The displacement of a particle is represented by the equation $y = \sin^{3}\omega t$. The motion is (a) non-periodic. (b) periodic but not simple harmonic. (c) simple harmonic with period $2\pi/\omega$. (d) simple harmonic with period $\pi/\omega$.

> [!TLDR] > The given displacement $y = \sin^3 \omega t$ is periodic but not simple harmonic because it cannot be expressed as a single sine or cosine term with a constant angular frequency — its period is $2\pi/\omega$, but it contains higher harmonics. The key to this question lies in understanding what makes a motion simple harmonic. Simple harmonic motion (SHM) requires the restoring force (and hence acceleration) to be directly proportional to the displacement from equilibrium, and the displacement must be a pure sinusoidal function of time — something like $y = A \sin(\omega t + \phi)$ or $y = A \cos(\omega t + \phi)$. The given function $y = \sin^3 \omega t$ is a cube of a sine, not a pure sine. That alone should raise suspicion. Let’s break it down step by step. 1. **Check if the motion is periodic.** A function is periodic if $f(t + T) = f(t)$ for some finite $T$. Since $\sin \omega t$ has period $2\pi/\omega$, any integer power of it will also repeat after $2\pi/\omega$. Indeed, \[ y(t + 2\pi/\omega) = \sin^3[\omega(t + 2\pi/\omega)] = \sin^3(\omega t + 2\pi) = \sin^3 \omega t = y(t). \] So the motion is periodic with period $T = 2\pi/\omega$. This eliminates option (A). 2. **Test if it is simple harmonic.** For SHM, the displacement must satisfy $\frac{d^2y}{dt^2} = -\omega_0^2 y$ for some constant $\omega_0$. Let’s compute the acceleration. First derivative: \[ \frac{dy}{dt} = 3 \sin^2 \omega t \cdot \omega \cos \omega t = 3\omega \sin^2 \omega t \cos \omega t. \] Second derivative (using product rule): \[ \frac{d^2y}{dt^2} = 3\omega \left[ 2\sin \omega t \cos \omega t \cdot \omega \cos \omega t + \sin^2 \omega t \cdot (-\omega \sin \omega t) \right]. \] Simplify: \[ \frac{d^2y}{dt^2} = 3\omega^2 \left[ 2\sin \omega t \cos^2 \omega t - \sin^3 \omega t \right]. \] Using $\cos^2 \omega t = 1 - \sin^2 \omega t$, \[ \frac{d^2y}{dt^2} = 3\omega^2 \left[ 2\sin \omega t (

The displacement of a particle is represented by the equation y = sin³ω t. The motion is (a) non-periodic. (b)

NCERT Exemplar · Q2

y = \sin^{3}\omega t

(a) non-periodic.

(b) periodic but not simple harmonic.

2\pi/\omega

(d) simple harmonic with period

\pi/\omega

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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).

From $\omega$ , you get the period: $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}$ .

Watch out

Do not confuse angular frequency $\omega$ (rad/s) with ordinary frequency $f$ (Hz). They are related by $\omega = 2\pi f$ . Many exam errors come from mixing these up.

Real-World Examples

SHM is an idealization — a perfect model. But many real systems approximate it beautifully:

A mass on a spring (horizontal or vertical) — the classic textbook example.
A simple pendulum — but only for small angles (less than about $15^\circ$ ). For large swings, the restoring force is no longer proportional to displacement, and the motion is not simple harmonic.
The vibration of atoms in a solid — each atom is held in place by bonds that act like tiny springs.
A tuning fork — the prongs vibrate in SHM, producing a pure tone.

The Bottom Line

Simple Harmonic Motion is any motion driven by a restoring force that is proportional to and opposite the displacement. It produces a sinusoidal oscillation with a constant period that is independent of amplitude. Everything else — the equations, the graphs, the energy transformations — is just unpacking that single, elegant idea.

Looking up "Simple Harmonic Motion: Definition, Formula & Real-World Examples" or "Simple Harmonic Motion important questions 11" is a common way students land here, and rightly so — simple harmonic motion is a core part of the Class 11 Physics NCERT/CBSE curriculum. Expect it to reappear, often in a slightly disguised form, across JEE Main, NEET and state engineering/medical entrance exams.

The given displacement $y = \sin^3 \omega t$ is periodic but not simple harmonic because it cannot be expressed as a single sine or cosine term with a constant angular frequency — its period is $2\pi/\omega$ , but it contains higher harmonics.

