Water Column Resonance Time Interval Calculation

[utkarshakash]

Homework Statement

There is a cylindrical vessel of cross-sectional area 20cm^2 and length 1m is initially filled with water to a certain height as shown. There is a very small pinhole of cross-sectional area 0.01cm^2 at the bottom of the cylinder which is initially plugged with a pin cork. The air column in the cylinder is resonating in fundamental mode with a tuning fork of frequency 340Hz. Suddenly the pin cork is pulled out and water starts flowing out of the cylinder. Find the time interval after which resonance occurs again.

The Attempt at a Solution

The setup is equivalent to a closed organ pipe. The fundamental frequency is given by nv/4L.
Let the initial height of water column be h1. At second resonance let the height change to h2.

h1-h2=v/4*340

The velocity of water coming out of hole is given by \sqrt{2gx}

Thus, rate of decrease of water column = 0.01v/20.

\dfrac{dx}{dt} = \dfrac{0.01\sqrt{2gx}}{20}

If I integrate this from h1 to h2, I am left with the expression \sqrt{h_1}-\sqrt{h_2}.

I already know the difference between the heights but not the difference between their square roots. How do I calculate it?

 

[TSny]

h1-h2=v/4*340

I don't think this is correct. If you haven't already done so, draw a diagram showing the nodes and antinodes of the standing sound wave for each of the two resonances.

\dfrac{dx}{dt} = \dfrac{0.01\sqrt{2gx}}{20}

This looks good except you should think about whether or not a negative sign needs to be included (x is decreasing as time increases).

If I integrate this from h1 to h2, I am left with the expression \sqrt{h_1}-\sqrt{h_2}.

I already know the difference between the heights but not the difference between their square roots. How do I calculate it?

From your diagrams of nodes and antinodes, you should be able to determine the heights h1 and h2.

By the way, what are you supposed to use for the speed of sound?

 

[Vibhor]

From your diagrams of nodes and antinodes, you should be able to determine the heights h1 and h2.

By the way, what are you supposed to use for the speed of sound?

Here is my attempt at the problem .

I am using g = 9.8ms^-2 and speed of sound = 330 m/s

The length of air column l = nλ/4 = nv/4f , n=1,3,5...

l₁ = 0.24m
l₂ = 0.72m

Consequently h₁ = 0.76m and h₂ = 0.28m

$\dfrac{dx}{dt} = - \dfrac{0.01\sqrt{2gx}}{20}$

Integrating , t = 902.93 (√h₁ - √h₂)

Putting values of h₁ and h₂ , I am getting t ≈ 307 sec .

TSny ,are you getting the same answer ?

Many Thanks

 

[TSny]

That looks good. I get slightly different results for some of the numbers in your calculation. My final answer for t = 315 s ≈ 320 s to 2 significant figures.

 

[Vibhor]

Thanks a lot