A point mass oscillates along the x-axis according to the law `x=x_(0) cos(moegat-pi//4). If the acceleration of the particle is written as `a=A cos(omegat+delta), the .

Options

A. (a) `A=x_(0)omega^(2),delta=3pi//4`
B. (b) `A=x_(0),delta=-pi//4`
C. © `A=x_(0)omega^(2), delta=pi//4`
D. (d)`A=x_(0)omega^(2), delta=-pi//4`

Explanation

To solve the problem, we need to find the values of \( A \) and \( \delta \) from the given displacement equation of a point mass oscillating along the x-axis:

\[ x = x_0 \cos(\omega t - \frac{\pi}{4}) \]

### Step 1: Differentiate the displacement to find velocity
To find the velocity \( v \), we differentiate the displacement \( x \) with respect to time \( t \):

\[ v = \frac{dx}{dt} = -x_0 \omega \sin(\omega t - \frac{\pi}{4}) \]

### Step 2: Differentiate the velocity to find acceleration
Next, we differentiate the velocity \( v \) to find the acceleration \( a \):

\[ a = \frac{dv}{dt} = -x_0 \omega \cos(\omega t - \frac{\pi}{4}) \cdot \omega \]

This simplifies to:

\[ a = -x_0 \omega^2 \cos(\omega t - \frac{\pi}{4}) \]

### Step 3: Rewrite the acceleration in the standard form
We need to express the acceleration in the form \( a = A \cos(\omega t + \delta) \). To do this, we can use the identity for cosine:

\[ -\cos(\theta) = \cos(\theta + \pi) \]

Thus, we can rewrite the acceleration as:

\[ a = x_0 \omega^2 \cos(\omega t - \frac{\pi}{4} + \pi) = x_0 \omega^2 \cos(\omega t + \frac{3\pi}{4}) \]

### Step 4: Identify \( A \) and \( \delta \)
Now, we can compare this with the standard form \( a = A \cos(\omega t + \delta) \):

- From the equation, we see that \( A = x_0 \omega^2 \)
- Also, \( \delta = \frac{3\pi}{4} \)

### Final Answer
Thus, the values are:

\[ A = x_0 \omega^2 \quad \text{and} \quad \delta = \frac{3\pi}{4} \]

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