An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $$t = \frac{T}{4}$$ s starting from mean position. Assume that the initial phase of the oscillation is zero.

We have an object of mass $$m = 0.5\;{\rm kg}$$ executing simple harmonic motion (S.H.M.).

Its amplitude is given as $$A = 5\;{\rm cm}$$. Converting centimetres to metres,

$$A = 5\;{\rm cm} = 5 \times 10^{-2}\;{\rm m} = 0.05\;{\rm m}.$$

The time-period is $$T = 0.2\;{\rm s}.$$ For S.H.M. the angular frequency is related to the period by the formula

$$\omega = \frac{2\pi}{T}.$$

Substituting the given value of $$T$$, we get

$$\omega = \frac{2\pi}{0.2} = 10\pi\;{\rm rad\,s^{-1}}.$$

Since the motion starts from the mean position with zero initial phase, the displacement as a function of time is

$$x(t) = A \sin(\omega t).$$

At the instant $$t = \dfrac{T}{4}$$, we have

$$\omega t = \omega \left(\frac{T}{4}\right) = \frac{2\pi}{T}\,\frac{T}{4} = \frac{2\pi}{4} = \frac{\pi}{2}.$$

So the displacement becomes

$$x\!\left(\frac{T}{4}\right) = A \sin\!\left(\frac{\pi}{2}\right) = A \times 1 = A = 0.05\;{\rm m}.$$

Thus, at $$t = \dfrac{T}{4}$$ the particle has reached the extreme position; its kinetic energy is zero and its potential energy equals the total mechanical energy of the oscillation.

The total mechanical energy $$E$$ stored in S.H.M. is given by the formula

$$E = \frac{1}{2} k A^{2},$$

where $$k$$ is the force constant. Using the relation $$k = m\omega^{2},$$ the energy can also be written as

$$E = \frac{1}{2} m \omega^{2} A^{2}.$$

Substituting the known values step by step:

$$\omega^{2} = (10\pi)^{2} = 100\pi^{2} \approx 100 \times 9.8696 = 986.96,$$

$$A^{2} = (0.05)^{2} = 0.0025,$$

$$m \omega^{2} A^{2} = 0.5 \times 986.96 \times 0.0025 = 0.5 \times 2.4674 = 1.2337.$$

Now applying the factor $$\dfrac{1}{2}$$ in the formula,

$$E = \frac{1}{2} \times 1.2337 = 0.6169\;{\rm J}.$$

This is approximately $$0.62\;{\rm J}.$$ Because all the mechanical energy at this instant is stored as potential energy, we have

$$U\!\left(t = \frac{T}{4}\right) \approx 0.62\;{\rm J}.$$

Hence, the correct answer is Option A.