The potential energy of a particle (Ux) executing S.H.M. is given by

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  1. \(\mathrm{U}_{\mathrm{x}}=\frac{\mathrm{k}}{2}(\mathrm{x}-\mathrm{a})^{2}\)
  2. Ux = k1x + k2x2 + k3x
  3. Ux = Ae−bx
  4. Ux = a constant 

Answer (Detailed Solution Below)

Option 1 : \(\mathrm{U}_{\mathrm{x}}=\frac{\mathrm{k}}{2}(\mathrm{x}-\mathrm{a})^{2}\)
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Detailed Solution

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Concept:

In Simple Harmonic Motion (S.H.M.), the potential energy of a particle is given by the formula:

Ux = (1/2) k (x - a)2

where k is the force constant and (x - a) is the displacement from the mean position.

Calculation:

Given the potential energy function, we compare it with the standard form of potential energy in S.H.M.

Option 1: Ux = (k/2)(x - a)2

⇒ This matches the standard form Ux = (1/2) k (x - a)2

∴ The correct answer is Option 1: Ux = (k/2)(x - a)2.

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