A particle of mass 0.5 kg executes SHM. If its period of oscillation is `pi` seconds and total energy is 0.04 J, then the amplitude of oscillation will be

Options

A. 40 cm
B. 20 cm
C. 15 cm
D. 10 cm

Explanation

To find the amplitude of oscillation for a particle executing Simple Harmonic Motion (SHM), we can follow these steps:

### Step 1: Identify the given values
- Mass of the particle, \( m = 0.5 \, \text{kg} \)
- Period of oscillation, \( T = \pi \, \text{s} \)
- Total energy, \( E = 0.04 \, \text{J} \)

### Step 2: Relate the period to angular frequency
The relationship between the period \( T \) and angular frequency \( \omega \) is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( T \):
\[ \pi = \frac{2\pi}{\omega} \]
To find \( \omega \), rearranging gives:
\[ \omega = \frac{2\pi}{\pi} = 2 \, \text{rad/s} \]

### Step 3: Use the formula for total energy in SHM
The total energy \( E \) in SHM is given by:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
Where \( A \) is the amplitude. We can substitute the known values into this equation:
\[ 0.04 = \frac{1}{2} \times 0.5 \times (2)^2 \times A^2 \]

### Step 4: Simplify the equation
Calculating the right-hand side:
\[ 0.04 = \frac{1}{2} \times 0.5 \times 4 \times A^2 \]
\[ 0.04 = 1 \times A^2 \]
Thus, we have:
\[ A^2 = 0.04 \]

### Step 5: Solve for amplitude \( A \)
Taking the square root of both sides gives:
\[ A = \sqrt{0.04} = 0.2 \, \text{m} \]

### Step 6: Convert to centimeters
To convert meters to centimeters:
\[ A = 0.2 \, \text{m} = 20 \, \text{cm} \]

### Final Answer
The amplitude of oscillation is \( 20 \, \text{cm} \).

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