A particle executing SHM. The phase difference between velocity and displacement is

Options

A. 0
B. `(pi)/(2)`
C. `pi`
D. `2pi`

Explanation

To solve the problem of finding the phase difference between velocity and displacement in simple harmonic motion (SHM), we can follow these steps:

### Step 1: Understand the equations of SHM
In SHM, the displacement \( x \) of a particle can be expressed as:
\[ x(t) = A \cos(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

### Step 2: Derive the expression for velocity
The velocity \( v \) of the particle is the time derivative of displacement \( x \):
\[ v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi) \]
This can be rewritten as:
\[ v(t) = -A \omega \sin(\omega t + \phi) \]

### Step 3: Identify the phase of displacement and velocity
From the equations:
- Displacement \( x(t) = A \cos(\omega t + \phi) \) has a phase of \( \omega t + \phi \).
- Velocity \( v(t) = -A \omega \sin(\omega t + \phi) \) can be rewritten using the identity \( \sin(\theta) = \cos(\theta - \frac{\pi}{2}) \):
\[ v(t) = -A \omega \cos\left(\omega t + \phi - \frac{\pi}{2}\right) \]
This shows that the phase of the velocity is \( \omega t + \phi - \frac{\pi}{2} \).

### Step 4: Calculate the phase difference
The phase difference \( \Delta \phi \) between velocity and displacement is given by:
\[ \Delta \phi = \text{Phase of velocity} - \text{Phase of displacement} \]
Substituting the phases:
\[ \Delta \phi = \left(\omega t + \phi - \frac{\pi}{2}\right) - (\omega t + \phi) = -\frac{\pi}{2} \]
Since phase difference is often expressed as a positive angle, we can write:
\[ \Delta \phi = \frac{\pi}{2} \]

### Conclusion
The phase difference between velocity and displacement in SHM is \( \frac{\pi}{2} \).