The displacement of an object oscillating on a spring is given by `x(t) = x_(m) cos (omega t+ phi)`. If the initial displacement is zero and the initial velocity is in the negative x direction. Then the phase constant `phi` is

Options

A. 0 rad
B. `(pi)/(2)` rad
C. `pi` rad
D. `(3pi)/(2)` rad

Answer

To solve the problem, we need to find the phase constant \( \phi \) given the conditions of the oscillating object. The displacement is described by the equation:

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

### Step 1: Analyze Initial Conditions
We are given that the initial displacement is zero. This means:

\[ x(0) = 0 \]

Substituting \( t = 0 \) into the displacement equation:

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

For this equation to hold true, \( \cos(\phi) \) must equal zero. Therefore, the possible values of \( \phi \) are:

\[ \phi = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

### Step 2: Analyze Initial Velocity
We are also given that the initial velocity is in the negative x direction. The velocity \( v(t) \) is the derivative of the displacement \( x(t) \):

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

At \( t = 0 \):

\[ v(0) = -x_m \omega \sin(\phi) \]

Since the initial velocity is negative, we have:

\[ v(0) < 0 \implies -x_m \omega \sin(\phi) < 0 \]

Given that \( x_m \) and \( \omega \) are positive constants, this implies:

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

### Step 3: Determine the Quadrant for \( \phi \)
From the conditions derived:

1. \( \cos(\phi) = 0 \) gives us \( \phi = \frac{\pi}{2} + n\pi \).
2. \( \sin(\phi) > 0 \) indicates that \( \phi \) must be in the first or second quadrant.

The only value that satisfies both conditions is:

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

### Conclusion
Thus, the phase constant \( \phi \) is:

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