For a body executing S.H.M.:
(a) Potential energy is always equal to its K.E.
(b) Average potential and kinetic energy over any given time interval are always equal.
(c) Sum of the kinetic and potential energy at any point of time is constant.
(d) Average K.E. in one time period is equal to average potential energy in one time period.
Choose the most appropriate option from the options given below:

We begin by recalling the standard mathematical description of a particle performing simple harmonic motion (S.H.M.).

Let the particle of mass $$m$$ have an amplitude $$A$$ and an angular frequency $$\omega$$. We take its displacement from the mean (equilibrium) position at any instant $$t$$ as

$$y = A \sin \omega t.$$

From this displacement we first obtain the velocity by differentiating with respect to time, because the definition of velocity is the time-derivative of displacement:

$$v = \frac{dy}{dt} = A \omega \cos \omega t.$$

Correspondingly, the speed squared is

$$v^{2} = A^{2}\omega^{2}\cos^{2}\omega t.$$

Now, the mechanical energies are written with the help of two well-known formulae:

1. Kinetic energy: $$K = \tfrac12 m v^{2}.$$

2. Potential energy stored in the “spring” (or elastic) system: $$U = \tfrac12 k y^{2},$$ where $$k$$ is the spring constant. Because for S.H.M. we have $$\omega^{2}=k/m,$$ we may also write $$U=\tfrac12 m\omega^{2}y^{2}.$$

Using the velocity already found, we can write the kinetic energy explicitly:

$$\begin{aligned} K &= \tfrac12 m v^{2} \\ &= \tfrac12 m \bigl(A^{2}\omega^{2}\cos^{2}\omega t\bigr) \\ &= \tfrac12 m\omega^{2}A^{2}\cos^{2}\omega t. \end{aligned}$$

Likewise, substituting $$y = A\sin\omega t$$ into the potential-energy formula gives

$$\begin{aligned} U &= \tfrac12 m\omega^{2}y^{2} \\ &= \tfrac12 m\omega^{2}\bigl(A^{2}\sin^{2}\omega t\bigr) \\ &= \tfrac12 m\omega^{2}A^{2}\sin^{2}\omega t. \end{aligned}$$

With these explicit forms of $$K$$ and $$U$$ in hand, let us examine each of the four statements.

Statement (a): “Potential energy is always equal to kinetic energy.”

Comparing the expressions we have just obtained, the ratio of the two energies at an arbitrary time is

$$\frac{U}{K} = \frac{\sin^{2}\omega t}{\cos^{2}\omega t} = \tan^{2}\omega t.$$

The equality $$U=K$$ requires $$\tan^{2}\omega t = 1,$$ or $$\tan\omega t = \pm1,$$ which is satisfied only at specific instants where $$\omega t = 45^{\circ},\,135^{\circ},\,225^{\circ},\ldots$$ Therefore the two energies coincide only at those special instants and are not always equal. So statement (a) is false.

Statement (b): “The average potential and kinetic energies over any given time interval are always equal.”

To test this, we note that the average of a time-dependent function $$f(t)$$ over an interval $$\Delta t$$ is

$$\langle f \rangle = \frac{1}{\Delta t}\int_{t_{0}}^{t_{0}+\Delta t} f(t)\,dt.$$

If the chosen interval is not an integer multiple of the time period $$T = \frac{2\pi}{\omega},$$ the integrals of $$\sin^{2}\omega t$$ and $$\cos^{2}\omega t$$ will, in general, not produce the same value. Hence there exist intervals for which the two averages differ. Therefore statement (b) is also false.

Statement (c): “The sum of the kinetic and potential energies at any point of time is constant.”

Adding our explicit expressions for $$K$$ and $$U$$ we obtain

$$\begin{aligned} K + U &= \tfrac12 m\omega^{2}A^{2}\cos^{2}\omega t + \tfrac12 m\omega^{2}A^{2}\sin^{2}\omega t \\ &= \tfrac12 m\omega^{2}A^{2}\bigl(\cos^{2}\omega t + \sin^{2}\omega t\bigr) \\ &= \tfrac12 m\omega^{2}A^{2}\times 1 \\ &= \text{constant}. \end{aligned}$$

Because $$\cos^{2}\theta + \sin^{2}\theta = 1$$ for every $$\theta,$$ the total mechanical energy $$E = \tfrac12 m\omega^{2}A^{2}$$ is indeed time-independent. Thus statement (c) is true.

Statement (d): “The average kinetic energy in one full time period is equal to the average potential energy in one full time period.”

To check, we compute the time average of, say, the kinetic energy over one period $$T$$:

$$\begin{aligned} \langle K \rangle &= \frac{1}{T}\int_{0}^{T} \tfrac12 m\omega^{2}A^{2}\cos^{2}\omega t \, dt. \end{aligned}$$

The integral of $$\cos^{2}\omega t$$ over a complete cycle is well known:

$$\int_{0}^{T} \cos^{2}\omega t\,dt = \frac{T}{2}.$$

Substituting this result we get

$$\langle K \rangle = \frac{1}{T}\Bigl[\tfrac12 m\omega^{2}A^{2}\times \frac{T}{2}\Bigr] = \tfrac14 m\omega^{2}A^{2}.$$

Performing the same steps for the potential energy (with $$\sin^{2}\omega t$$ whose integral over one period is equally $$T/2$$) yields

$$\langle U \rangle = \tfrac14 m\omega^{2}A^{2}.$$

Hence $$\langle K \rangle = \langle U \rangle$$ over one complete time period. Statement (d) is therefore true.

We have found that statements (c) and (d) are correct, while (a) and (b) are incorrect. Inspecting the answer choices, option A lists exactly (c) and (d).

Hence, the correct answer is Option A.