MHT-CET Physics · Oscillations

Energy in Simple Harmonic Motion

The total energy of SHM, ½mω²A² (= ½kA²), stays fixed while it passes back and forth between potential energy ½kx² and kinetic energy ½k(A² − x²); so the split at any point depends only on x/A, and the total grows as the square of both the amplitude and the frequency.

All Oscillations notes MHT-CET Physics strategy

Why this matters

17 PYQs, 2 HARD. Thirteen are the split between kinetic and potential energy — the ratio at half the amplitude, at T/6 or T/12 after the mean position, where they are equal, where one is 8 times the other, and the potential energy at x + y from its values at x and y (the HARD pair); four are the total energy — of a pendulum at its extreme, when its length is changed, and of two sources of different frequency. Two cards.

Concept 1 of 2: Kinetic and Potential Energy at a Point

With the total fixed at ½kA², potential energy is the fraction (x/A)² of it and kinetic energy the rest, 1 − (x/A)². So at x = A/2 the PE is a quarter and KE:PE = 3:1; they are equal at x = A/√2; PE = 8 KE at x = (2√2/3)A. In time from the mean position x = A sin ωt, so PE:KE = tan² ωt — at T/12 (30°) it is 1:3, at T/6 (60°) 3:1. Because PE goes as x², potential energies at x and y combine at x + y as (√E₁ + √E₂)².

Definition

$U = \tfrac{1}{2}kx^2$ , $K = \tfrac{1}{2}k(A^2 - x^2)$ , $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}m\omega^2A^2$ .
$\dfrac{U}{E} = \left(\dfrac{x}{A}\right)^2$ : $x = \tfrac{A}{2}$ ⇒ K:U = 3:1; $x = \tfrac{A}{\sqrt{2}}$ ⇒ equal; $x = \tfrac{\sqrt{3}}{2}A$ ⇒ E = 4K.
In time from the mean: $\dfrac{U}{K} = \tan^2\omega t$ ; KE 75% of E at $\tfrac{T}{12}$ , 50% at $\tfrac{T}{8}$ .
K = 1.25 U with A = 3 cm ⇒ x = 2 cm.
PE at x + y: $\left(\sqrt{E_x} + \sqrt{E_y}\right)^2 = E_x + E_y + 2\sqrt{E_xE_y}$ .

Energy split

\frac{U}{E} = \frac{x^2}{A^2}, \qquad \frac{K}{E} = 1 - \frac{x^2}{A^2}

Worked example

At what displacement is the kinetic energy three times the potential energy, amplitude 6 cm?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q31Moderate

Example 1 · Oscillations · SHM Energy — Kinetic, Potential, and Total

A particle is executing linear S.H.M. starting from mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is

Putting the energies equal at half the amplitude

At A/2 the potential energy is only a quarter of the total. The half-and-half point is A/√2 ≈ 0.71A.

Drill 12 more on kinetic and potential energy at a point

Concept 2 of 2: Total Energy and What It Depends On

E = ½mω²A² grows with the square of the amplitude AND the square of the frequency. For a pendulum ω² = g/L, so at the same amplitude a quarter-length pendulum holds four times the energy, and its energy at the extreme, all potential, is mgA²/(2L). Two sources of the same energy with frequencies in the ratio 4 : 1 must have amplitudes in the ratio 1 : 4.

Definition

$E = \tfrac{1}{2}m\omega^2A^2 \propto n^2A^2$ .
Pendulum: $E = \dfrac{mgA^2}{2L}$ ; length ÷ 4 at the same amplitude ⇒ E × 4; length ÷ 3 ⇒ × 3.
Equal energies, frequencies n and n/4: amplitudes A and 4A.

Total energy

E = \tfrac{1}{2}m\omega^2A^2

Worked example

A 0.2 kg mass oscillates with amplitude 5 cm at 2 Hz. Total energy (π² = 10)?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q9Moderate

Example 2 · Oscillations · SHM Energy — Kinetic, Potential, and Total

At a place, the length of the oscillating simple pendulum is made

\frac{1}{4}

times keeping amplitude same then the total energy will be

Forgetting the frequency in the energy

Same amplitude does not mean same energy. Energy goes as ω²A², so a stiffer or shorter oscillator at the same amplitude carries more.

Drill 3 more on total energy and what it depends on

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 Point
Energy split
$\frac{U}{E} = \frac{x^2}{A^2}, \qquad \frac{K}{E} = 1 - \frac{x^2}{A^2}$
Total Energy and What It Depends On
Total energy
$E = \tfrac{1}{2}m\omega^2A^2$

Watch out for (2)

Putting the energies equal at half the amplitude
→ Kinetic and Potential Energy at a Point
Forgetting the frequency in the energy
→ Total Energy and What It Depends On

Test yourself on Oscillations

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Oscillations chapter test Drill all 17 questions