1. Given data

• Mass: $m = 0.5\,\text{kg}$
• Amplitude: $A = 5\,\text{cm} = 0.05\,\text{m}$
• Time period: $T = 0.2\,\text{s}$
• Time at which potential energy is required: $t = \dfrac{T}{4}$
• Initial phase is zero and motion starts from mean position.

2. Equation of SHM

Since the particle starts from mean position with zero initial phase, we can write

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

where

$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi\,\text{rad/s}$

3. Position at $t = \dfrac{T}{4}$

$x = A\sin\left(\omega \cdot \frac{T}{4}\right)$

Now,

$\omega \cdot \frac{T}{4} = \frac{2\pi}{T}\cdot \frac{T}{4} = \frac{\pi}{2}$

So,

$x = A\sin\frac{\pi}{2} = A$

Thus at $t=\dfrac{T}{4}$, the object is at the extreme position.

4. Potential energy in SHM

Potential energy at displacement $x$ is

$U = \frac{1}{2}m\omega^2 x^2$

Since $x=A$ here,

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

Substitute the values:

$U = \frac{1}{2}(0.5)(10\pi)^2(0.05)^2$

$U = 0.25 \times 100\pi^2 \times 0.0025$

$U = 0.0625\pi^2$

Using $\pi^2 \approx 9.87$,

$U \approx 0.0625 \times 9.87 \approx 0.617\,\text{J}$

$U \approx 0.62\,\text{J}$

5. Option check

• A: $0.62\,\text{J}$ ✅
• B: $6.2 \times 10^{-3}\,\text{J}$ ❌
• C: $1.2 \times 10^3\,\text{J}$ ❌
• D: $6.2 \times 10^3\,\text{J}$ ❌

Therefore, the correct option is A.