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=(T)/(4)s` starting from mean position. Assume that the initial phase of the oscillation is zero.

Options

A. `0.62 J`
B. `6.2xx10^(-3)J`
C. `1.2xx10^(3)J`
D. `6.2xx10^(3)J`

Answer

To solve the problem of finding the potential energy of an object executing simple harmonic motion (SHM) at a specific time, we can follow these steps:

### Step-by-Step Solution

1. **Identify Given Values**:
- Mass of the object, \( m = 0.5 \, \text{kg} \)
- Amplitude, \( A = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Time period, \( T = 0.2 \, \text{s} \)

2. **Calculate Angular Frequency (\( \omega \))**:
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi \, \text{rad/s} \]

3. **Determine the Displacement at \( t = \frac{T}{4} \)**:
- At \( t = \frac{T}{4} \), the object is at its maximum displacement (amplitude) since it starts from the mean position.
\[ x = A = 0.05 \, \text{m} \]

4. **Use the Formula for Potential Energy in SHM**:
The potential energy (\( PE \)) at a displacement \( x \) in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant. We can relate \( k \) to \( \omega \) and \( m \):
\[ k = m \omega^2 \]

5. **Calculate \( k \)**:
\[ k = m \omega^2 = 0.5 \times (10\pi)^2 = 0.5 \times 100\pi^2 = 50\pi^2 \, \text{N/m} \]

6. **Calculate Potential Energy**:
Substitute \( k \) and \( x \) into the potential energy formula:
\[ PE = \frac{1}{2} (50\pi^2) (0.05)^2 \]
\[ PE = \frac{1}{2} (50\pi^2) (0.0025) \]
\[ PE = 25\pi^2 \times 0.0025 = 0.0625\pi^2 \, \text{J} \]

7. **Calculate Numerical Value**:
Using \( \pi \approx 3.14 \):
\[ PE \approx 0.0625 \times (3.14)^2 \approx 0.0625 \times 9.8596 \approx 0.616 \, \text{J} \]

### Final Answer
Thus, the potential energy of the object at \( t = \frac{T}{4} \) is approximately:
\[ PE \approx 0.62 \, \text{J} \]