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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Worked Example W13D1-1: U-tube Oscillations Solution
A U-tube open at both ends to atmospheric pressure P0 is filled with an incompressible
fluid of density  . The cross-sectional area A of the tube is uniform and the total length
of the column of fluid is L . A piston is used to depress the height of the liquid column on
one side by a distance x0 , and then is quickly removed. What is the frequency of the
ensuing simple harmonic motion? Assume streamline flow and no drag at the walls of the
U-tube. (Hint: use conservation of energy).
Solution: We shall use conservation of energy. First choose as a zero for gravitational
potential energy when the water levels are equal on both sides of the tube. When the fluid
is depressed by the piston on one side, it rises on the other.
At a given instant in time when a portion of the fluid of mass m   Ax is a height x
above the equilibrium height, the potential energy of the fluid is given by
U  mgx  ( Ax)gx   Agx 2 .
At that same instant the entire fluid of length L and mass m   AL is moving with
speed v , so the kinetic energy is
1
1
K  mv 2   ALvx2 .
2
2
Thus the total energy is
1
E  K  U   ALv 2   Agx 2 .
2
By assuming streamline flow and no drag at the walls of the U-tube the mechanical
energy of the fluid is constant therefore
0
dE
dv
dx
  ALv  2  Agx .
dt
dt
dt
If we just consider the top of the fluid above the equilibrium position on the right arm in
the above figure, we can rewrite the above equation as
0
dv
dE
dx
  ALvx x  2  Agx
dt
dt
dt
where vx  dx / dt , we can rewrite the energy condition using dvx / dt  d 2 x / dt 2 as
 d 2x

0  vx  A  L 2  2gx  .
 dt

This condition is satisfied when v x  0 , i.e. the equilibrium condition or when
0L
d 2x
 2 gx .
dt 2
This last condition can be written as
d 2 x 2g
0 2 
x.
dt
L
This last equation is an equation describing simple harmonic motion. Using the same
mathematical techniques as we used for the spring-block system, the solution for the
height of the fluid above the equilibrium position is given by
x(t)  Bcos( 0t)  C sin( 0t)
where
0 
2g
L
is the angular frequency of oscillation. The x -component of the velocity of the top of the
fluid on the right hand side of the u-tube is given by
vx (t) 
dx(t)
  0 Bsin( 0t)   0C cos( 0t) .
dt
The coefficients B and C are determined by the initial conditions. At t  0 , the height of
the fluid was x(t  0)  B  x0 . At t  0 , the speed is zero so vx (t  0)   0C  0 , hence
C  0 . So the height of the fluid above the equilibrium position as a function of time is
thus
 2g 
x(t )  x0 cos 
t  .
 L 