Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial ($t = 0$) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: ($x$ is in cm and $t$ is in s). (a) $x = -2\sin(3t + \pi/3)$ (b) $x = \cos(\pi/6 - t)$ (c) $x = 3\sin(2\pi t + \pi/4)$ (d) $x = 2\cos\pi t$

> [!TLDR] > The reference circle method maps SHM to uniform circular motion. For each equation, we rewrite it in the standard form $x = A\cos(\omega t + \phi)$, then read off amplitude $A$ (radius), angular speed $\omega$, and initial phase $\phi$. The initial position is $x_0 = A\cos\phi$. The sense of rotation is fixed anticlockwise. ## The Concept: Why a Circle? Simple harmonic motion is the projection of uniform circular motion onto a diameter. Imagine a particle moving anticlockwise on a circle of radius $A$ with constant angular speed $\omega$. At time $t$, its angular displacement from the positive $x$-axis is $\omega t + \phi$. The $x$-coordinate of this particle is $A\cos(\omega t + \phi)$ — that's our SHM. So every SHM of the form $x = A\cos(\omega t + \phi)$ corresponds to: - **Radius** = $A$ (the amplitude) - **Angular speed** = $\omega$ - **Initial phase** = $\phi$ (the angle at $t=0$ from the positive $x$-axis) - **Initial position** = $A\cos\phi$ The trick is to get every given equation into that exact cosine form. Sine can be converted using $\sin\theta = \cos(\theta - \pi/2)$. --- ## (a) $x = -2\sin(3t + \pi/3)$ **Step 1: Convert to cosine form.** First, handle the minus sign: $-2\sin\theta = 2\sin(\theta + \pi)$ because $\sin(\theta + \pi) = -\sin\theta$. So $x = 2\sin(3t + \pi/3 + \pi) = 2\sin(3t + 4\pi/3)$. Now convert sine to cosine: $\sin\alpha = \cos(\alpha - \pi/2)$. Thus $x = 2\cos(3t + 4\pi/3 - \pi/2) = 2\cos(3t + 5\pi/6)$. **Step 2: Read off parameters.** - Amplitude $A = 2$ cm → radius of circle = 2 cm - Angular speed $\omega = 3$ rad/s - Initial phase $\phi = 5\pi/6$ rad (150°) **Step 3: Initial position.** $x_0 = A\cos\phi = 2\cos(5\pi/6) = 2(-\sqrt{3}/2) = -\sqrt{3}$ cm ≈ -1.73 cm. > [!TIP] > The minus sign in front of the sine is not the same as a phase shift of $\pi$ — it is exactly that. Adding $\pi$ to the angle flips the sign. Always handle the amplitude sign before converting sine to cosine. --- ## (b)

Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initia

Exercises · 13.12

t = 0

(a)

x = -2\sin(3t + \pi/3)

(b)

x = \cos(\pi/6 - t)

(c)

x = 3\sin(2\pi t + \pi/4)

(d)

x = 2\cos\pi t

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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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Concept: Simple Harmonic Motion — each SHM can be represented as the projection of uniform circular motion on a diameter. The standard form is $x = A\cos(\omega t + \phi)$ , where $A$ is the radius, $\omega$ the angular speed, and $\phi$ the initial phase.

Step 1 — Rewrite each in cosine form

$x = -2\sin(3t + \pi/3) = 2\cos(3t + \pi/3 + \pi/2) = 2\cos(3t + 5\pi/6)$
$x = \cos(\pi/6 - t) = \cos(t - \pi/6)$
$x = 3\sin(2\pi t + \pi/4) = 3\cos(2\pi t + \pi/4 - \pi/2) = 3\cos(2\pi t - \pi/4)$
$x = 2\cos\pi t$ (already in standard form) Step 2 — Extract parameters

Case	$A$ (cm)	$\omega$ (rad/s)	$\phi$ (rad)	$x(0) = A\cos\phi$
(a)	2	3	$5\pi/6$	$2\cos(5\pi/6) = -\sqrt{3}$
(b)	1	1	$-\pi/6$	$\cos(-\pi/6) = \sqrt{3}/2$
(c)	3	$2\pi$	$-\pi/4$	$3\cos(-\pi/4) = 3\sqrt{2}/2$
(d)	2	$\pi$	0	$2\cos 0 = 2$

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The reference circle method maps SHM to uniform circular motion. For each equation, we rewrite it in the standard form $x = A\cos(\omega t + \phi)$ , then read off amplitude $A$ (radius), angular speed $\omega$ , and initial phase $\phi$ . The initial position is $x_0 = A\cos\phi$ . The sense of rotation is fixed anticlockwise.

