The amplitude of a particle executing SHM about `O` is `10 cm`. Then

Options

A. a. when the `KE` is `0.64` times of its maximum`KE`, its displacement is `6cm` from `O`
B. b. its speed is half the maximum speed when its displacement is half the maximum displacement
C. c. Both (a) and (b) are correct
D. d. Both (a) and (b) are wrong

Explanation

To solve the problem step by step, we need to analyze the given information and apply the formulas related to Simple Harmonic Motion (SHM).

### Step 1: Understand the Given Information
We are given that the amplitude \( A \) of a particle executing SHM is \( 10 \, \text{cm} \).

### Step 2: Recall the Formula for Kinetic Energy in SHM
The kinetic energy \( KE \) of a particle in SHM is given by the formula:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - y^2) \]
where:
- \( m \) is the mass of the particle,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude,
- \( y \) is the displacement from the mean position.

### Step 3: Determine the Maximum Kinetic Energy
The maximum kinetic energy \( KE_{max} \) occurs when the displacement \( y = 0 \):
\[ KE_{max} = \frac{1}{2} m \omega^2 A^2 \]

### Step 4: Set Up the Equation for Kinetic Energy
According to the problem, the kinetic energy is \( 0.64 \) times the maximum kinetic energy:
\[ KE = 0.64 \times KE_{max} \]
Substituting the expressions for kinetic energy, we have:
\[ \frac{1}{2} m \omega^2 (A^2 - y^2) = 0.64 \times \frac{1}{2} m \omega^2 A^2 \]

### Step 5: Simplify the Equation
We can cancel \( \frac{1}{2} m \omega^2 \) from both sides (assuming \( m \neq 0 \) and \( \omega \neq 0 \)):
\[ A^2 - y^2 = 0.64 A^2 \]
Rearranging gives:
\[ y^2 = A^2 - 0.64 A^2 = 0.36 A^2 \]

### Step 6: Solve for Displacement \( y \)
Taking the square root of both sides:
\[ y = \sqrt{0.36} A = 0.6 A \]
Given that \( A = 10 \, \text{cm} \):
\[ y = 0.6 \times 10 \, \text{cm} = 6 \, \text{cm} \]

### Conclusion
The displacement \( y \) when the kinetic energy is \( 0.64 \) times the maximum kinetic energy is \( 6 \, \text{cm} \).