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A drinking straw is dipped in a pan of water to depth $d$ from the surface (see figure below). Now water is sucked into it up to an initial height $h_0$ and then left to oscillate. As a result, its height $y$ from the surface of the water varies periodically. Ignoring damping, the second order differential equation for $y$ is ($g$ is the acceleration due to gravity):

enter image description here

Why will the liquid oscillate? From what we have learnt about capillary action, the liquid just climbs up to a specific height and stays there as long as that height is not larger than the tube-length. So doesn’t it make sense for the water to just drop/climb to the required height? Why oscillate? I’m sure this has something to do with pressure and height, but I cannot pinpoint it.

Also, why should the differential equation depend on $d$? (The answer says it does) The length of tube below the surface of water shouldn’t matter right? Because the pressure on the liquid (at the level of the surface outside the tube), both inside and outside the tube, are already same, so there should be no difference in the bulk of the liquid (according to me).

Also, a derivation of the Differential Equation would be nice.

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    $\begingroup$ I think the capillar effect for a drinking straw is small compared to the hight you suck it. If the water in the tube oscillates the water in the container also does. Maybe you first think of a U tube, water on one side of the tube higher than on the other in the beginning . $\endgroup$
    – trula
    Jul 18, 2022 at 13:40

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The capillary action is not the concept at play here, the height $h_0$ of water inside the straw is initially caused by some external agent. Our consideration begins after this height is attained.


The momentum $\vec{p} $ of the mass of water inside the straw (I have assumed the straw to have an area of cross section $S$ and $\hat{j}$ to be a unit vector pointing upwards along the length of the straw) is given as

$$\vec{p}=(\rho S(d+y)\dot{y} )\hat{j}$$ $$\implies\frac{d\vec{p}}{dt}=\rho S[(d+y)\ddot{y}+(\dot{y})\dot{y}]\hat{j}$$

Calculating the net force on the mass inside the straw (caused by the weight of the extra height of water $y$), we have $$\vec{F}=-(\rho Sgy)\hat{j}$$

Now since $\vec{F}=\frac{d\vec{p}}{dt}$, we have

$$-(\rho Sgy )\hat{j}=\rho S[(d+y)\ddot{y}+\dot{y}^2]\hat{j}\implies \ddot{y}(d+y)+\dot{y}^2+gy=0$$

which is the required differential equation.

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  • $\begingroup$ Why not $\vec F=-\rho Sg(y+d)\hat j$? $\endgroup$ Jul 18, 2022 at 15:01
  • $\begingroup$ Consider the force at the bottom of the straw. The weight of the water above it (inside the straw) is $\vec{F_1}=-\rho Sg(y+d) \hat{j}$ as you mentioned. However, the water outside the straw has a pressure excess of $\rho g d$ at the bottom of the straw and thus exerts a force $\vec{F_2}=(\rho S g d) \hat{j}$ on the water inside the straw leaving us with a total force $\vec{F}$ given as $$\vec{F}=\vec{F_1}+\vec{F_2}=-(\rho Sgy) \hat{j}$$ $\endgroup$
    – Cathedral
    Jul 18, 2022 at 15:32

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