The displacement of a particle is represented by the equation $y = 3\cos\left(\dfrac{\pi}{4} - 2\omega t\right)$. The motion of the particle is (a) simple harmonic with period 2p/w. (b) simple harmonic with period $\pi/\omega$. (c) periodic but not simple harmonic. (d) non-periodic.

> [!TLDR] > A displacement of the form $y = A\cos(\alpha - \omega't)$ is simple harmonic motion; rewriting reveals angular frequency $2\omega$ and hence period $\frac{2\pi}{2\omega} = \frac{\pi}{\omega}$. ## Why this is simple harmonic motion Simple harmonic motion (SHM) requires the displacement to be a sinusoidal function of time. The standard forms are $y = A\sin(\omega't + \phi)$ or $y = A\cos(\omega't + \phi)$, where $A$ is amplitude, $\omega'$ is angular frequency, and $\phi$ is a phase constant. The given equation $y = 3\cos\left(\frac{\pi}{4} - 2\omega t\right)$ might look unusual because of the subtraction inside the cosine, but cosine is an even function: $\cos(-\theta) = \cos(\theta)$. This means we can rewrite the argument. ## Step-by-step analysis 1. **Rewrite the displacement equation** Start with $y = 3\cos\left(\frac{\pi}{4} - 2\omega t\right)$. Factor out the negative sign: $$y = 3\cos\left[-\left(2\omega t - \frac{\pi}{4}\right)\right]$$ Using $\cos(-\theta) = \cos(\theta)$: $$y = 3\cos\left(2\omega t - \frac{\pi}{4}\right)$$ 2. **Identify the form** This is now clearly in the standard SHM form $y = A\cos(\omega't + \phi)$, where: - Amplitude $A = 3$ - Angular frequency $\omega' = 2\omega$ - Phase constant $\phi = -\frac{\pi}{4}$ 3. **Calculate the period** The period $T$ of SHM is related to angular frequency by: $$T = \frac{2\pi}{\omega'}$$ Substituting $\omega' = 2\omega$: $$T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$$ > [!WARNING] > Don't confuse the parameter $\omega$ in the problem with the angular frequency of the motion. The angular frequency is $2\omega$, not $\omega$. > [!FORMULA] > $$T = \frac{2\pi}{\text{angular frequency}} = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}$$ The motion is indeed simple harmonic (a pure cosine function of time) with period $\pi/\omega$. > [!ANSWER] > The correct option is **(B)**: simple harmonic with period $\pi/\omega$.

The displacement of a particle is represented by t – Class 11 Physics

NCERT Exemplar · Q1

y = 3\cos\left(\dfrac{\pi}{4} - 2\omega t\right)

(a) simple harmonic with period 2p/w.

(b) simple harmonic with period

\pi/\omega

(d) non-periodic.

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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.

A displacement of the form $y = A\cos(\alpha - \omega't)$ is simple harmonic motion; rewriting reveals angular frequency $2\omega$ and hence period $\frac{2\pi}{2\omega} = \frac{\pi}{\omega}$ .

Why this is simple harmonic motion

Simple harmonic motion (SHM) requires the displacement to be a sinusoidal function of time. The standard forms are $y = A\sin(\omega't + \phi)$ or $y = A\cos(\omega't + \phi)$ , where $A$ is amplitude, $\omega'$ is angular frequency, and $\phi$ is a phase constant.

The given equation $y = 3\cos\left(\frac{\pi}{4} - 2\omega t\right)$ might look unusual because of the subtraction inside the cosine, but cosine is an even function: $\cos(-\theta) = \cos(\theta)$ . This means we can rewrite the argument.

Step-by-step analysis

Rewrite the displacement equation Start with $y = 3\cos\left(\frac{\pi}{4} - 2\omega t\right)$ . Factor out the negative sign:

y = 3\cos\left[-\left(2\omega t - \frac{\pi}{4}\right)\right]

Using $\cos(-\theta) = \cos(\theta)$ :

y = 3\cos\left(2\omega t - \frac{\pi}{4}\right)

Identify the form

This is now clearly in the standard SHM form $y = A\cos(\omega't + \phi)$ , where:
- Amplitude $A = 3$
- Angular frequency $\omega' = 2\omega$
- Phase constant $\phi = -\frac{\pi}{4}$
Calculate the period

The period $T$ of SHM is related to angular frequency by:

T = \frac{2\pi}{\omega'}

Substituting $\omega' = 2\omega$ :

T = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}

Watch out

Don't confuse the parameter $\omega$ in the problem with the angular frequency of the motion. The angular frequency is $2\omega$ , not $\omega$ .

T = \frac{2\pi}{\text{angular frequency}} = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}

The motion is indeed simple harmonic (a pure cosine function of time) with period $\pi/\omega$ .

✓Final answer

\pi/\omega

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 21 on this concept.

KEAM 2026Set eng-2026-04184 marksMCQ
$x$
›Reveal solutionSolution
Simple harmonic motion demands the acceleration be proportional to displacement and directed oppositely: $a=-\omega^2 x$ . The equation $a=-10x$ is the only linear, restoring form.

