At two particular closest instant of time `t_1` and `t_2` the displacements of a particle performing SHM are equal. At these instant

Options

A. instantaneaous speed are equal
B. instantaneous acceleration are equal
C. phase of the motion are unequal
D. kinetic energies are equal

Explanation

To solve the problem, we need to analyze the characteristics of a particle performing Simple Harmonic Motion (SHM) at two different instances, \( t_1 \) and \( t_2 \), where the displacements are equal. We will determine the instantaneous speed, instantaneous acceleration, phase of the motion, and kinetic energy at these two instances.

### Step-by-Step Solution:

1. **Understanding SHM Basics**:
- In SHM, the displacement \( x \) of a particle from its mean position can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
- Here, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

2. **Equal Displacements**:
- Given that at two times \( t_1 \) and \( t_2 \), the displacements are equal, we have:
\[ x(t_1) = x(t_2) \]
- This implies that:
\[ A \sin(\omega t_1 + \phi) = A \sin(\omega t_2 + \phi) \]
- Since \( A \) is non-zero, we can simplify this to:
\[ \sin(\omega t_1 + \phi) = \sin(\omega t_2 + \phi) \]

3. **Finding Instantaneous Speed**:
- The instantaneous speed \( v \) in SHM is given by the derivative of displacement:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]
- At \( t_1 \) and \( t_2 \), since the displacements are equal, we can analyze the speed:
\[ v(t_1) = A \omega \cos(\omega t_1 + \phi) \]
\[ v(t_2) = A \omega \cos(\omega t_2 + \phi) \]
- The values of \( \cos(\omega t_1 + \phi) \) and \( \cos(\omega t_2 + \phi) \) may not be equal, hence the instantaneous speeds may not be equal.

4. **Finding Instantaneous Acceleration**:
- The instantaneous acceleration \( a \) is given by:
\[ a(t) = \frac{d^2x}{dt^2} = -A \omega^2 \sin(\omega t + \phi) \]
- Since \( \sin(\omega t_1 + \phi) = \sin(\omega t_2 + \phi) \), we find that:
\[ a(t_1) = -A \omega^2 \sin(\omega t_1 + \phi) \]
\[ a(t_2) = -A \omega^2 \sin(\omega t_2 + \phi) \]
- Thus, the instantaneous accelerations are equal because they depend on the same sine value.

5. **Phase of the Motion**:
- The phase of the motion at \( t_1 \) and \( t_2 \) can be expressed as:
\[ \phi_1 = \omega t_1 + \phi \]
\[ \phi_2 = \omega t_2 + \phi \]
- Since \( t_1 \) and \( t_2 \) are different, \( \phi_1 \) and \( \phi_2 \) will also be different.

6. **Kinetic Energy**:
- The kinetic energy \( KE \) in SHM is given by:
\[ KE = \frac{1}{2} mv^2 \]
- Since the speeds at \( t_1 \) and \( t_2 \) are not necessarily equal, the kinetic energies may also differ.

### Summary of Results:
- **Instantaneous Speed**: Not equal.
- **Instantaneous Acceleration**: Equal.
- **Phase of Motion**: Not equal.
- **Kinetic Energy**: Not necessarily equal.