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=Acos(omegat+delta)`, the .

Options

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

Explanation

To solve the problem, we need to find the acceleration of a point mass oscillating along the x-axis given by the equation:

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

We want to express the acceleration in the form:

\[ a = A \cos(\omega t + \delta) \]

### Step-by-Step Solution:

1. **Identify the displacement function**:
The displacement of the particle is given by:
\[ x = x_0 \cos(\omega t - \frac{\pi}{4}) \]

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

3. **Differentiate to find acceleration**:
Now, we differentiate the velocity \( v \) to find the acceleration \( a \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(-x_0 \omega \sin(\omega t - \frac{\pi}{4})) \]
Again, using the chain rule:
\[ a = -x_0 \omega^2 \cos(\omega t - \frac{\pi}{4}) \]

4. **Rewrite the acceleration**:
We can rewrite the cosine term using the angle addition formula:
\[ a = -x_0 \omega^2 \cos\left(\omega t - \frac{\pi}{4}\right) = x_0 \omega^2 \cos\left(\omega t + \frac{3\pi}{4}\right) \]
This is because:
\[ \cos(\theta) = -\cos(\theta + \pi) \]

5. **Identify A and δ**:
Now, we can compare this with the given form \( a = A \cos(\omega t + \delta) \):
\[ A = x_0 \omega^2 \quad \text{and} \quad \delta = \frac{3\pi}{4} \]

6. **Final result**:
Therefore, we have:
\[ a = x_0 \omega^2 \cos(\omega t + \frac{3\pi}{4}) \]

### Conclusion:
From the above steps, we conclude that:
- \( A = x_0 \omega^2 \)
- \( \delta = \frac{3\pi}{4} \)

The correct option is:
- \( a = x_0 \omega^2 \) and \( \delta = \frac{3\pi}{4} \)

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