The key to this question lies in understanding what makes a motion simple harmonic. Simple harmonic motion (SHM) requires the restoring force (and hence acceleration) to be directly proportional to the displacement from equilibrium, and the displacement must be a pure sinusoidal function of time — something like $y = A \sin(\omega t + \phi)$ or $y = A \cos(\omega t + \phi)$ . The given function $y = \sin^3 \omega t$ is a cube of a sine, not a pure sine. That alone should raise suspicion.

Let’s break it down step by step.

Check if the motion is periodic. A function is periodic if $f(t + T) = f(t)$ for some finite $T$ . Since $\sin \omega t$ has period $2\pi/\omega$ , any integer power of it will also repeat after $2\pi/\omega$ . Indeed,

y(t + 2\pi/\omega) = \sin^3[\omega(t + 2\pi/\omega)] = \sin^3(\omega t + 2\pi) = \sin^3 \omega t = y(t).

So the motion is periodic with period $T = 2\pi/\omega$ . This eliminates option (A).

Test if it is simple harmonic. For SHM, the displacement must satisfy $\frac{d^2y}{dt^2} = -\omega_0^2 y$ for some constant $\omega_0$ . Let’s compute the acceleration. First derivative:

\frac{dy}{dt} = 3 \sin^2 \omega t \cdot \omega \cos \omega t = 3\omega \sin^2 \omega t \cos \omega t.

Second derivative (using product rule):

\frac{d^2y}{dt^2} = 3\omega \left[ 2\sin \omega t \cos \omega t \cdot \omega \cos \omega t + \sin^2 \omega t \cdot (-\omega \sin \omega t) \right].

Simplify:

\frac{d^2y}{dt^2} = 3\omega^2 \left[ 2\sin \omega t \cos^2 \omega t - \sin^3 \omega t \right].

Using $\cos^2 \omega t = 1 - \sin^2 \omega t$ ,

\frac{d^2y}{dt^2} = 3\omega^2 \left[ 2\sin \omega t (1 - \sin^2 \omega t) - \sin^3 \omega t \right] = 3\omega^2 \left[ 2\sin \omega t - 2\sin^3 \omega t - \sin^3 \omega t \right].

\frac{d^2y}{dt^2} = 3\omega^2 \left[ 2\sin \omega t - 3\sin^3 \omega t \right] = 6\omega^2 \sin \omega t - 9\omega^2 \sin^3 \omega t.

This is not proportional to $y = \sin^3 \omega t$ alone — there is an extra $\sin \omega t$ term. Hence the motion is not SHM.

A cleaner way: rewrite $\sin^3 \omega t$ using a trigonometric identity. Recall the triple-angle formula:

\sin 3\theta = 3\sin\theta - 4\sin^3\theta \quad \Rightarrow \quad \sin^3\theta = \frac{3\sin\theta - \sin 3\theta}{4}.

With $\theta = \omega t$ ,

y = \frac{3\sin \omega t - \sin 3\omega t}{4}.

This expresses $y$ as a sum of two sine waves: one with angular frequency $\omega$ and another with $3\omega$ . A simple harmonic oscillator can only vibrate at a single frequency. The presence of the $3\omega$ term means the motion is a superposition of two harmonics — it is periodic but not simple harmonic.

Tip

The identity $\sin^3\theta = \frac{3\sin\theta - \sin 3\theta}{4}$ is a powerful shortcut. It immediately reveals that the motion contains a frequency component at $3\omega$ , which disqualifies it from being SHM.

Determine the period from the rewritten form. The term $\sin \omega t$ has period $2\pi/\omega$ , and $\sin 3\omega t$ has period $2\pi/(3\omega)$ . The overall period is the least common multiple of these two periods, which is $2\pi/\omega$ . So the period is $2\pi/\omega$ , not $\pi/\omega$ .

Watch out

A common mistake is to think that because $\sin^3 \omega t$ looks like a sine wave, it must be SHM. But SHM requires a linear restoring force — a cubic term like $\sin^3$ introduces nonlinearity. Also, don’t confuse the period of $\sin^3 \omega t$ with that of $\sin \omega t$ ; they are the same here, but that doesn’t make it SHM.

Thus the motion is periodic with period $2\pi/\omega$ , but it is not simple harmonic.

✓Final answer

The correct option is (B) — periodic but not simple harmonic.

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