A particle is executing SHM, whose equation is given below
`x=A cos (omegat+phi)`. If t=0 , particle is mean position going towards positive x-direction then `phi` is equal to

Options

A. 0
B. `pi//2`
C. `pi`
D. noone of the above

Answer

To find the value of \(\phi\) in the equation of motion for a particle executing Simple Harmonic Motion (SHM), we start with the given equation:

\[ x = A \cos(\omega t + \phi) \]

### Step 1: Analyze the condition at \(t = 0\)

At \(t = 0\), the position of the particle is:

\[ x = A \cos(\phi) \]

Given that the particle is at the mean position (which is \(x = 0\)) and moving towards the positive x-direction, we can set up the equation:

\[ 0 = A \cos(\phi) \]

### Step 2: Solve for \(\phi\)

Since \(A\) (the amplitude) is not zero, we can divide both sides by \(A\):

\[ \cos(\phi) = 0 \]

The cosine function equals zero at specific angles:

\[ \phi = \frac{\pi}{2} + n\pi \quad \text{where } n \text{ is an integer} \]

### Step 3: Determine the correct value of \(\phi\)

To find the correct value of \(\phi\) that matches the condition of the particle moving towards the positive x-direction, we need to consider the velocity of the particle. The velocity \(v\) in SHM is given by:

\[ v = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi) \]

At \(t = 0\):

\[ v = -A \omega \sin(\phi) \]

Since the particle is moving towards the positive x-direction, the velocity must be positive:

\[ -A \omega \sin(\phi) > 0 \]

This implies:

\[ \sin(\phi) < 0 \]

### Step 4: Find the correct \(\phi\)

From the condition \(\cos(\phi) = 0\), we have:

- \(\phi = \frac{\pi}{2}\) (which gives \(\sin(\phi) = 1\), not valid since it is not less than 0)
- \(\phi = \frac{3\pi}{2}\) (which gives \(\sin(\phi) = -1\), valid since it is less than 0)

Thus, the correct value of \(\phi\) that satisfies both conditions is:

\[ \phi = \frac{3\pi}{2} \]

### Final Answer

\(\phi = \frac{3\pi}{2}\)

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