A particle is executing SHM with amplitude A. At displacement `x=(-A/4)`, force acting on the particle is F, potential energy of the particle is U, velocity of particle is v and kinetic energy is K. Assuming potential energyh to be zero at mean position. At displacement x=A/2

Options

A. Force acting on the particle will be 2F
B. potential energy of the particle will be 4U.
C. velocity of particle must be `sqrt(4/5)`v
D. kinetic energy of particle will be 0.8 K

Answer

To solve the problem step by step, we will analyze the motion of a particle executing Simple Harmonic Motion (SHM) at two different displacements: \( x = -\frac{A}{4} \) and \( x = \frac{A}{2} \). We will calculate the force, potential energy, velocity, and kinetic energy at each position.

### Step 1: Analyze the particle at \( x = -\frac{A}{4} \)

1. **Force (F)**:
The force acting on a particle in SHM is given by:
\[ F = -k x \]
At \( x = -\frac{A}{4} \):
\[ F = -k \left(-\frac{A}{4}\right) = \frac{kA}{4} \]

2. **Potential Energy (U)**:
The potential energy in SHM is given by:
\[ U = \frac{1}{2} k x^2 \]
Substituting \( x = -\frac{A}{4} \):
\[ U = \frac{1}{2} k \left(-\frac{A}{4}\right)^2 = \frac{1}{2} k \frac{A^2}{16} = \frac{kA^2}{32} \]

3. **Velocity (v)**:
The velocity of the particle can be calculated using the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
Here, \( \omega \) is the angular frequency. Substituting \( x = -\frac{A}{4} \):
\[ v = \omega \sqrt{A^2 - \left(-\frac{A}{4}\right)^2} = \omega \sqrt{A^2 - \frac{A^2}{16}} = \omega \sqrt{\frac{15A^2}{16}} = \frac{\sqrt{15}}{4} \omega A \]

4. **Kinetic Energy (K)**:
The kinetic energy is given by:
\[ K = \frac{1}{2} m v^2 \]
Substituting for \( v \):
\[ K = \frac{1}{2} m \left(\frac{\sqrt{15}}{4} \omega A\right)^2 = \frac{15}{32} m \omega^2 A^2 \]

### Step 2: Analyze the particle at \( x = \frac{A}{2} \)

1. **Force (F')**:
At \( x = \frac{A}{2} \):
\[ F' = -k \left(\frac{A}{2}\right) = -\frac{kA}{2} \]

2. **Potential Energy (U')**:
At \( x = \frac{A}{2} \):
\[ U' = \frac{1}{2} k \left(\frac{A}{2}\right)^2 = \frac{1}{2} k \frac{A^2}{4} = \frac{kA^2}{8} \]

3. **Velocity (v')**:
At \( x = \frac{A}{2} \):
\[ v' = \omega \sqrt{A^2 - \left(\frac{A}{2}\right)^2} = \omega \sqrt{A^2 - \frac{A^2}{4}} = \omega \sqrt{\frac{3A^2}{4}} = \frac{\sqrt{3}}{2} \omega A \]

4. **Kinetic Energy (K')**:
\[ K' = \frac{1}{2} m v'^2 = \frac{1}{2} m \left(\frac{\sqrt{3}}{2} \omega A\right)^2 = \frac{3}{8} m \omega^2 A^2 \]

### Summary of Results:
- At \( x = -\frac{A}{4} \):
- Force: \( F = \frac{kA}{4} \)
- Potential Energy: \( U = \frac{kA^2}{32} \)
- Velocity: \( v = \frac{\sqrt{15}}{4} \omega A \)
- Kinetic Energy: \( K = \frac{15}{32} m \omega^2 A^2 \)

- At \( x = \frac{A}{2} \):
- Force: \( F' = -\frac{kA}{2} \)
- Potential Energy: \( U' = \frac{kA^2}{8} \)
- Velocity: \( v' = \frac{\sqrt{3}}{2} \omega A \)
- Kinetic Energy: \( K' = \frac{3}{8} m \omega^2 A^2 \)