In SHM , the acceleration is ahead of velocity by a phase angle

Options

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

Explanation

To solve the question regarding the phase relationship between acceleration and velocity in Simple Harmonic Motion (SHM), we can follow these steps:

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

### Step 2: Find the expression for velocity
The velocity \( v \) is the derivative of displacement with respect to time:
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(A \sin(\omega t + \phi)) = A \omega \cos(\omega t + \phi) \]

### Step 3: Find the expression for acceleration
The acceleration \( a \) is the derivative of velocity with respect to time:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(A \omega \cos(\omega t + \phi)) = -A \omega^2 \sin(\omega t + \phi) \]

### Step 4: Analyze the phase relationship
Now we have:
- Velocity: \( v(t) = A \omega \cos(\omega t + \phi) \)
- Acceleration: \( a(t) = -A \omega^2 \sin(\omega t + \phi) \)

To express both in terms of sine, we can use the identity \( \cos(\theta) = \sin(\theta + \frac{\pi}{2}) \):
\[ v(t) = A \omega \sin\left(\omega t + \phi + \frac{\pi}{2}\right) \]

### Step 5: Determine the phase difference
From the equations:
- The phase of velocity \( \phi_v = \omega t + \phi + \frac{\pi}{2} \)
- The phase of acceleration \( \phi_a = \omega t + \phi \)

The phase difference \( \Delta \phi \) between acceleration and velocity is:
\[ \Delta \phi = \phi_a - \phi_v = \left(\omega t + \phi\right) - \left(\omega t + \phi + \frac{\pi}{2}\right) = -\frac{\pi}{2} \]

This indicates that the acceleration is ahead of the velocity by \( \frac{\pi}{2} \) radians or 90 degrees.

### Conclusion
Thus, in Simple Harmonic Motion, the acceleration is ahead of the velocity by a phase angle of \( \frac{\pi}{2} \) radians.

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