The Concept: Why a Circle?

Simple harmonic motion is the projection of uniform circular motion onto a diameter. Imagine a particle moving anticlockwise on a circle of radius $A$ with constant angular speed $\omega$ . At time $t$ , its angular displacement from the positive $x$ -axis is $\omega t + \phi$ . The $x$ -coordinate of this particle is $A\cos(\omega t + \phi)$ — that's our SHM.

So every SHM of the form $x = A\cos(\omega t + \phi)$ corresponds to:

Radius = $A$ (the amplitude)
Angular speed = $\omega$
Initial phase = $\phi$ (the angle at $t=0$ from the positive $x$ -axis)
Initial position = $A\cos\phi$

The trick is to get every given equation into that exact cosine form. Sine can be converted using $\sin\theta = \cos(\theta - \pi/2)$ .

(a) $x = -2\sin(3t + \pi/3)$

Step 1: Convert to cosine form.

First, handle the minus sign: $-2\sin\theta = 2\sin(\theta + \pi)$ because $\sin(\theta + \pi) = -\sin\theta$ .

So $x = 2\sin(3t + \pi/3 + \pi) = 2\sin(3t + 4\pi/3)$ .

Now convert sine to cosine: $\sin\alpha = \cos(\alpha - \pi/2)$ .

Thus $x = 2\cos(3t + 4\pi/3 - \pi/2) = 2\cos(3t + 5\pi/6)$ .

Step 2: Read off parameters.

Amplitude $A = 2$ cm → radius of circle = 2 cm
Angular speed $\omega = 3$ rad/s
Initial phase $\phi = 5\pi/6$ rad (150°)

Step 3: Initial position.

$x_0 = A\cos\phi = 2\cos(5\pi/6) = 2(-\sqrt{3}/2) = -\sqrt{3}$ cm ≈ -1.73 cm.

Tip

The minus sign in front of the sine is not the same as a phase shift of $\pi$ — it is exactly that. Adding $\pi$ to the angle flips the sign. Always handle the amplitude sign before converting sine to cosine.

(b) $x = \cos(\pi/6 - t)$

Step 1: Rewrite as $\cos(\omega t + \phi)$ .

Cosine is even: $\cos(\theta) = \cos(-\theta)$ . So $\cos(\pi/6 - t) = \cos(t - \pi/6)$ .

Now it's in standard form: $x = 1\cdot\cos(1\cdot t - \pi/6)$ .

Step 2: Read off parameters.

$A = 1$ cm
$\omega = 1$ rad/s
$\phi = -\pi/6$ rad (-30°)

Step 3: Initial position.

$x_0 = \cos(-\pi/6) = \cos(\pi/6) = \sqrt{3}/2$ cm ≈ 0.866 cm.

Watch out

A common mistake is to read $\phi$ as $+\pi/6$ from the original form. But $\cos(\pi/6 - t)$ is not $\cos(t + \pi/6)$ — the sign of $t$ matters. Always rearrange so the $t$ term is positive.

(c) $x = 3\sin(2\pi t + \pi/4)$

Step 1: Convert sine to cosine.

$\sin\theta = \cos(\theta - \pi/2)$ .

So $x = 3\cos(2\pi t + \pi/4 - \pi/2) = 3\cos(2\pi t - \pi/4)$ .

Step 2: Read off parameters.

$A = 3$ cm
$\omega = 2\pi$ rad/s
$\phi = -\pi/4$ rad (-45°)

Step 3: Initial position.

