The equation describing the motion of a simple harmonic oscillator along the X axis is given as `x = A cos (omega t + phi)`. If at time t = 0, the oscillator is at x = 0 and moving in the negative x direction, then the phase angle `phi` is

Options

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

Explanation

To find the phase angle \(\phi\) for the simple harmonic oscillator described by the equation \(x = A \cos(\omega t + \phi)\), we will follow these steps:

### Step 1: Substitute the initial conditions
At time \(t = 0\), we know that \(x = 0\). Therefore, we can substitute these values into the equation:
\[ 0 = A \cos(\omega \cdot 0 + \phi) \]
This simplifies to:
\[ 0 = A \cos(\phi) \]

### Step 2: Analyze the equation
Since \(A\) (the amplitude) is not zero, we can conclude that:
\[ \cos(\phi) = 0 \]
The cosine function is equal to zero at specific angles:
\[ \phi = \frac{\pi}{2} + n\pi \quad (n \text{ is an integer}) \]
This means \(\phi\) could be \(\frac{\pi}{2}\) or \(\frac{3\pi}{2}\) (or any odd multiple of \(\frac{\pi}{2}\)).

### Step 3: Determine the direction of motion
We are also given that the oscillator is moving in the negative x-direction at \(t = 0\). To find the velocity, we differentiate the position function with respect to time:
\[ v = \frac{dx}{dt} = -A \omega \sin(\omega t + \phi) \]
At \(t = 0\), this becomes:
\[ v = -A \omega \sin(\phi) \]
Since the oscillator is moving in the negative x-direction, we have:
\[ v < 0 \implies -A \omega \sin(\phi) < 0 \]
This implies:
\[ \sin(\phi) > 0 \]

### Step 4: Determine the correct phase angle
From the values of \(\phi\) we found earlier:
- \(\phi = \frac{\pi}{2}\) gives \(\sin(\phi) = 1\) (which is positive).
- \(\phi = \frac{3\pi}{2}\) gives \(\sin(\phi) = -1\) (which is negative).

Since we need \(\sin(\phi) > 0\), the only valid solution is:
\[ \phi = \frac{\pi}{2} \]

### Final Answer
Thus, the phase angle \(\phi\) is:
\[ \phi = \frac{\pi}{2} \]

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