One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Let us consider a U-tube containing mercury of density ρ. Let A be the area of cross-section of the tube and L be the total length of the mercury column. Initially, a small pressure difference is maintained between the two arms by the suction pump, creating a height difference between the mercury levels in the two columns.
When the suction pump is removed, let us assume that the mercury column in one arm is displaced by a small distance y above its equilibrium position. Consequently, the mercury in the other arm is depressed by the same distance y below its equilibrium position. This creates a height difference of 2y between the two columns.
The difference in height of 2y creates a pressure difference between the two arms. This pressure difference is given by ΔP = ρg(2y) = 2ρgy. This excess pressure acts on the entire mercury column and tends to restore it to the equilibrium position.
The restoring force acting on the mercury column is: F = Pressure × Area = -2ρgy × A = -2ρgAy. The negative sign indicates that the force is directed opposite to the displacement y, i.e., towards the equilibrium position.
The mass of the entire mercury column is m = ρAL. Applying Newton's second law: ma = -2ρgAy. Therefore, the acceleration is: a = -(2ρgA/m)y = -(2ρgA/ρAL)y = -(2g/L)y. This can be written as a = -ω²y, where ω² = 2g/L.
Since the acceleration is proportional to displacement and is always directed towards the mean position, the mercury column executes simple harmonic motion.
Explanation
This question tests the student's understanding of the condition for simple harmonic motion. The key insight from the textbook is that 'The force acting in a simple harmonic motion is proportional to the displacement and is always directed towards the centre of motion.' The derivation shows that when the pump is removed, the hydrostatic pressure difference creates a restoring force proportional to displacement, satisfying the condition F = -kx (Hooke's law), which is the defining characteristic of SHM.
Solution Steps
Step 1: Define variables - mercury density ρ, cross-sectional area A, total length L of mercury column, displacement y from equilibrium
Step 2: Establish that height difference between columns = 2y, creating pressure difference ΔP = 2ρgy
Step 3: Calculate restoring force F = -2ρgAy (negative sign indicates restoring nature)
Step 4: Apply Newton's second law: ma = -2ρgAy, where m = ρAL
Step 5: Simplify to get a = -(2g/L)y, which is of the form a = -ω²y, proving SHM