The defining condition of SHM is
$a=-\omega^2 x$
with $\omega^2>0$ , so $a$ must be (i) directly proportional to $x$ (first power only) and (ii) of opposite sign.

(A) $a=0.9x^2$ — non-linear (quadratic). Not SHM.

(B) $a=-10x$ — linear, negative coefficient $\Rightarrow \omega^2=10$ . SHM.

(C) $a=100x^3$ — non-linear (cubic). Not SHM.

(D) $a=1.2x$ — linear but positive sign (repulsive, unstable). Not SHM.

(E) $a^2=12x$ — non-linear relation. Not SHM.

✓Final answer
$The correct option is (B) .$
KEAM 2026Set eng-2026-04194 marksMCQ
$y = 2(\sin 2\pi t + \sqrt{3}\cos 2\pi t)$
›Reveal solutionSolution

Combine $a\sin\theta+b\cos\theta$ into a single sinusoid of amplitude $\sqrt{a^2+b^2}$ .

Given $y=2\sin2\pi t+2\sqrt3\cos2\pi t$ , this is of the form $a\sin\theta+b\cos\theta$ with $a=2$ and $b=2\sqrt3$ .

The resultant amplitude is
$A=\sqrt{a^2+b^2}=\sqrt{2^2+(2\sqrt3)^2}=\sqrt{4+12}=\sqrt{16}=4\ \text{cm}.$
✓Final answer
$The correct option is (A) .$
KEAM 2026Set eng-2026-04204 marksMCQ
$x_1 = a\sin\omega t$
›Reveal solutionSolution

Because $\cos\omega t=\sin(\omega t+\pi/2)$ , the two motions differ in phase by $\pi/2$ .

To compare phases, express both displacements in the same (sine) form. Using the identity $\cos\theta=\sin\!\left(\theta+\tfrac{\pi}{2}\right)$ :
$x_2=a\cos\omega t=a\sin\!\left(\omega t+\frac{\pi}{2}\right).$
Comparing the arguments with $x_1=a\sin\omega t$ , the phase of $x_2$ exceeds that of $x_1$ by
$\Delta\phi=\left(\omega t+\frac{\pi}{2}\right)-\omega t=\frac{\pi}{2}.$
So the two particles are a quarter-period out of step.

✓Final answer
$The correct option is (C) .$
KEAM 2026Set eng-2026-04214 marksMCQ
$A$
›Reveal solutionSolution

Mean-to-extreme is one quarter cycle $=T/4$ .

Write $x=A\sin(\omega t)$ with $\omega=2\pi/T$ . At $x=0$ , $t=0$ ; at $x=A$ , $\sin(\omega t)=1\Rightarrow\omega t=\pi/2$ , so
$t=\frac{\pi/2}{\omega}=\frac{\pi/2}{2\pi/T}=\frac{T}{4}.$
This is the standard quarter-period from the mean position to the amplitude extreme.

✓Final answer
$The correct option is (C) .$
KEAM 2026Set eng-2026-04224 marksMCQ
$Q. Which one of the following motions is not simple harmonic motion? (A) Rotation of earth about its axis (B) Small oscillations of a mass attached to a spring (C) Oscillation of mercury column in a U tube (D) Oscillation of a second's pendulum (E) The projection of uniform circular motion on the diameter of the circle$
›Reveal solutionSolution

SHM requires an oscillatory, to-and-fro motion with restoring force $\propto$ displacement; steady rotation is none of these.

Rotation of the earth about its axis is uniform rotational motion — it is periodic but not oscillatory and has no linear restoring force, so it is not SHM.

The spring–mass system, mercury in a U-tube, the second's pendulum, and the projection of uniform circular motion on a diameter are all SHM.

✓Final answer
$The correct option is (A) .$
KEAM 2026Set pha-2026-0418F4 marksMCQ
$Q. The magnitude and direction of acceleration change in the case of an object (A) executing simple harmonic motion (B) in circular motion at constant speed (C) falling under gravity from lower altitudes (D) falling under gravity from higher altitudes (E) falling in a viscous liquid medium$
›Reveal solutionSolution

Only in SHM do BOTH the magnitude and direction of acceleration change continuously.

In simple harmonic motion $a = -\omega^2 x$ , so the acceleration varies in magnitude (maximum at extremes, zero at mean position) and reverses direction as the body oscillates — both magnitude and direction change. In uniform circular motion the centripetal acceleration $v^2/r$ is constant in magnitude and only its direction changes. Free fall (near-uniform g) keeps acceleration essentially constant. Hence the case where both change is SHM.

✓Final answer
$The correct option is (A) .$
KEAM 2026Set pha-2026-0419F4 marksMCQ
$^{-2}$
›Reveal solutionSolution

Using $a=\omega^2 x$ : $\omega=\sqrt{48/3}=4\ \mathrm{s^{-1}}$ .

