Simple Harmonic Motion: Energy Analysis
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Direct answer
In simple harmonic motion the restoring force is proportional to displacement, F = −k x, giving a = −omega^2 x; the total energy E = (1/2) m omega^2 A^2 = (1/2) k A^2 stays constant while energy shuttles between kinetic and potential forms. Kinetic energy peaks at the mean position (equal to E) and vanishes at the extremes; potential energy mirrors it; the two are equal at displacement x = A/sqrt(2).
What you must remember
- Displacement x = A sin(omega t + phi); speed v = omega sqrt(A^2 − x^2) — maximum omega A at the mean position, zero at the extremes.
- Energies: KE = (1/2) m omega^2 (A^2 − x^2), PE = (1/2) m omega^2 x^2, total E = (1/2) m omega^2 A^2 — constant, and proportional to the squares of both amplitude and frequency.
- KE equals PE at x = A/sqrt(2); at x = A/2 the split is 3:1 in favour of kinetic energy.
- Averaged over a cycle, mean KE = mean PE = (1/4) m omega^2 A^2, exactly half of the total each.
- Time periods: spring–mass 2 pi sqrt(m/k); simple pendulum 2 pi sqrt(l/g); physical pendulum 2 pi sqrt(I/(m g d)); liquid column of total length L in a U-tube 2 pi sqrt(L/(2g)).
- Springs in series: 1/k = 1/k1 + 1/k2; in parallel: k = k1 + k2; cutting a spring in half doubles its stiffness.
- Effective g shifts in accelerating lifts — g + a upward, g − a downward, zero in free fall (no oscillation).
Common confusion
The recurring error is assuming potential energy must vanish at the mean position always; that holds only when the reference is the equilibrium point (natural length of a spring, lowest point of a pendulum), since PE depends on the chosen zero. Students also forget that energy scales as amplitude squared — halving the amplitude quarters the energy — and misstate that acceleration, not speed, is greatest at the extremes.
Exam-focused takeaway
JEE Main asks the KE–PE ratio at a given displacement, the equality position, the energy fraction at x = A/2, and the effect of changing amplitude as numerical-value questions — clean marks from two formulas. JEE Advanced embeds SHM in larger setups: a body oscillating in a liquid, a sphere rolling in a hollow (rotation adds to effective inertia, lengthening the period), two blocks tied by a cut thread, and composite pendulums whose I must first be found. Identify the restoring force or torque, confirm it is proportional to displacement, read off omega^2, then apply the energy relations.
Frequently asked questions
Where does kinetic energy equal potential energy in SHM?
At x = A/sqrt(2) from the mean position, where each form holds exactly half of the total energy.
What is the speed at the mean position?
The maximum value omega A, since v = omega sqrt(A^2 − x^2) peaks at x = 0 and falls to zero at the extremes.
Why is the total energy constant?
The restoring force is conservative and no dissipation acts, so energy only converts between forms; E = (1/2) m omega^2 A^2 depends on amplitude alone.
What happens to the energy if amplitude is doubled?
It becomes four times — total energy is proportional to A^2.
Why is a pendulum's swing SHM only when small?
The restoring torque is proportional to sin(theta), which approximates theta only for small angles in radians; beyond that the period grows with amplitude.
What is the period of a liquid oscillating in a U-tube?
2 pi sqrt(L/(2g)) for total column length L, because a height difference 2x drives the restoring pressure head — a standard JEE variant.