Show, starting from the expressions for kinetic and potential energy of a particle executing S.H.M., that its

Exercise · Q14

Q. Show, starting from the expressions for kinetic and potential energy of a particle executing S.H.M., that its total mechanical energy remains constant at all times and at all displacements.

West Bengal WbchseTextbookSubjectiveImportance★★★★★

52% · 14/27 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 Energy

Simple Harmonic Motion Energy: From Intuition to Precision

Imagine a pendulum swinging, or a mass bouncing on a spring. You push it once, and it keeps moving back and forth. Where does that energy go? It doesn't vanish — it just changes form. That's the core idea.

The Intuition: A Trade Between Two Forms

Think of a child on a swing. At the highest point, the swing is momentarily still — all the energy is stored as potential energy (the height you could fall from). At the lowest point, the swing is moving fastest — all that stored energy has turned into kinetic energy (the energy of motion). In between, it's a mix of both.

For a spring-mass system (the simplest SHM), the same trade happens:

When the mass is at the extreme position (maximum displacement), it's momentarily at rest — all energy is potential.
When the mass passes through the equilibrium position (the centre), it's moving fastest — all energy is kinetic.
Everywhere else, it's a blend.

The key insight: total mechanical energy stays constant (if no friction). Energy is never created or destroyed — it just shifts between potential and kinetic.

The Precise Statement

For a particle of mass $m$ executing SHM with angular frequency $\omega$ and amplitude $A$ :

E_{\text{total}} = \frac{1}{2} m \omega^2 A^2

This is a constant. At any displacement $x$ from equilibrium:

Kinetic energy: $K = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 (A^2 - x^2)$
Potential energy: $U = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2$ (since $k = m\omega^2$ )
Total energy: $E = K + U = \frac{1}{2} m \omega^2 A^2$

Notice: when $x = \pm A$ , $K = 0$ and $U = E$ . When $x = 0$ , $K = E$ and $U = 0$ .

Why This Matters for Exams

Three things to remember:

Total energy depends only on amplitude and frequency — not on the mass's position or speed at any instant. It's a fixed number for a given oscillation.
Energy is proportional to the square of amplitude: double the amplitude, quadruple the energy. This is a common exam trap — students think doubling amplitude doubles energy. It doesn't.
The potential energy curve is a parabola: $U = \frac{1}{2} k x^2$ . This is why SHM is called "harmonic" — the restoring force ( $F = -kx$ ) comes from this parabolic potential well.

Watch out

A frequent mistake: writing $U = \frac{1}{2} m \omega^2 x^2$ but forgetting that $\omega^2 = k/m$ . Both forms are equivalent — use whichever is given in the problem.

A Quick Check

A mass of 0.5 kg oscillates on a spring with $k = 8$ N/m and amplitude 0.1 m. Find total energy. …

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

For a particle executing S.H.M. with amplitude $A$ and angular frequency $\omega$ , the kinetic energy at displacement $x$ is, using $v=\omega\sqrt{A^2-x^2}$ ,

K(x) = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2-x^2)

and the potential energy, taking $U=0$ at the mean position and using $k=m\omega^2$ , is

U(x) = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2x^2

Adding the two together,

$K(x)+U(x) = \frac{1}{2}m\omega^2(A^2-x^2) + \frac{1}{2}m\omega^2x^2 = \frac{1}{2}m\omega^2A^2 - \frac{1}{2}m\omega^2x^2 + \frac{1}{2}m\omega^2x^2 = \frac{1}{2}m\omega^2A^2$ …

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.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The potential energy of a particle executing simple harmonic motion at a distance x from the mean position is proportional to$
(a) x
(b) √x
(c) x^2
(d) x^3
›Reveal solutionSolution

PE in SHM is proportional to x^2. Answer (C).

In SHM the restoring force is F = -k x, and the associated potential energy is:

U = (1/2) k x^2 = (1/2) m omega^2 x^2.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The kinetic energy of a pendulum bob is maximum when$
(a) kinetic energy remains constant in a simple pendulum
(b) it is at the lowest point of its swing
(c) it is at the midpoint of its swing
(d) it is at the highest point of its swing
›Reveal solutionSolution

By conservation of mechanical energy, a pendulum bob has maximum speed (and thus maximum kinetic energy) at the lowest point of its swing, where its height (and potential energy) is minimum.

As a pendulum bob swings, its total mechanical energy (KE + PE) stays constant (ignoring air resistance).

At the extreme ends of the swing (highest points), the bob is momentarily at rest (v = 0), so KE = 0 and all the energy is potential energy (maximum height).

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. When the hanging bob of a pendulum crosses its equilibrium position, then its energy is (A) zero (B) completely potential (C) completely kinetic (D) partially potential and partially kinetic$
›Reveal solutionSolution

At the equilibrium position, a pendulum's energy is entirely kinetic.

For a simple pendulum executing SHM, potential energy $\propto x^2$ (displacement from equilibrium) and is zero exactly at the mean/equilibrium position, while speed — and hence kinetic energy — is maximum there (total mechanical energy stays constan …

Start your 14-day free trial to see the full solution →
CBSE 2024Set SET-AP55001 markMCQ
$Q. At the mean position of simple harmonic motion, there will be:$
(a) Kinetic energy maximum and potential energy minimum
(b) Kinetic energy minimum and potential energy maximum
(c) Both kinetic energy and potential energy maximum
(d) Both kinetic energy and potential energy minimum
›Reveal solutionSolution

At the mean position of SHM, displacement x = 0, so PE (∝ x^2) is minimum (zero), while speed is maximum, so KE is maximum.

