An object of mass `0.2 kg` executes SHm along the X-axis with frequency of `(25//pi) Hz`. At the point `X=0.4 m` the object has `KE 0.5 J` and `PE 0.4 J`. The amplitude of oscilation is-

Options

A. 0.06 m
B. 0.04 m
C. 0.05 m
D. 0.25 m

Answer

To solve the problem step by step, we need to find the amplitude of oscillation for an object executing Simple Harmonic Motion (SHM).

### Given Data:
- Mass of the object, \( m = 0.2 \, \text{kg} \)
- Frequency, \( f = \frac{25}{\pi} \, \text{Hz} \)
- Kinetic Energy, \( KE = 0.5 \, \text{J} \)
- Potential Energy, \( PE = 0.4 \, \text{J} \)
- Position, \( x = 0.4 \, \text{m} \)

### Step 1: Calculate Total Energy
The total energy \( E \) in SHM is the sum of kinetic energy and potential energy:
\[ E = KE + PE \]
Substituting the values:
\[ E = 0.5 \, \text{J} + 0.4 \, \text{J} = 0.9 \, \text{J} \]

### Step 2: Calculate Angular Frequency \( \omega \)
The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:
\[ \omega = 2\pi f \]
Substituting the value of \( f \):
\[ \omega = 2\pi \left(\frac{25}{\pi}\right) = 50 \, \text{rad/s} \]

### Step 3: Use the Total Energy Formula
The total energy in SHM can also be expressed as:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
Where \( A \) is the amplitude. Rearranging this formula to find \( A^2 \):
\[ A^2 = \frac{2E}{m \omega^2} \]

### Step 4: Substitute Values to Find Amplitude
Now we can substitute the values of \( E \), \( m \), and \( \omega \):
\[ A^2 = \frac{2 \times 0.9 \, \text{J}}{0.2 \, \text{kg} \times (50 \, \text{rad/s})^2} \]
Calculating \( (50)^2 \):
\[ (50)^2 = 2500 \]
Now substituting this back:
\[ A^2 = \frac{1.8}{0.2 \times 2500} = \frac{1.8}{500} = 0.0036 \]

### Step 5: Calculate Amplitude \( A \)
Taking the square root to find \( A \):
\[ A = \sqrt{0.0036} = 0.06 \, \text{m} \]

### Final Answer
The amplitude of oscillation is:
\[ A = 0.06 \, \text{m} \quad \text{or} \quad 6 \, \text{cm} \]