaT]

Homework Statement

Calculate the room temperature by using the speed of sound formula and using the given values.

Known Data:
Frequency = 480 Hz
2nd Resonant length = 54cm or 0.54m

Homework Equations

v = 331 + (0.60)T
T = (v - 331)/0.60

v = fλ

(Open-Closed air column)
L = (3/4)λ

The Attempt at a Solution

(1) v = fλ
= 480Hz (0.53m)
= 254.4m/s

(2) T = (254.4m/s) - 331 / 0.60
= -127.6 degrees celsius[/B]

I highly doubt the temperature is this low. I realized that I used the wavelength as the resonant length so, I decided to solve for λ using the resonant length. I did the same process again but using the calculated λ but it didnt work either.

 

[haruspex]

I don't understand how you got 0.53m. You don't seem to have used the $L=(3/4)\lambda$ formula.

 

[SelenaT]

I don't understand how you got 0.53m. You don't seem to have used the $L=(3/4)\lambda$ formula.

Sorry, forgot to give a little context. I will remember next time.

This is a follow up question for an experiment that I did. Using an open-closed air column, I found that the second resonant length was 0.53m

 

[haruspex]

Sorry, forgot to give a little context. I will remember next time.

This is a follow up question for an experiment that I did. Using an open-closed air column, I found that the second resonant length was 0.53m

OK, but that's L, not $\lambda$. You plugged that value into your $v=f\lambda$ formula.