Q.The total energy of a particle in SHM is E. Find the kinetic and potential energy in terms of the total energy at the instant when the displacement is half of the amplitude.

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.

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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).

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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 …

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.

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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:

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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

1. 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.
2. 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. …

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:

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

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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 $K = \tfrac{1}{2}k(A^2 - x^2)$.

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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.

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