A particle of mass `m` excuting `SHM` make `f` oscillation per second. The difference of its kinetic energy when at the centre, and when at distance x from the centre is

Options

A. `pi^(2)f^(2)x^(2)m`
B. `2pi^(2)f^(2)x^(2)m`
C. `(1)/(2)pi^(2)f^(2)x^(2)m`
D. `f^(2)x^(2)m`

Explanation

To solve the problem, we need to find the difference in kinetic energy of a particle executing simple harmonic motion (SHM) when it is at the center and when it is at a distance \( x \) from the center.

### Step-by-Step Solution:

1. **Understanding Kinetic Energy at the Center:**
- When the particle is at the center of its motion (equilibrium position), it has maximum kinetic energy. The kinetic energy \( KE_1 \) at the center can be expressed as:
\[ KE_1 = \frac{1}{2} m v^2 \]
- Here, \( v \) is the velocity of the particle at the center. The velocity can be related to the angular frequency \( \omega \) and amplitude \( A \) of the motion.

2. **Relating Velocity to Angular Frequency:**
- The angular frequency \( \omega \) is given by:
\[ \omega = 2 \pi f \]
- The maximum velocity \( v_{\text{max}} \) at the center is:
\[ v_{\text{max}} = \omega A \]
- Therefore, substituting this into the kinetic energy expression, we get:
\[ KE_1 = \frac{1}{2} m (\omega A)^2 = \frac{1}{2} m \omega^2 A^2 \]

3. **Understanding Kinetic Energy at Distance \( x \):**
- When the particle is at a distance \( x \) from the center, its kinetic energy \( KE_2 \) is given by:
\[ KE_2 = \frac{1}{2} m v^2 \]
- The velocity at this position can be found using the conservation of energy principle. The total mechanical energy \( E \) in SHM is constant and is given by:
\[ E = KE + PE \]
- The potential energy \( PE \) at a distance \( x \) is:
\[ PE = \frac{1}{2} k x^2 \]
- The total energy at maximum displacement (amplitude) is:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
- Thus, at position \( x \):
\[ KE_2 = E - PE = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} k x^2 \]
- Since \( k = m \omega^2 \):
\[ KE_2 = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 (A^2 - x^2) \]

4. **Finding the Difference in Kinetic Energies:**
- Now, we need to find the difference between the kinetic energies:
\[ \Delta KE = KE_1 - KE_2 = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 (A^2 - x^2) \]
- Simplifying this gives:
\[ \Delta KE = \frac{1}{2} m \omega^2 x^2 \]

5. **Substituting for Angular Frequency:**
- Substitute \( \omega = 2 \pi f \):
\[ \Delta KE = \frac{1}{2} m (2 \pi f)^2 x^2 = 2 m \pi^2 f^2 x^2 \]

### Final Answer:
The difference in kinetic energy when the particle is at the center and at a distance \( x \) from the center is:
\[ \Delta KE = 2 m \pi^2 f^2 x^2 \]