For a particle in SHM with displacement x = A sin ωt, the restoring-force potential energy is PE = ½kx^2 = ½mω^2x^2. At the mean position x = 0, so PE = 0 — its minimum value.

The velocity is v = Aω cos ωt, which is maximum in magnitude (= Aω) exactly when x = 0 (mean position), since cos ωt = ±1 there. So kinetic energy KE = ½mv^2 is at its maximum at the mean position.

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. The energy in simple harmonic motion is:$
(a) KA²/2
(b) KA²
(c) KA²/4
(d) 2KA²
›Reveal solutionSolution

The total energy of a simple harmonic oscillator is E = (1/2) K A², where K is the force constant and A the amplitude — constant throughout the motion, continuously exchanged between kinetic and potential forms.

For SHM with displacement x = A sin(ωt), the restoring force is F = −Kx (K = force constant = m ω²). The potential energy is U(x) = (1/2) K x², and the kinetic energy is KE = (1/2) m v² = (1/2) K (A² − x²) (using v² = ω²(A²−x²) and K = mω²). Adding them:

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set annual1 mark
$Q. At what point is the energy of a Harmonic Oscillator entirely:$
(i) Kinetic
(ii) Potential?
›Reveal solutionSolution
A harmonic oscillator's energy is purely kinetic at the mean position and purely potential at the extreme positions.

For a particle executing SHM with displacement x = A sin(wt), amplitude A and angular frequency w:

Kinetic Energy: KE = (1/2) m v^2 = (1/2) m w^2 (A^2 - x^2)

Potential Energy: PE = (1/2) m w^2 x^2

At the mean position, x = 0, so v is maximum (v = Aw). Then PE = (1/2) m w^2 (0)^2 = 0, and KE = (1/2) m w^2 A^2, which is the maximum possible value and equal to the total energy. So the energy is entirely kinetic at the mean position.

At the extreme positions, x = +/-A, the particle momentarily stops, so v = 0. Then KE = 0, and PE = (1/2) m w^2 A^2, the total energy. So the energy is entirely potential at the extreme positions. …
Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$Q. The potential energy of a particle in SHM at a distance x from the equilibrium position is:$
(a) ½ mω²x²
(b) ½ mω²a²
(c) ½ mω²(a² − x²)
(d) Zero
›Reveal solutionSolution

The PE of a particle in SHM at displacement x from the mean position is $\tfrac12 m\omega^2x^2$ .

Derivation. In SHM, the restoring force is $F = -m\omega^2 x$ . The potential energy is obtained by integrating the work done against this force:

$U(x) = -\int_0^x F\,dx = -\int_0^x(-m\omega^2x)\,dx = \tfrac12 m\omega^2 x^2$

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$Q. When the oscillation bob is crossing its mean position its energy is —$
(a) Zero
(b) Wholly P.E.
(c) Wholly K.E.
(d) Partly P.E. & partly K.E.
›Reveal solutionSolution

At the mean position the energy of an oscillating bob is entirely kinetic.

In simple harmonic motion the potential energy is $U = \tfrac{1}{2}kx^2$ and the kinetic energy is $K = \tfrac{1}{2}k(A^2 - x^2)$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2018Set sz1 markMCQ
$Q. If x is the displacement of the particle from the mean position while executing S.H.M., then the total energy of a particle executing S.H.M. is:$
(a) proportional to x
(b) proportional to x^2
(c) independent of x
(d) proportional to x^(1/2)
›Reveal solutionSolution

While the kinetic and potential energies of a particle in S.H.M. individually depend on displacement x, their SUM (the total mechanical energy) is always constant, independent of x.

For a particle of mass m executing simple harmonic motion with amplitude A and angular frequency omega, at displacement x from the mean position:

Kinetic energy: KE = (1/2) m omega^2 (A^2 - x^2) - this decreases as x increases (maximum at x = 0, the mean position; zero at x = A, the extreme position).

Potential energy: PE = (1/2) m omega^2 x^2 - this increases as x increases (zero at x = 0; maximum at x = A).

Total energy: Adding these,

E = KE + PE = (1/2) m omega^2 (A^2 - x^2) + (1/2) m omega^2 x^2

= (1/2) m omega^2 A^2 - (1/2) m omega^2 x^2 + (1/2) m omega^2 x^2

= (1/2) m omega^2 A^2

The x^2 terms cancel exactly, leaving a result that depends only on the mass, angular frequency, and amplitude - all constants for a given oscillation - and NOT on the instantaneous displacement x. This reflects the conservation of mechanical energy in S.H.M.: as the particle moves, kinetic energy continuously converts into potential energy and back, but their sum never changes.

…

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

←Previous questionQ13 — Two springs of force constants k₁ and k₂ are connected side by side (i…→Next questionQ15 — Derive the expression for the time period of a simple pendulum, T = 2π…

Q.Show, starting from the expressions for kinetic and potential energy of a particle executing S.H.M., that its total mechanical energy remains constant at all times and at all displacements.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days