Doubt regarding SHM in a U-Tube using energy conservation

Question

While finding the SHM equation for liquid in a U tube using energy conservation, we have:$$E=K+U tag{1}$$
$$K=frac12Mv^2$$
where $v=dot{y}$

For claculation of  $U$, it is given in "Vibrations and waves" by AP French, that:

The increase in gravitational potential energy in the situation corresponds to taking a coloumn of liquid of length $y$ from left-hand tube, raising it through the distance $y$, and placing it on the top of the right-hand column. Thus we can put:$$U=grho Ay^2tag{2}$$

I think that in eq(1), we should have used the value of potential energy $U$, not $Delta{U}$, the increase in potential energy, as given in the book. So the Eq(2) according to me should be :$$ U= U( ofspace Waterspace abovespace equilibrium) + U(ofspace waterspace belowspace equilibrium)$$ which can be calculated by arbitrarily defining point of $U=0$ as the equilibrium point:
$$U=int_0^{y}dmgy + int_0^{-y}dmgy + U(ofspace arc)$$
Why is the method given in the book justified? There are no increase/decrease terms in eq 1. Conservation of Mechanical energy deals with energies at a particular instant. Sum of $U$ and  $K$at an instant is constant. Increases/decreases in $U$ are not referenced. But the author talks about increase in $U$ and substitutes the increase in eq(1). I don't think it is right.

gandalf61 · Accepted Answer

The potential energy $U$ is a function of the displacement $y$ from the liquid's equilibrium position. Although the final answer will be symmetric in $y$, for the sake of definiteness let's take $y$ to be positive when the liquid in the right hand side of the tube is raised by a distance $y$. If the absolute potential energy at the equilibrium position is $U_E$ (relative to some zero point, which we shall see is arbitrary) then we have
$$U(y) = U_E + g rho A y^2$$
We also know that the kinetic energy $K$ is a function of the liquid's velocity $v$:
$$K(v) = frac 1 2 M v^2$$
Finally, we know that the total energy $E=U(y)+K(v)$ is constant. So
$displaystyle frac {dE}{dt} = 0
 displaystyle  Rightarrow frac{dU}{dt} + frac{dK}{dt}=0
 displaystyle  Rightarrow frac{dU}{dy} frac{dy}{dt} + frac{dK}{dv} frac{dv}{dt}=0$
Since we are not interested in $U$ itself, but only in its derivative $frac {dU}{dy}$, we can see that the value of $U_E$ is irrelevant since this constant term will disappear when we differentiate $U$. The book is implicitly setting $U_E$ to $0$ without explaining why this is acceptable.

Chet Miller · Answer

If the bottom section were horizontal and the datum for potential energy were taken as the bottom, then the total potential energy would be $$U=rho gA frac{(H+y)^2}{2}+rho gA frac{(H-y)^2}{2}=rho gA(2H^2+y^2)$$where H is the equilibrium height; and the total kinetic energy would be $$K=frac{M}{2}left(frac{dy}{dt}right)^2$$So the total mechanical energy would be:
$$E=rho gA(2H^2+y^2)+frac{M}{2}left(frac{dy}{dt}right)^2$$The rate of change of kinetic energy would then be $$dfrac{dE}{dt}=2rho gAyleft(frac{dy}{dt}right)+Mleft(frac{dy}{dt}right)frac{d^2y}{dt^2}=0 $$The equation would then reduce to $$frac{d^2y}{dt^2}+frac{2rho gA}{M}y=0$$
This result would not change if the bottom were curved.

Doubt regarding SHM in a U-Tube using energy conservation

2 Answers

Add your own answers!

Ask a Question