A body of mass $m$ is situated in a potential field $U(x) = U_0(1 - \cos\alpha x)$ when $U_0$ and $\alpha$ are constants. Find the time period of small oscillations.

A body of mass m is situated in a potential field – Class 11 Physics

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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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For small displacements from equilibrium, the cosine potential behaves like a harmonic oscillator with effective spring constant $k = U_0 \alpha^2$ . The period is $T = 2\pi\sqrt{\frac{m}{U_0\alpha^2}}$ .

Why this works: the harmonic approximation

When a particle sits in any smooth potential well, small oscillations about the equilibrium point are always approximately simple harmonic. The reason is mathematical: near a minimum, any smooth function looks like a parabola to leading order (Taylor's theorem). The curvature of the potential at equilibrium determines the restoring force, which in turn sets the oscillation frequency.

The potential $U(x) = U_0(1 - \cos\alpha x)$ has the shape of a cosine wave flipped upside-down and shifted. The particle naturally settles at the bottom of the well, where $\cos\alpha x$ is maximum.

Finding the period

Locate equilibrium

Equilibrium occurs where the force vanishes, i.e., where $\frac{dU}{dx} = 0$ .

\frac{dU}{dx} = U_0 \alpha \sin\alpha x = 0

This gives $\alpha x = 0, \pm\pi, \pm 2\pi, \ldots$ , so $x = 0$ is one equilibrium point (and the others are equivalent by periodicity). At $x = 0$ , we have $U(0) = 0$ , the minimum of the potential.

Expand the potential near equilibrium

For small displacements $x$ from equilibrium, we Taylor-expand $\cos\alpha x$ :

\cos\alpha x \approx 1 - \frac{(\alpha x)^2}{2} + O(x^4)

Substituting into the potential:

U(x) = U_0\left(1 - \cos\alpha x\right) \approx U_0\left(1 - 1 + \frac{\alpha^2 x^2}{2}\right) = \frac{1}{2}U_0\alpha^2 x^2

Identify the effective spring constant

The approximate potential has the form $U(x) = \frac{1}{2}kx^2$ , where

$k = U_0\alpha^2$ …

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Q.A body of mass mmm is situated in a potential field U(x)=U0(1−cos⁡αx)U(x) = U_0(1 - \cos\alpha x)U(x)=U0​(1−cosαx) when U0U_0U0​ and α\alphaα are constants. Find the time period of small oscillations.

Why this works: the harmonic approximation

Finding the period

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Q.A body of mass $m$ is situated in a potential field $U(x) = U_0(1 - \cos\alpha x)$ when $U_0$ and $\alpha$ are constants. Find the time period of small oscillations.

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