Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not

Exercises · 13.2

Q. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

(a) the rotation of earth about its axis.

(b) motion of an oscillating mercury column in a U-tube.

(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.

(d) general vibrations of a polyatomic molecule about its equilibrium position.

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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.

Simple harmonic motion requires a restoring force exactly proportional to displacement (linear). Among the given examples, only the U-tube mercury column and the ball bearing in a smooth bowl are nearly SHM for small amplitudes; Earth’s rotation is periodic but not SHM, and polyatomic molecular vibrations are periodic but generally not simple harmonic.

The key to recognising simple harmonic motion (SHM) is the linear restoring force — the force must be proportional to the displacement from equilibrium and directed opposite to it. That gives the equation $F = -kx$ , or equivalently $a = -\omega^2 x$ . If the force law is different (e.g., constant, or nonlinear), the motion is periodic but not simple harmonic.

Let’s examine each case.

Rotation of Earth about its axis

The Earth spins with a nearly constant angular velocity. There is no restoring force at all — it’s uniform circular motion, not oscillatory. The motion repeats every 24 hours, so it is periodic, but there is no force proportional to displacement. Hence it is periodic but not SHM.
Oscillating mercury column in a U-tube

When you push the mercury down on one side, the imbalance in height creates a pressure difference. The restoring force is proportional to the height difference $h$ (since $F \propto \rho g h$ ), and for small displacements this gives $a \propto -h$ . That’s exactly the SHM condition.

Tip

The U-tube oscillator is a classic example of SHM — the restoring force is linear because the weight of the unbalanced column is directly proportional to the displacement.

So this represents nearly SHM (exactly SHM for small amplitudes, ignoring friction).
Ball bearing inside a smooth curved bowl

For a bowl that is spherical (or parabolic near the bottom), the restoring force along the tangent is $F = -mg \sin\theta$ . For small angles, $\sin\theta \approx \theta$ , and the displacement along the arc is $s = R\theta$ , so $F \approx -(mg/R)s$ , which is linear. Hence the motion is nearly SHM for small releases.

Watch out

For large amplitudes, $\sin\theta \neq \theta$ , and the motion becomes periodic but anharmonic (not SHM). The problem says “nearly” — so small oscillations qualify.
General vibrations of a polyatomic molecule

A polyatomic molecule has many vibrational modes. For very small displacements from equilibrium, the potential energy is approximately quadratic (Taylor expansion), so each normal mode is SHM. But the problem says “general vibrations” — that includes large-amplitude motions where anharmonic terms matter. Also, the molecule has many coupled oscillators; the overall motion is a superposition of many frequencies. It is periodic (if all frequencies are commensurate) but not simple harmonic because the net motion is not a single sine wave.

Note

If the question meant a single normal mode at small amplitude, it would be SHM. But “general vibrations” implies the full, possibly anharmonic, motion — so it’s periodic but not SHM.

✓Final answer

Periodic but not SHM; nearly SHM; nearly SHM; periodic but not SHM.

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.

This is exactly what makes an oscillator's motion "simple harmonic" -- any motion where the restoring force is proportional to displacement (and directed opposite to it) will be SHM, with acceleration a = -(k/m)x also being proportional to displacement (which is why "acceleration" is sometimes tempting to pick, but the FUNDAMENTAL defining proportionality, stated in the force law itself, is with displacement).

✓Final answer
$(b) Displacement.$
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.

At the extreme positions (x = +-A), sin(omega t) = 0, so velocity is zero there (the object momentarily stops before reversing direction) -- exactly the opposite of "maximum displacement." This makes physical sense energetically too: all the mechanical energy is kinetic at the mean position (zero elastic PE there) and all potential at the extremes (zero KE there), by conservation of energy in SHM.

✓Final answer
$(a) The equilibrium position.$
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 the cycle -- for example, on the way back toward the mean position, both the force and the velocity point in the same direction, toward equilibrium. But the basic, always-true relationship examined here is that the restoring force continually acts against the object's displacement/outward motion from equilibrium, which is why it restores the oscillation.)

✓Final answer
$(b) opposite to the direction of motion.$
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; when displacement is zero (at the mean position), velocity is maximum — this quarter-cycle offset is exactly a phase difference of π/2.

✓Final answer
$The correct option is (c) π/2.$
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 time period even starts depending slightly on the amplitude). This is why the familiar SHM formulas (like T = 2π√(l/g) for a pendulum) are valid only in the small-amplitude limit.

✓Final answer
$Oscillatory motion is simple harmonic only for small amplitude.$
CBSE 2026Set ANNUAL1 mark
$Q. If the amplitude of a particle performing simple harmonic motion is doubled, then which quantity will become double?$
›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 energy of the oscillator, E = ½mω²A², which is proportional to A² and therefore becomes four times as large (not double) when A is doubled.

✓Final answer
$The maximum velocity and maximum acceleration both become double; the total energy becomes four times (not double).$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The position of the body is given by x = A sin ωt. The time at which the displacement would be maximum is:$
(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 $\sin\omega t = 1$ .

$\sin\theta = 1$ first occurs at $\theta = \pi/2$ , so:

$\omega t = \frac{\pi}{2} \quad\Rightarrow\quad t=\frac{\pi}{2\omega}$

✓Final answer
$\pi/2\omega$
CBSE 2026Set ANNUAL1 mark
$Q. Write true or false: Every oscillatory motion is periodic but every periodic motion is not oscillatory.$
›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.

An oscillatory motion, by contrast, is a to-and-fro motion about a fixed mean (equilibrium) position, repeated over and over — for example, a simple pendulum or a mass on a spring. Since this to-and-fro motion also repeats after a fixed interval, every oscillatory motion is automatically periodic. But as the circular-motion example shows, the converse is not true.

✓Final answer
$True — every oscillatory motion is periodic, but every periodic motion need not be oscillatory.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. A particle executes simple harmonic motion with a time period of 6 second and amplitude of 3 cm. Its maximum speed in cm/sec will be$
(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:

omega = 2 pi/6 = pi/3 rad/s.

v_max = A omega = 3 x (pi/3) = pi cm/s.

✓Final answer
$(C) pi cm/s.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The displacement of a particle moving in simple harmonic motion (SHM) at any instant is given by y = a cosωt. The acceleration after time t = T/4 (where T is time period) is$
(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.

At t = T/4: omega t = (2 pi/T)(T/4) = pi/2, and cos(pi/2) = 0.

So y = 0 (the particle is at the mean position) and the acceleration = -omega^2 (0) = 0.

✓Final answer
$(A) 0.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. A simple harmonic oscillator has an amplitude a and time period T. The time required by it to travel from x = a to x = a/2 is$
(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:

a/2 = a cos(omega t) -> cos(omega t) = 1/2 -> omega t = pi/3.

Since omega = 2 pi/T:

(2 pi/T) t = pi/3 -> t = (pi/3)(T/2 pi) = T/6.

✓Final answer
$(D) T/6.$
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:

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

The negative sign means acceleration is always directed opposite to displacement (toward the mean position), which corresponds to a phase difference of $\pi$ radians (180°) between the two.

✓Final answer
$(B) π.$

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Q.Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

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