The displacement of a particle executing simple harmonic motion is given by `y=A_(0)+A sin omegat+B cos omegat`. Then the amplitude of its oscillation is given by

Options

A. `A+B`
B. `A_(0)+sqrt(A^(2)+B^(2))`
C. `sqrt(A^(2)+B^(2))`
D. `sqrt(A_(0)^(2)+(A+B)^(2))`

Explanation

To solve the problem, we need to find the amplitude of the oscillation given the displacement equation of a particle executing simple harmonic motion (SHM):

\[ y = A_0 + A \sin(\omega t) + B \cos(\omega t) \]

### Step-by-Step Solution:

1. **Identify the Terms**:
The displacement equation consists of three terms: a constant term \( A_0 \), a sine term \( A \sin(\omega t) \), and a cosine term \( B \cos(\omega t) \).

2. **Rewrite the Cosine Term**:
We can express the cosine term in terms of sine to combine the sine and cosine components. The cosine function can be rewritten as:
\[ B \cos(\omega t) = B \sin\left(\omega t + \frac{\pi}{2}\right) \]
This allows us to express both components in terms of sine.

3. **Use the Principle of Superposition**:
The total displacement can be viewed as the superposition of two waves:
\[ y = A_0 + A \sin(\omega t) + B \sin\left(\omega t + \frac{\pi}{2}\right) \]
This indicates that we have two oscillatory components.

4. **Calculate the Resultant Amplitude**:
The resultant amplitude \( R \) of the two sine components can be calculated using the formula:
\[ R = \sqrt{A^2 + B^2} \]
This formula arises from the Pythagorean theorem, considering the two components as perpendicular vectors in a phase space.

5. **Final Amplitude Expression**:
The amplitude of the oscillation is given by the resultant amplitude \( R \):
\[ R = \sqrt{A^2 + B^2} \]
The constant term \( A_0 \) does not affect the amplitude of the oscillation but represents a shift in the equilibrium position.

### Conclusion:
Thus, the amplitude of the oscillation is:
\[ \text{Amplitude} = \sqrt{A^2 + B^2} \]