Q.Derive an expression for the total energy of a particle oscillates in simple harmonic motion. OR Derive the Newton's formula to find the speed of a longitudinal wave in an ideal gas and apply Laplace's correction.

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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