For simple harmonic motion the magnitude of acceleration is related to displacement by
$|a| = \omega^2 x.$
Solving for the angular frequency,
$\omega = \sqrt{\frac{a}{x}} = \sqrt{\frac{48.0}{3.0}} = \sqrt{16} = 4\\ \text{s}^{-1}.$
✓Final answer
$The correct option is (E) .$
KEAM 2026Set pha-2026-0420F4 marksMCQ
$T$
›Reveal solutionSolution

$a_{max}=\omega^2 A$ ; doubling it doubles $\omega^2$ , so $\omega\to\sqrt2\,\omega$ and $T\to T/\sqrt2$ .

In SHM the maximum acceleration is $a_{max} = \omega^2 A$ . With the amplitude $A$ constant, doubling $a_{max}$ requires
$\omega'^2 = 2\omega^2 \Rightarrow \omega' = \sqrt{2}\,\omega.$
Since $T = \dfrac{2\pi}{\omega}$ ,
$T' = \frac{2\pi}{\omega'} = \frac{2\pi}{\sqrt{2}\,\omega} = \frac{T}{\sqrt{2}}.$
✓Final answer
$The correct option is (D) .$
KEAM 2025Set eng-2025-04234 marksMCQ
$\frac{1}{\sqrt{2}}$
›Reveal solutionSolution
The time period of the SHM is 16 seconds.

Concept and Intuition

Starting from the mean position, displacement follows $x = A\sin(\omega t)$ . Given $x = A/\sqrt{2}$ at $t = 2\,s$ , we solve for $\omega$ and then the period.

Step-by-Step Solution

$x = A\sin(\omega t)$ , with $x = \frac{A}{\sqrt{2}}$ at $t = 2\,s$ .

$\sin(2\omega) = \frac{1}{\sqrt{2}} \Rightarrow 2\omega = \frac{\pi}{4}$ .

$\omega = \frac{\pi}{8}$ .

$T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi/8} = 16\,s$ .

Common Mistakes

Treating the displacement as measured from an extreme position (cosine) instead of from the mean position (sine).

✓Final answer
$The correct option is (E) — 16.$

ANSWER: E
KEAM 2025Set eng-2025-04234 marksMCQ
$A$
›Reveal solutionSolution
After the bigger pendulum completes one oscillation, the phase difference is $\pi$ .

Concept and Intuition

In the time the larger pendulum (period $5T/2$ ) finishes one oscillation, the smaller pendulum (period $T$ ) completes 2.5 oscillations. Phase difference is $2\pi$ times the difference in the number of oscillations, reduced modulo $2\pi$ .

Step-by-Step Solution

Time elapsed $= \frac{5T}{2}$ .

Oscillations of A (period $T$ ): $\frac{5T/2}{T} = 2.5$ ; phase $= 2.5\times 2\pi = 5\pi$ .

Oscillations of B: 1; phase $= 2\pi$ .

Phase difference $= 5\pi - 2\pi = 3\pi \equiv \pi$ (mod $2\pi$ ).

Common Mistakes

Forgetting to reduce $3\pi$ modulo $2\pi$ to get $\pi$ .

✓Final answer
$\pi$

ANSWER: E
KEAM 2025Set eng-2025-04294 marksMCQ
$A$
›Reveal solutionSolution

$a=-\omega^2 x$ , so at $O$ ( $x=0$ ) acceleration is zero; at the extremes it is maximum.

SHM relations. For displacement $x$ from the mean position,
$a = -\omega^2 x, \qquad v = \omega\sqrt{A^2 - x^2}.$
At the mean position $O$ ( $x=0$ ). Acceleration $a=-\omega^2(0)=0$ , while speed $v=\omega A$ is maximum. At the extreme positions $A$ and $B$ ( $x=\pm A$ ), acceleration is maximum and velocity is zero.

Checking the options: "at $O$ , $a=0$ " is the correct statement.

✓Final answer
$The correct option is (E) .$
KEAM 2025Set eng-2025-04294 marksMCQ
$x$
›Reveal solutionSolution

Substitute $t=3$ s into $x(t)=5\cos 4\pi t$ . Since $4\pi\times3=12\pi$ is a whole multiple of $2\pi$ , the cosine equals $1$ and $x=5$ m.

The displacement is $x(t)=5\cos(4\pi t)$ . At $t=3$ s the phase is $4\pi(3)=12\pi$ rad. Cosine is periodic with period $2\pi$ , and $12\pi=6\times2\pi$ , so $\cos(12\pi)=\cos 0=1$ . Therefore $x=5\times1=5$ m — the particle is back at its amplitude position.

✓Final answer
$The correct option is (B) .$

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Q.The displacement of a particle is represented by the equation y=3cos⁡(π4−2ωt)y = 3\cos\left(\dfrac{\pi}{4} - 2\omega t\right)y=3cos(4π​−2ωt). The motion of the particle is

Why this is simple harmonic motion

Step-by-step analysis

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Q.The displacement of a particle is represented by the equation $y = 3\cos\left(\dfrac{\pi}{4} - 2\omega t\right)$ . The motion of the particle is

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