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

Options

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

Explanation

To solve the problem step by step, we need to derive the expressions for velocity and acceleration from the given position function and then match the form of acceleration to find the values of \( A \) and \( \delta \).

### Step 1: Write down the position function
The position of the particle is given by:
\[ x = x_0 \cos(\omega t - \frac{\pi}{4}) \]

### Step 2: Differentiate the position function to find velocity
To find the velocity \( V \), we differentiate the position function \( x \) with respect to time \( t \):
\[ V = \frac{dx}{dt} = \frac{d}{dt}[x_0 \cos(\omega t - \frac{\pi}{4})] \]
Using the chain rule, we get:
\[ V = -x_0 \omega \sin(\omega t - \frac{\pi}{4}) \]

### Step 3: Differentiate the velocity function to find acceleration
Next, we differentiate the velocity function \( V \) to find the acceleration \( a \):
\[ a = \frac{dV}{dt} = \frac{d}{dt}[-x_0 \omega \sin(\omega t - \frac{\pi}{4})] \]
Again applying the chain rule:
\[ a = -x_0 \omega^2 \cos(\omega t - \frac{\pi}{4}) \]

### Step 4: Rewrite the acceleration in the desired form
The problem states that the acceleration can be expressed as:
\[ a = A \cos(\omega t + \delta) \]
We need to rewrite our expression for acceleration to match this form. We have:
\[ a = -x_0 \omega^2 \cos(\omega t - \frac{\pi}{4}) \]
Using the cosine angle subtraction identity, we can rewrite this as:
\[ a = -x_0 \omega^2 \cos\left(\omega t + \left(-\frac{\pi}{4}\right)\right) \]
This can be expressed as:
\[ a = x_0 \omega^2 \cos\left(\omega t + \left(\pi - \frac{\pi}{4}\right)\right) \]
This simplifies to:
\[ a = x_0 \omega^2 \cos\left(\omega t + \frac{3\pi}{4}\right) \]

### Step 5: Identify \( A \) and \( \delta \)
From the expression \( a = A \cos(\omega t + \delta) \), we can identify:
\[ A = x_0 \omega^2 \]
\[ \delta = \frac{3\pi}{4} \]

### Final Answer
Thus, the values are:
\[ A = x_0 \omega^2, \quad \delta = \frac{3\pi}{4} \]