Calculating Speed of Sound: 480 Hz & 33.34 cm

[Alec]

Hello, I'm new here and I'm not sure wether this is the correct forum or not, ohwell here it goes.
I'm wondering how to calculate the speed of sound by using the formula:
Lamba = v * T
The frequency was 480 Hz and the distance was 33.34 cm.

This was done in a tube with sand so you could distinguish the soundwaves in the sand thus measuring the distance of one wave length.
Any help would be appreciated!

 

[LeonhardEuler]

T here means the period, which is one over the frequency:
T=\frac{1}{f}
f=\frac{1}{T}
So you can calculate the period from the frequency. You can solve for "v" in the equation you gave and use the values of T and \lambda to get a number. Just remember that your \lambda is in centimeters so you should convert if you want a speed in m/s.

 

[Alec]

So basicly the speed is 0,3334 * 480 = 160.03 m/s ??
Sorry, but I'm a bit slow and this isn't my primary language either.

Another question, what if I wish to calculate the theoretical resonance wave length for the tube, (75 centimeters of lenght, closed in one end, open in the other end).

 

[LeonhardEuler]

So basicly the speed is 0,3334 * 480 = 160.03 m/s ??

Right.

Another question, what if I wish to calculate the theoretical resonance wave length for the tube, (75 centimeters of lenght, closed in one end, open in the other end).

Since it is closed on one end, there will be a node at that end (a point of zero amplitude). Sice it is open at the other, there will be a maximum there. So there will be resonence when the lenth of the tube is 1/2\lambda, 3/2\lambda, 5/2\lambda...

 

[Alec]

Thank you for your answers so far, but how do I calculate the resonance? (if that's possible)

 

[LeonhardEuler]

There will be resonence when the length of the tube is (1/2)\lambda, (3/2)\lambda... and so on. So the longest wavelength at which there will be resonence is when
\frac{1}{2}\lambda=L
(L is the length of the tube). So you just have to solve for \lambda.

 

[Alec]

Alright
So I calculated it like this: (meters)
L1 = 1/2 * 1.5
L2 = 3/2 * 0.5
L3 = 5/2 * 0.3
And it checks out.
Any comments?
I'm also supposed to compare this wave length to the ones I noticed during my experiment, the biggest one I got was 0.3334 meters and the smallest one was 0.07 meters, what does this say?
So here it would be the wave length of 1.5 meters as the biggest and 0.3 meters as the smallest. Why is it so much bigger than the wave length that I got?

 

[BananaMan]

So basicly the speed is 0,3334 * 480 = 160.03 m/s ??
Sorry, but I'm a bit slow and this isn't my primary language either.

remember that f=1/T when doing this calculation

 

[LeonhardEuler]

remember that f=1/T when doing this calculation

He did. He moved it over to the other side.

 

[BananaMan]

ooops sorry, its late and I am tired, should probably get myself off to bed, just one more crack at my optics and i will i think

 

[Alec]

It's okay, those calculations are fine now, although I'd appreciate some feedback on my previous post!