The kinetic energy of a particle, executing S.H.M. is 16 J when it is in its mean position. IF the amplitude of oscillation is 25 cm and the mass of the particle is 5.12 Kg, the time period of its oscillations is

Options

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

Explanation

To solve the problem step by step, we will use the formulas related to kinetic energy in simple harmonic motion (SHM) and the relationship between angular frequency and time period.

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

### Step 2: Write down the formula for kinetic energy in SHM
The kinetic energy (KE) of a particle in SHM at the mean position is given by the formula:
\[ KE = \frac{1}{2} m \omega^2 A^2 \]
where:
- \( \omega \) is the angular frequency in radians per second,
- \( A \) is the amplitude in meters.

### Step 3: Substitute the known values into the kinetic energy formula
Substituting the known values into the formula:
\[ 16 = \frac{1}{2} \times 5.12 \times \omega^2 \times (0.25)^2 \]

### Step 4: Simplify the equation
Calculate \( (0.25)^2 \):
\[ (0.25)^2 = 0.0625 \]
Now substitute this value into the equation:
\[ 16 = \frac{1}{2} \times 5.12 \times \omega^2 \times 0.0625 \]

### Step 5: Rearrange the equation to solve for \( \omega^2 \)
Multiply both sides by 2 to eliminate the fraction:
\[ 32 = 5.12 \times \omega^2 \times 0.0625 \]
Now divide both sides by \( (5.12 \times 0.0625) \):
\[ \omega^2 = \frac{32}{5.12 \times 0.0625} \]

### Step 6: Calculate \( \omega^2 \)
Calculate \( 5.12 \times 0.0625 \):
\[ 5.12 \times 0.0625 = 0.32 \]
Now substitute this back into the equation:
\[ \omega^2 = \frac{32}{0.32} = 100 \]

### Step 7: Find \( \omega \)
Take the square root of both sides to find \( \omega \):
\[ \omega = \sqrt{100} = 10 \, \text{rad/s} \]

### Step 8: Calculate the time period \( T \)
The time period \( T \) is related to angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \omega = 10 \):
\[ T = \frac{2\pi}{10} = \frac{\pi}{5} \, \text{seconds} \]

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

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