One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small

Exercises · 13.18

Q. 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.

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Concept understanding — Simple Harmonic Motion

Simple Harmonic Motion: The Natural Rhythm of Things

Imagine a ball placed at the bottom of a perfectly smooth, U-shaped bowl. If you give it a gentle push, what happens? It rolls up one side, slows down, stops for an instant, then rolls back down, past the bottom, up the other side, stops, and returns. Left alone, it keeps doing this — back and forth, back and forth — in a steady, repeating rhythm.

That rhythm is the heart of Simple Harmonic Motion (SHM). It's the most fundamental kind of oscillatory (back-and-forth) motion in physics.

The Intuition: A Restoring Force That Fights Displacement

The key idea is this: the further you push the object from its resting (equilibrium) position, the stronger the force that tries to pull it back.

In the bowl, when the ball is at the bottom (equilibrium), gravity pulls straight down, and the bowl pushes straight up — no sideways force. But when you push the ball up the side, gravity now has a component that pulls it down the slope. The higher up the side you push it, the steeper the slope, and the stronger that pull-back force becomes.

This is a restoring force — it always points toward equilibrium. And crucially, in SHM, this restoring force is directly proportional to the displacement from equilibrium. Double the displacement, double the restoring force.

F = -k x

The Precise Statement

Simple Harmonic Motion is the motion of an object where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction.

That's it. That single condition — $F = -kx$ — is the entire definition. Everything else (the sine waves, the formulas for period and frequency) follows mathematically from this one law.

What Does This Motion Look Like?

If you track the ball's position over time, you get a beautiful, smooth wave — a sine wave (or cosine wave). It's the same shape as the shadow of a spinning wheel cast on a wall.

The motion has three key descriptors:

Amplitude ( $A$ ): The maximum displacement from equilibrium. How far you initially pushed the ball up the side of the bowl.
Period ( $T$ ): The time it takes to complete one full back-and-forth cycle (e.g., from the leftmost point, back to the leftmost point).
Frequency ( $f$ ): How many cycles happen per second. $f = 1/T$ .

Note

A remarkable fact: for a given system (fixed $k$ and fixed mass $m$ ), the period and frequency do not depend on the amplitude. A big push and a tiny push take exactly the same time to complete one cycle. This is called isochronism — and it's why pendulums were used to keep time in clocks.

The Mathematical Description (Derived from $F = -kx$ )

Using Newton's second law ( $F = ma$ ) and the definition of acceleration ( $a = \frac{d^2x}{dt^2}$ ), the condition $F = -kx$ becomes:

m \frac{d^2x}{dt^2} = -k x

This is a differential equation. Its solution — the position as a function of time — is:

x(t) = A \cos(\omega t + \phi)

Where:

$\omega = \sqrt{\frac{k}{m}}$ is the angular frequency (radians per second). It tells you how fast the oscillation is.
$\phi$ is the phase constant (determines where in the cycle you start measuring time). …

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When the suction pump is removed, the excess mercury height on one side creates a restoring pressure difference proportional to the displacement, so $\dfrac{d^2x}{dt^2}=-\dfrac{2g}{L}x$ -- the standard SHM equation. The mercury column therefore executes SHM with angular frequency $\omega=\sqrt{\dfrac{2g}{L}}$ and time period $T=2\pi\sqrt{\dfrac{L}{2g}}$ , where $L$ is the total length of the mercury column.

Setting up the problem

Let the U-tube have uniform cross-sectional area $A$ and hold a total length $L$ of mercury (density $\rho$ ). At equilibrium, the mercury levels in both arms are equal. While the pump runs, it creates a small pressure difference, displacing mercury by $y$ in each arm -- one arm rises by $y$ , the other falls by $y$ , so the total height difference between the two arms is $2y$ .

Deriving the restoring force

When the pump is removed, this height difference drives the mercury back toward equilibrium:

\Delta P = \rho g(2y)

Acting over the cross-section $A$ , this produces a net restoring force:

F = -A\Delta P = -2\rho gAy

Applying Newton's second law

The total mass of mercury is $m=\rho AL$ . With $F=ma$ and $a=\dfrac{d^2y}{dt^2}$ :

\rho AL\frac{d^2y}{dt^2} = -2\rho gAy \quad\Rightarrow\quad \frac{d^2y}{dt^2} = -\frac{2g}{L}y

This is exactly the standard SHM equation $\dfrac{d^2y}{dt^2}=-\omega^2y$ , with:

\omega = \sqrt{\frac{2g}{L}}

\frac{d^2y}{dt^2} + \frac{2g}{L}y = 0

Interpreting the result

Since the acceleration is directly proportional to displacement and always directed opposite to it, the motion is confirmed to be SHM. The time period is:

$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{2g}}$ …

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Q.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.

Setting up the problem

Deriving the restoring force

Applying Newton's second law

Interpreting the result

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