- #1
Const@ntine
- 285
- 18
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?
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!