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Step-by-Step Solution
Step 1: Write down the known quantities
• Mass of the object, $m = 0.5 \,\text{kg}$
• Amplitude of oscillation, $A = 5\,\text{cm} = 0.05 \,\text{m}$
• Time period, $T = 0.2\,\text{s}$
• Time at which we want the potential energy, $t = \frac{T}{4} = 0.05\,\text{s}$
• Initial phase of oscillation, $\phi = 0$
Step 2: Use the formula for time period to find $k$
For a simple harmonic oscillator, the time period $T$ is given by
$$T = 2\pi \sqrt{\frac{m}{k}}.$$
Substituting $T = 0.2\,\text{s}$ and $m = 0.5\,\text{kg}$:
$$
0.2 = 2\pi \sqrt{\frac{0.5}{k}}.
$$
Step 3: Solve for the spring constant $k$
Rearrange and square both sides:
$$
\frac{0.2}{2\pi} = \sqrt{\frac{0.5}{k}}
\quad\Longrightarrow\quad
\left(\frac{0.2}{2\pi}\right)^2 = \frac{0.5}{k}.
$$
Hence,
$$
k = \frac{0.5}{\left(\frac{0.2}{2\pi}\right)^2}.
$$
Numerically, this evaluates approximately to
$$
k \approx 50\pi^2 \approx 500\,\text{N/m}.
$$
Step 4: Determine the displacement at $t = \frac{T}{4}$
In simple harmonic motion, the displacement $x(t)$ (assuming the initial phase $\phi = 0$) can be written as
$$
x(t) = A \sin(\omega t),
$$
where $\omega = \frac{2\pi}{T}$. At $t = \frac{T}{4}$,
$$
x\left(\frac{T}{4}\right) = A \sin\left(\omega \cdot \frac{T}{4}\right) = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{4}\right) = A \sin\left(\frac{\pi}{2}\right).
$$
Since $\sin\left(\frac{\pi}{2}\right) = 1$,
$$
x\left(\frac{T}{4}\right) = A = 0.05\,\text{m}.
$$
Step 5: Calculate the potential energy at this displacement
The potential energy stored in the spring (or in SHM) at displacement $x$ is:
$$
PE = \frac{1}{2} k x^2.
$$
Substituting $k \approx 500\,\text{N/m}$ and $x = 0.05\,\text{m}$,
$$
PE = \frac{1}{2} \times 500 \times \left(0.05\right)^2
= 250 \times 0.0025
= 0.625\,\text{J}.
$$
This is approximately $0.62\,\text{J}$.
Final Answer
$\boxed{0.62\,\text{J}}$
