Δt between consecutive frequency configurations

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Homework Statement



At a long, vertical tube, water is poured in, with an inflow R = 1.00 L/min = 1.67 * 10-5 m3/s. The radius, r = 5.00 cm. At the open end, a tuning fork is oscillating with a frequency of f = 512 Hz. What's the time difference between two consecutive configurations, while the water slowly rises?

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Homework Equations



f = n*v/4L

The Attempt at a Solution



Alright, in this case, we have a constant f that doesn't change, an L that keeps getting shorter, and we're in a "tube with one end open, one end closed" environment.

>Volume of the water: V = π*r2*h = 7.85 * 10-3m2*h
>hstart = 0 & Lstart = max
>h keeps going up, while L keeps going "down"
>every 1s, a volume of V = 1.67 * 10-5 m3 is added. So, (using the V formula from the first >), we have h = 2.13 * 10-3 m. So the height of the water goes up by said h every 1s.

And that's where I get stuck. Initially I figured I'd say that the given f is the harmonic, put n=1, v = 343 m/s, and find the initial/max length L. But as I kept upping the ante on n (3,5,7), the L kept growing, which wasn't realistic, since it should go down. So that's one out.

Any help is appreciated!
 
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haruspex said:
Why do you assume the first resonance is at the fundamental frequency?
Well, at first I did so because I thought that I'd use it to find the entire length of the tube. Obviously that was wrong. I'm out of ideas though.
 
haruspex said:
But you do not need to find that.
What L values do you get for n=1, 2, 3?

It's an open/closed tube, so f = n*v/4L & n = 1,3,5,7..., right?

>n = 1: L = 0.168 m

>n = 3: L = 0.502 m

>n = 5: L = 0.837 m
 
haruspex said:
So how big is each step?
From 1 to 3 it's 0.334 m & from 3 to 5 it's 0.335 m.
 
haruspex said:
Within rounding error, they're the same, right?

My problem is that if I go higher (5, 7, 9), then the steps change a bit. From 5 to 7 it's 0.333. From 7 to 9 it's 0.331. I guess it's to be attributed to the rounding error, but there is a difference.

haruspex said:
Can you generalise to the step from n to n+1?

Something like this?

>fn = n*v/4Ln
>fn+1 = (n+1)*v/4Ln+1
>fn = fn+1
___________________________
fn+1/fn = ((n+1)*v/4Ln+1)/(n*v/4Ln) <=> n*Ln+1 = (n+1)*Ln <=> Ln+1 = ((n+1)/n)*Ln
haruspex said:
Can you relate those distance steps to the time intervals?

I'm a bit stuck on that.
 
haruspex said:
The frequency is fixed.
Yeah, it's stable, 512 Hz, right? I factored that in (fn = fn+1) and ended up with Ln+1 = ((n+1)/n)*Ln
 
haruspex said:
True, but not as useful as finding Ln+1-Ln.
Well, in that case: Ln+1 - Ln = Ln/n
 
haruspex said:
Still not that useful. Try to get an expression that has no L term on the right. It will have v.
Yeah, I forgot about that: f = n*v/4*Ln <=> Ln = n*v/4*f || So we have: Ln+1 - Ln = (n*v/4*f)*(1/n) = v/4*f = 343 m/s / 4*512 Hz = 0.167 m
 
Darthkostis said:
Yeah, I forgot about that: f = n*v/4*Ln <=> Ln = n*v/4*f || So we have: Ln+1 - Ln = (n*v/4*f)*(1/n) = v/4*f = 343 m/s / 4*512 Hz = 0.167 m
Right. How long will it take for the air column to change in length by that mch?
 
haruspex said:
Right. How long will it take for the air column to change in length by that mch?
Based on what I found previously (first post):

In 1 s, the water rises by 2.13 * 10-3 m, so the Length decreases by so.
How much time, x, do we need for the Length to change by 0.167 m ?

By doing the Rule of Three I end up with 78.4 s, which is about half of the book's answer (158 s).
 
Darthkostis said:
Based on what I found previously (first post):

In 1 s, the water rises by 2.13 * 10-3 m, so the Length decreases by so.
How much time, x, do we need for the Length to change by 0.167 m ?

By doing the Rule of Three I end up with 78.4 s, which is about half of the book's answer (158 s).
Sorry, I did not check this equation:
Darthkostis said:
f = n*v/4L
That is not right for a half open tube.
 
haruspex said:
Sorry, I did not check this equation:

That is not right for a half open tube.
Isn't it a tube with one end closed, and one end open? The top is the open part, and the bottom is the closed one. That's what my book has, with n being able to take the values of 1,3,5,7...
 
Darthkostis said:
Isn't it a tube with one end closed, and one end open? The top is the open part, and the bottom is the closed one. That's what my book has, with n being able to take the values of 1,3,5,7...
Oh ok, if that is how n is being defined... But you took consecutive values as n, n+1 instead of n and n+2.
 
haruspex said:
Oh ok, if that is how n is being defined... But you took consecutive values as n, n+1 instead of n and n+2.
Oh yes, darn it. Yeah, rookie mistake. Taking that into account:

>fn = n*v/4Ln
>fn+2 = (n+2)*v/4Ln+2
>fn = fn+2
___________________________
fn+2/fn = ((n+2)*v/4Ln+2)/(n*v/4Ln) <=> n*Ln+2 = (n+2)*Ln <=> Ln+2 = ((n+2)/n)*Ln <=> Ln+2 - Ln = 2*Ln/n
Ln = n*v/4*f
_____________________________

Ln+2 - Ln = 2/n * nv/4f = 2v/4f = v/2f = 343 m/s / 2*512 Hz = 0.335 m

Following the Rule of Three, same as before:

In 1 s, the water rises by 2.13 * 10-3 m, so the Length decreases by so.
How much time, x, do we need for the Length to change by 0.335 m ?

And so we get x = 157.3 s || A bit off from the book's answer, but that's due to all the multiplications/divisions, Significant Digits and whatnot.

Thanks a ton for the help and the patience!
 

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