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

Options

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

Answer

To solve the problem of determining the range of the phase constant \( \phi \) for an object oscillating on a spring with given initial conditions, we can follow these steps:

### Step 1: Understand the displacement equation
The displacement of the object is given by:
\[ x(t) = x_m \cos(\omega t + \phi) \]
where \( x_m \) is the maximum displacement (amplitude), \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

### Step 2: Analyze the initial conditions
We know that:
- The object is initially displaced in the negative x direction, meaning \( x(0) < 0 \).
- The object has a negative initial velocity, meaning \( v(0) < 0 \).

### Step 3: Evaluate the displacement at \( t = 0 \)
At \( t = 0 \):
\[ x(0) = x_m \cos(\phi) \]
Since \( x(0) < 0 \), it follows that:
\[ \cos(\phi) < 0 \]
This implies that \( \phi \) must be in the second or third quadrant (where cosine is negative).

### Step 4: Evaluate the velocity at \( t = 0 \)
The velocity is given by the derivative of displacement:
\[ v(t) = \frac{dx}{dt} = -x_m \omega \sin(\omega t + \phi) \]
At \( t = 0 \):
\[ v(0) = -x_m \omega \sin(\phi) \]
Since \( v(0) < 0 \), it follows that:
\[ -\sin(\phi) < 0 \quad \Rightarrow \quad \sin(\phi) > 0 \]
This implies that \( \phi \) must be in the first or second quadrant (where sine is positive).

### Step 5: Combine the results
From the analysis:
- From the displacement condition, \( \phi \) is in the second or third quadrant.
- From the velocity condition, \( \phi \) is in the first or second quadrant.

The only quadrant that satisfies both conditions is the second quadrant. Therefore, we conclude:
\[ 90^\circ < \phi < 180^\circ \quad \text{or} \quad \frac{\pi}{2} < \phi < \pi \]

### Final Answer
Thus, the phase constant \( \phi \) is between:
\[ \frac{\pi}{2} \text{ and } \pi \]

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