$x_0 = 3\cos(-\pi/4) = 3(\sqrt{2}/2) = \frac{3\sqrt{2}}{2}$ cm ≈ 2.12 cm. …

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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 52 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. In SHM, the restoring force is directly proportional to$
(a) Velocity
(b) Displacement
(c) Acceleration
(d) Time
›Reveal solutionSolution
The defining equation of simple harmonic motion is F = -k x -- the restoring force is proportional to displacement x, with the negative sign showing it always acts to pull the object back toward equilibrium.

Simple Harmonic Motion (SHM) is defined precisely by this relationship between restoring force and displacement:

F = -k x

where x is the displacement from the mean/equilibrium position and k is a positive constant (the force constant). This means:

The magnitude of the restoring force grows linearly with how far the object is displaced from equilibrium.

The negative sign shows the force always points back toward the equilibrium position, opposing the displacement. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. In SHM, the maximum velocity occurs at$
(a) The equilibrium position
(b) Half of the amplitude
(c) Maximum displacement from equilibrium
(d) It is constant throughout the motion
›Reveal solutionSolution

Velocity in SHM is v(t) = -Aomegasin(omega t), which has its maximum magnitude (A*omega) exactly when displacement x = 0, i.e., at the equilibrium position.

For SHM with displacement x(t) = A cos(omega t), the velocity is:

v(t) = dx/dt = -A omega sin(omega t)

The magnitude of velocity is maximum when |sin(omega t)| = 1, which occurs precisely when cos(omega t) = 0, i.e., when x = 0 -- the equilibrium (mean) position.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The restoring force in SHM is always$
(a) in the direction of motion
(b) opposite to the direction of motion
(c) independent of motion
(d) zero
›Reveal solutionSolution

The restoring force F = -kx always points opposite to the displacement x (toward equilibrium), which is why it is described as always acting opposite to the object's motion away from the mean position -- this restoring tendency is what continually pulls the oscillator back and sustains the oscillation.

In SHM, F = -k x: whichever side of equilibrium the object is displaced to, the restoring force always points back toward the centre (equilibrium), i.e., in the direction opposite to the displacement vector. Practically, this means the restoring force is always directed so as to oppose (resist) the object's continued displacement away from the mean position -- described in this basic sense as "opposite to the direction of motion" away from equilibrium. (A more precise statement, taught alongside this idea, is that the restoring force is always opposite to displacement, not literally opposite to instantaneous velocity at every point of t …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. In simple harmonic motion, what is the phase difference between displacement and velocity?$
(a) 3π/4
(b) π
(c) π/2
(d) 0
›Reveal solutionSolution

In simple harmonic motion, velocity is always π/2 (90°) out of phase with (ahead of) displacement.

For SHM, x(t) = A sin(ωt + φ). Differentiating, v(t) = dx/dt = Aω cos(ωt + φ) = Aω sin(ωt + φ + π/2). Comparing the arguments of the sine functions for x and v, velocity has an extra phase of +π/2 relative to displacement. Physically: when displacement is maximum (at the extreme position), velocity is zero; w …

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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: Oscillatory motion is ________ only for small amplitude.$
›Reveal solutionSolution

General oscillatory motion behaves as simple harmonic motion (SHM) only when the amplitude is small.

SHM requires a restoring force (or torque) that is exactly proportional to displacement from equilibrium, F = −kx. For many real oscillators — e.g. a simple pendulum, where the restoring torque involves sin θ — this proportionality is only approximate: sin θ ≈ θ holds accurately only for small angles θ (small amplitude). For larger amplitudes, the restoring force is no longer linear in the displacement; the motion is still periodic but is no longer strictly simple harmonic (the ti …

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$Q. If the amplitude of a particle performing simple harmonic motion is doubled, then which quantity will become double?$
›Reveal solutionSolution

Doubling the amplitude of an SHM particle doubles its maximum velocity and maximum acceleration (both are ∝ A); the total energy, however, becomes four times as large, since E ∝ A².

