The kinetic energy of a particle executing SHM is 16J. When it is in its mean position. If the amplitude of oscillation is 25cm and the mass of the particle is 5.12 kg, the itme period of its oscillation in second is

Options

A. `pi//5`
B. `2pi`
C. `5pi`
D. `20pi`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the given information
- Kinetic Energy (KE) = 16 J
- Amplitude (A) = 25 cm = 0.25 m (convert to meters)
- Mass (m) = 5.12 kg
- Displacement (y) at mean position = 0 m

### Step 2: Write the formula for kinetic energy in SHM
The kinetic energy of a particle in simple harmonic motion (SHM) is given by the formula:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - y^2) \]
Since the particle is at the mean position, \(y = 0\). Therefore, the formula simplifies to:
\[ KE = \frac{1}{2} m \omega^2 A^2 \]

### Step 3: Substitute the known values into the kinetic energy equation
Substituting the known values into the equation:
\[ 16 = \frac{1}{2} \times 5.12 \times \omega^2 \times (0.25)^2 \]
Calculating \( (0.25)^2 \):
\[ (0.25)^2 = 0.0625 \]
Now substituting this back into the equation:
\[ 16 = \frac{1}{2} \times 5.12 \times \omega^2 \times 0.0625 \]

### Step 4: Simplify the equation
First, calculate \( \frac{1}{2} \times 5.12 \times 0.0625 \):
\[ \frac{1}{2} \times 5.12 = 2.56 \]
\[ 2.56 \times 0.0625 = 0.16 \]
Now, we have:
\[ 16 = 0.16 \omega^2 \]

### Step 5: Solve for \( \omega^2 \)
Rearranging the equation to solve for \( \omega^2 \):
\[ \omega^2 = \frac{16}{0.16} = 100 \]

### Step 6: Calculate \( \omega \)
Taking the square root of both sides:
\[ \omega = \sqrt{100} = 10 \text{ rad/s} \]

### Step 7: Calculate the time period \( T \)
The time period \( T \) is related to \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{10} = \frac{\pi}{5} \text{ seconds} \]

### Final Answer
The time period of the oscillation is:
\[ T = \frac{\pi}{5} \text{ seconds} \]

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