A particle is acted simultaneously by mutually perpendicular simple hormonic motions $x = a\cos\omega t$ and $y = a\sin\omega t$. The trajectory of motion of the particle will be (a) an ellipse. (b) a parabola. (c) a circle. (d) a straight line.

A particle is acted simultaneously by mutually perpendicular simple hormonic motions x = acosω t and y = asinω

NCERT Exemplar · Q5

x = a\cos\omega t

(a) an ellipse.

(b) a parabola.

(d) a straight line.

Kerala DhseMCQ· 1mImportance★★★★★est

47% · 31/66 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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). …

Start your 14-day free trial to unlock the full solution →

The two perpendicular SHMs have equal amplitude and a phase difference of $\frac{\pi}{2}$ , which produces a circular trajectory. The correct option is (C).

When two simple harmonic motions act on a particle along perpendicular axes, the resulting path depends on three things: the amplitudes, the frequencies, and the phase difference between them. Here, both motions have the same amplitude $a$ and the same angular frequency $\omega$ , but one is a cosine and the other a sine — that’s a phase difference of $\frac{\pi}{2}$ .

The key insight: if you square and add the two equations, the time dependence cancels out, leaving a relation between $x$ and $y$ alone. That relation is the equation of the trajectory.

Write the given equations:

x = a\cos\omega t

y = a\sin\omega t

Square both:

x^2 = a^2\cos^2\omega t

y^2 = a^2\sin^2\omega t

Add them:

x^2 + y^2 = a^2(\cos^2\omega t + \sin^2\omega t) = a^2

This is the equation of a circle centered at the origin with radius $a$ .

Check the direction of motion. At $t=0$ , $x = a$ and $y = 0$ — the particle is at $(a,0)$ . A moment later, $\omega t$ becomes a small positive angle, so $\cos\omega t$ decreases slightly and $\sin\omega t$ becomes positive — meaning $x$ decreases and $y$ becomes positive. The particle moves anticlockwise along the circle. …

Start your 14-day free trial to unlock the full solution →

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. …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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

$t=\frac{\pi/2}{\omega}=\frac{\pi/2}{2\pi/T}=\frac{T}{4}.$ …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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, …
Start your 14-day free trial to see the full solution →
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

$\omega'^2 = 2\omega^2 \Rightarrow \omega' = \sqrt{2}\,\omega.$ …
Start your 14-day free trial to see the full solution →
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 $t = 2\,s$ .

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

$\omega = \frac{\pi}{8}$ . …
Start your 14-day free trial to see the full solution →
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 $= 2.5\times 2\pi = 5\pi$ . …
Start your 14-day free trial to see the full solution →
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,

$a = -\omega^2 x, \qquad v = \omega\sqrt{A^2 - x^2}.$ …
Start your 14-day free trial to see the full solution →
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. …

Start your 14-day free trial to see the full solution →

←Previous questionQ4 — Motion of an oscillating liquid column in a U-tube is (a) periodic but…→Next questionQ6 — The displacement of a particle varies with time according to the relat…

Q.A particle is acted simultaneously by mutually perpendicular simple hormonic motions x=acos⁡ωtx = a\cos\omega tx=acosωt and y=asin⁡ωty = a\sin\omega ty=asinωt. The trajectory of motion of the particle will be

🎓Unlock everything free for 14 days

Q.A particle is acted simultaneously by mutually perpendicular simple hormonic motions $x = a\cos\omega t$ and $y = a\sin\omega t$ . The trajectory of motion of the particle will be

Unlock everything free for 14 days