For a particle in SHM, x = A sin(ωt), v = Aω cos(ωt), a = −Aω² sin(ωt). The maximum values are x_max = A, v_max = Aω, and a_max = Aω². Since ω (angular frequency, set by the system's mass and restoring-force constant) does not depend on amplitude, both v_max and a_max are directly proportional to A — so if A is doubled, v_max and a_max each double as well. It's important to distinguish this from the total mechanical energ …

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$Q. The position of the body is given by x = A sin ωt. The time at which the displacement would be maximum is:$
(a) π/ω
(b) π/2ω
(c) πω
(d) None of the above
›Reveal solutionSolution

$x=A\sin\omega t$ reaches its peak value $A$ when the sine term equals 1, i.e. at $t=\pi/2\omega$ .

The position is $x(t)=A\sin\omega t$ . Since $\sin\omega t$ ranges only between $-1$ and $+1$ , the displacement $x$ is maximum (equal to $A$ ) exactly when $\sin\omega t = 1$ .

…

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$Q. Write true or false: Every oscillatory motion is periodic but every periodic motion is not oscillatory.$
›Reveal solutionSolution

The statement is TRUE: 'oscillatory' is a stricter condition than 'periodic'.

A periodic motion is any motion that repeats itself identically after a fixed time interval (the period) — for example, uniform circular motion (a particle going round and round a circle at constant speed) is periodic, but it is not oscillatory because the particle never reverses direction and moves back and forth about a mean position; it just keeps circulating the same way.

…

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$Q. A particle executes simple harmonic motion with a time period of 6 second and amplitude of 3 cm. Its maximum speed in cm/sec will be$
(a) 3π
(b) 2π
(c) π
(d) π/2
›Reveal solutionSolution

v_max = A omega = 3(2 pi/6) = pi cm/s. Answer (C).

In SHM the maximum speed is v_max = A omega, where omega = 2 pi/T.

Here A = 3 cm and T = 6 s, so:

…

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$Q. The displacement of a particle moving in simple harmonic motion (SHM) at any instant is given by y = a cosωt. The acceleration after time t = T/4 (where T is time period) is$
(a) 0
(b) − aω^2
(c) aω^2
(d) − aω
›Reveal solutionSolution

At t = T/4 the displacement is zero (mean position), so acceleration = -omega^2 y = 0. Answer (A).

Given y = a cos(omega t). Acceleration in SHM is:

a(t) = d^2y/dt^2 = -a omega^2 cos(omega t) = -omega^2 y.

…

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$Q. A simple harmonic oscillator has an amplitude a and time period T. The time required by it to travel from x = a to x = a/2 is$
(a) T/2
(b) T/4
(c) T/3
(d) T/6
›Reveal solutionSolution

Starting from the extreme (x = a), x = a/2 is reached at t = T/6. Answer (D).

Start timing from the extreme position, so x = a cos(omega t) (at t = 0, x = a).

When x = a/2: …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. In simple harmonic motion, the phase difference between acceleration and displacement is (A) 0° (B) π (C) π/2 (D) 2π$
›Reveal solutionSolution

Acceleration and displacement in SHM are 180° (π radian) out of phase.

For a particle in SHM, displacement is $x = A\sin(\omega t)$ and acceleration is:

$a = -\omega^2x = -\omega^2A\sin(\omega t) = \omega^2A\sin(\omega t + \pi)$

…

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←Previous question13.11 — Two particles each move uniformly on a circle lying in a plane, the ce…→Next question13.13 — Figure 13.21(a) shows a spring of force constant k clamped rigidly at …

The Concept: Why a Circle?

(a) x=−2sin⁡(3t+π/3)x = -2\sin(3t + \pi/3)x=−2sin(3t+π/3)

(b) x=cos⁡(π/6−t)x = \cos(\pi/6 - t)x=cos(π/6−t)

(c) x=3sin⁡(2πt+π/4)x = 3\sin(2\pi t + \pi/4)x=3sin(2πt+π/4)

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(a) $x = -2\sin(3t + \pi/3)$

(b) $x = \cos(\pi/6 - t)$

(c) $x = 3\sin(2\pi t + \pi/4)$

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