JEE Mains Physics · Oscillations

Energy in SHM, Amplitude Changes and Damping

The total energy ½kA² stays fixed while it passes between kinetic and potential, so the split at any point depends only on x/A; a sudden push, an added mass or damping changes the amplitude.

All Oscillations notes JEE Mains Physics strategy

Why this matters

Twenty-five PYQs, seventeen of them multiple choice, and two from 2026. Thirteen split the energy at a point between kinetic and potential; six pick an energy graph or an average over a period; six change the amplitude, by a sudden push, by a mass dropped on at the mean position, or by damping.

Concept 1 of 3: Kinetic and potential energy at a displacement in SHM

The total energy is fixed at

\tfrac{1}{2}kA^{2}

. The potential energy is the fraction

(x/A)^{2}

of it and the kinetic energy is the rest. So the split depends only on how far out the particle is compared with the amplitude.

Definition

$U = \tfrac{1}{2}kx^{2}$ , $K = \tfrac{1}{2}k(A^{2} - x^{2})$ , $E = \tfrac{1}{2}kA^{2} = \tfrac{1}{2}m\omega^{2}A^{2}$ .
$\dfrac{U}{E} = \dfrac{x^{2}}{A^{2}}$ and $\dfrac{K}{E} = 1 - \dfrac{x^{2}}{A^{2}}$ .
K = U at $x = A/\sqrt{2}$ . At $x = A/2$ : $U = E/4$ , $K = 3E/4$ . At $x = A/3$ : $K = 8E/9$ .
At any point, $E = K + U$ there; then $A^{2} = 2E/k$ .
For a spring, E depends on k and A only. Doubling the mass at the same amplitude leaves E unchanged; only ω falls.
Vertical spring: the energy stored in the spring at the lowest point includes the static stretch Δ: $\tfrac{1}{2}k(\Delta + A)^{2}$ , more than the oscillation energy $\tfrac{1}{2}kA^{2}$ .

Energy in SHM

E = \frac{1}{2}kA^{2} = \frac{1}{2}m\omega^{2}A^{2} \qquad U = \frac{1}{2}kx^{2} \qquad K = \frac{1}{2}k\left(A^{2} - x^{2}\right)

Worked example

A 0.5 kg block on a spring oscillates with

\omega = 20

rad/s. At

x = 3

cm its kinetic energy is 0.16 J. Find the potential energy there, the total energy and the amplitude.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 8 April 2024 · Q117Moderate

Example 1 · Oscillations · Energy in SHM, Amplitude Changes and Damping

An object of mass

0.2\text{ }kg

executes simple harmonic motion along

x

axis with frequency of

\left( \frac{25}{\pi} \right)Hz

. At the position

x= 0.04\text{ }m

the object has kinetic energy

0.5\text{ }J

and potential energy

0.4\text{ }J

. The amplitude of oscillation is_____

cm

The energies are equal at A/√2, not at A/2

At half the amplitude the potential energy is only a quarter of the total. Kinetic and potential are equal at x = A/√2, about 0.71A.

A spring's energy does not depend on the mass

E = ½kA² holds for any mass. A heavier block at the same amplitude has the same energy, a lower ω and a lower top speed.

"Energy of the block" at a point is the total

When a question gives the block's energy at some x, it means K + U, which equals ½kA². Do not set it equal to ½kx².

Drill 12 more on kinetic and potential energy at a displacement in shm

Concept 2 of 3: Graphs and averages of energy in SHM

Kinetic and potential energy go as

\cos^{2}

and

\sin^{2}

of the phase, so each repeats twice per oscillation and is never negative. Against displacement they are two parabolas that add to a flat line. Most graph questions are answered by the shape and by where each curve is zero.

Definition

From the mean, $U = \dfrac{E}{2}(1 - \cos 2\omega t)$ and $K = \dfrac{E}{2}(1 + \cos 2\omega t)$ .
Both energies oscillate with angular frequency $2\omega$ : frequency $2f$ , period $T/2$ .
Over one period, the average kinetic and potential energies are each $E/2 = \tfrac{1}{4}kA^{2}$ .
K equals U only at four instants per period (at $x = \pm A/\sqrt{2}$ ), not all the time.
$E - U$ is K, so its graph against x is the K graph.

Quantity	Against displacement x	Against time t (starting at the mean)	Average over a period
Potential energy U	Upward parabola $\tfrac{1}{2}kx^{2}$ : zero at x = 0, E at $x = \pm A$	$\tfrac{E}{2}(1 - \cos 2\omega t)$ : zero at t = 0, peak E at T/4, repeats every T/2	$E/2$
Kinetic energy K	Downward parabola $\tfrac{1}{2}k(A^{2} - x^{2})$ : E at x = 0, zero at $x = \pm A$	$\tfrac{E}{2}(1 + \cos 2\omega t)$ : E at t = 0, zero at T/4, repeats every T/2	$E/2$ E − U against x is this same downward parabola.
Total energy E	Horizontal line at $\tfrac{1}{2}kA^{2}$	Horizontal line at $\tfrac{1}{2}kA^{2}$	$E$
Velocity v	Ellipse $\dfrac{x^{2}}{A^{2}} + \dfrac{v^{2}}{A^{2}\omega^{2}} = 1$	$A\omega\cos\omega t$ , repeats every T	Zero (average speed $2A\omega/\pi$ )
Acceleration a	Straight line $a = -\omega^{2}x$ through the origin	$-\omega^{2}A\sin\omega t$ , repeats every T	Zero

Energies repeat at twice the oscillator's frequency and never go below zero.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 2 · Q8Moderate

Example 2 · Oscillations · Energy in SHM, Amplitude Changes and Damping

The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of

176rad/s

. The frequency of this simple harmonic oscillator is___ Hz.

