Δt between consecutive frequency configurations

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1. Jul 18, 2017

Techno_Knight

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

f = n*v/4L

3. 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!

2. Jul 18, 2017

haruspex

Why do you assume the first resonance is at the fundamental frequency?

3. Jul 18, 2017

Techno_Knight

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.

4. Jul 18, 2017

haruspex

But you do not need to find that.
What L values do you get for n=1, 2, 3?

5. Jul 18, 2017

Techno_Knight

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

6. Jul 18, 2017

haruspex

So how big is each step?

7. Jul 18, 2017

Techno_Knight

From 1 to 3 it's 0.334 m & from 3 to 5 it's 0.335 m.

8. Jul 18, 2017

haruspex

Within rounding error, they're the same, right? Can you generalise to the step from n to n+1?
Can you relate those distance steps to the time intervals?

9. Jul 19, 2017

Techno_Knight

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.

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

I'm a bit stuck on that.

10. Jul 19, 2017

haruspex

The frequency is fixed.

11. Jul 19, 2017

Techno_Knight

Yeah, it's stable, 512 Hz, right? I factored that in (fn = fn+1) and ended up with Ln+1 = ((n+1)/n)*Ln

12. Jul 19, 2017

haruspex

True, but not as useful as finding Ln+1-Ln.

13. Jul 19, 2017

Techno_Knight

Well, in that case: Ln+1 - Ln = Ln/n

14. Jul 19, 2017

haruspex

Still not that useful. Try to get an expression that has no L term on the right. It will have v.

15. Jul 19, 2017

Techno_Knight

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

16. Jul 19, 2017

haruspex

Right. How long will it take for the air column to change in length by that mch?

17. Jul 19, 2017

Techno_Knight

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).

18. Jul 19, 2017

haruspex

Sorry, I did not check this equation:
That is not right for a half open tube.

19. Jul 19, 2017

Techno_Knight

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...

20. Jul 19, 2017

haruspex

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.