A U-tube open at both ends to atmospheric pressure is filled with an incompressible fluid of density . The cross-sectional area of the tube is uniform and the total length of the column of fluid is . A piston is used to depress the height of the liquid column on one side by a distance , 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). Express your answer in terms of and acceleration due to gravity (enter L for and g for ).

To find the frequency of the ensuing simple harmonic motion, we need to use the concept of conservation of energy.

Initially, the liquid column has a height of L on both sides of the U-tube. The potential energy of the system is given by the product of the density (ρ), acceleration due to gravity (g), and the height of the liquid column (L). So, the initial potential energy is given by:

Initial Potential Energy = ρ * g * L * A

After depressing the height of the liquid column on one side by a distance d, the potential energy of the system becomes:

Final Potential Energy = ρ * g * (L - d) * A

Since the fluid is incompressible and there is no drag at the walls of the U-tube, the change in potential energy is converted into kinetic energy as the liquid column oscillates back and forth. At the extremes of the motion, all the potential energy is converted into kinetic energy and vice versa, resulting in simple harmonic motion.

The total energy of the system at any point during the oscillation can be expressed as the sum of potential energy and kinetic energy:

Total Energy = Initial Potential Energy + Kinetic Energy

Since energy is conserved, the total energy at any point during the oscillation remains constant. Therefore, we can equate the initial and final total energies:

Initial Potential Energy = Final Potential Energy + Final Kinetic Energy

Expanding this equation leads to:

ρ * g * L * A = ρ * g * (L - d) * A + Final Kinetic Energy

As the liquid column oscillates back and forth, the kinetic energy can be expressed as:

Final Kinetic Energy = (1/2) * m * v^2

where m is the mass of the liquid column and v is the velocity of the liquid column.

Considering that the mass of the liquid column is given by the product of the density (ρ), cross-sectional area (A), and the length of the liquid column (L-d), we can rewrite the equation as:

Final Kinetic Energy = (1/2) * ρ * A * (L - d) * v^2

Substituting this expression for the final kinetic energy back into the equation, we have:

ρ * g * L * A = ρ * g * (L - d) * A + (1/2) * ρ * A * (L - d) * v^2

Now, let's simplify the equation by canceling out ρ, A, and (L - d):

g * L = g * (L - d) + (1/2) * (L - d) * v^2

Dividing both sides of the equation by g * (L - d), we get:

L / (L - d) = 1 + (1/2) * (L - d) * (v^2 / g)

Using the small-angle approximation for small oscillations, we can assume v^2 / g ≈ 0. Hence, the equation becomes:

L / (L - d) ≈ 1

Simplifying further, we get:

L ≈ L - d

This implies that d is negligible compared to L. Hence, the length of the liquid column stays approximately constant during the oscillations.

Finally, the frequency of simple harmonic motion is given by the equation:

Frequency (f) = 1 / Time Period (T)

Since the length of the liquid column remains constant and the motion is simple harmonic, the time period can be expressed as:

T = 2π * √(L / g)

Therefore, the frequency of the ensuing simple harmonic motion is:

f = 1 / T = 1 / (2π * √(L / g))

Expressing it in terms of L and g as mentioned in the question, the frequency is:

f = g / (2πL)

Hence, the frequency of the ensuing simple harmonic motion is g / (2πL).