\left\lbrack take \right.

\left. \ \pi =\frac{22}{7} \right\rbrack

Energy oscillates at twice the frequency

If the kinetic energy varies at angular frequency Ω, the oscillator's own angular frequency is Ω/2. Taking them equal doubles the answer.

Potential energy is never negative

U = ½kx² is zero at the mean position and positive on both sides. A U–t graph that dips below zero, or a U–x graph that is a straight line, is wrong.

Drill 5 more on graphs and averages of energy in shm

Concept 3 of 3: New amplitude after a push or an added mass, and decay by damping

A sudden change keeps the particle where it is but changes its speed, and perhaps its ω. Use the new speed and ω at that point to find the new amplitude. Damping instead drains the energy slowly, and the amplitude falls by the same factor in each equal interval of time.

Definition

Sudden change at x: $A'^{2} = x^{2} + \dfrac{v'^{2}}{\omega'^{2}}$ , with the new speed and new ω.
Speed multiplied by n at x (same ω): $v'^{2} = n^{2}\omega^{2}(A^{2} - x^{2})$ .
Mass m placed gently on M at the mean: momentum is conserved, $(M + m)v' = Mv_{max}$ , and $\omega' = \sqrt{k/(M + m)}$ . So $A' = A\sqrt{\dfrac{M}{M + m}}$ .
Mass added at an extreme: the speed is already zero, so the amplitude stays A; only the period changes.
Damping $F = -bv$ : $A = A_0e^{-bt/2m}$ . The energy goes as $A^{2}$ , so $E = E_0e^{-bt/m}$ .
Time for the amplitude to halve: $t = \dfrac{2m\ln 2}{b}$ ; for the energy to halve: $\dfrac{m\ln 2}{b}$ .

Amplitude after a sudden change, and damped amplitude

A'^{2} = x^{2} + \frac{v'^{2}}{\omega'^{2}} \qquad A = A_0e^{-bt/2m}

Worked example

A particle does SHM with amplitude 5 cm. As it passes

x = 3

cm, its speed is suddenly doubled. Find the new amplitude.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 31 January 2024 · Q29Moderate

Example 3 · Oscillations · Energy in SHM, Amplitude Changes and Damping

A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is

\frac{2\text{ }A}{3}

. The new amplitude of motion is

\frac{nA}{3}

. The value of

n

is ______ .

Where the mass is added decides the new amplitude

Added at the mean, momentum is shared and the amplitude drops to A√(M/(M + m)). Added at an extreme, the block is at rest there, so the amplitude stays A.

Multiply the speed, not the energy

Tripling the speed at x multiplies the kinetic energy there by 9. The new amplitude comes from A'² = x² + v'²/ω², not from tripling A.

Energy decays twice as fast as amplitude

A = A₀e^(−bt/2m) but E = E₀e^(−bt/m). The energy halves in half the time the amplitude takes.

Drill 5 more on new amplitude after a push or an added mass, and decay by damping

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Kinetic and potential energy at a displacement in SHM
Energy in SHM
$E = \frac{1}{2}kA^{2} = \frac{1}{2}m\omega^{2}A^{2} \qquad U = \frac{1}{2}kx^{2} \qquad K = \frac{1}{2}k\left(A^{2} - x^{2}\right)$
New amplitude after a push or an added mass, and decay by damping
Amplitude after a sudden change, and damped amplitude
$A'^{2} = x^{2} + \frac{v'^{2}}{\omega'^{2}} \qquad A = A_0e^{-bt/2m}$

Reference tables (1)

Graphs and averages of energy in SHM5 rows

Quantity	Against displacement x	Against time t (starting at the mean)	Average over a period
Potential energy U	Upward parabola $\tfrac{1}{2}kx^{2}$ : zero at x = 0, E at $x = \pm A$	$\tfrac{E}{2}(1 - \cos 2\omega t)$ : zero at t = 0, peak E at T/4, repeats every T/2	$E/2$
Kinetic energy K	Downward parabola $\tfrac{1}{2}k(A^{2} - x^{2})$ : E at x = 0, zero at $x = \pm A$	$\tfrac{E}{2}(1 + \cos 2\omega t)$ : E at t = 0, zero at T/4, repeats every T/2	$E/2$ E − U against x is this same downward parabola.
Total energy E	Horizontal line at $\tfrac{1}{2}kA^{2}$	Horizontal line at $\tfrac{1}{2}kA^{2}$	$E$
Velocity v	Ellipse $\dfrac{x^{2}}{A^{2}} + \dfrac{v^{2}}{A^{2}\omega^{2}} = 1$	$A\omega\cos\omega t$ , repeats every T	Zero (average speed $2A\omega/\pi$ )
Acceleration a	Straight line $a = -\omega^{2}x$ through the origin	$-\omega^{2}A\sin\omega t$ , repeats every T	Zero

Energies repeat at twice the oscillator's frequency and never go below zero.

Watch out for (8)

The energies are equal at A/√2, not at A/2
→ Kinetic and potential energy at a displacement in SHM
A spring's energy does not depend on the mass
→ Kinetic and potential energy at a displacement in SHM
"Energy of the block" at a point is the total
→ Kinetic and potential energy at a displacement in SHM
Energy oscillates at twice the frequency
→ Graphs and averages of energy in SHM
Potential energy is never negative
→ Graphs and averages of energy in SHM
Where the mass is added decides the new amplitude
→ New amplitude after a push or an added mass, and decay by damping
Multiply the speed, not the energy
→ New amplitude after a push or an added mass, and decay by damping
Energy decays twice as fast as amplitude
→ New amplitude after a push or an added mass, and decay by damping

Test yourself on Oscillations

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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