The displacement of a particle varies with time according to the relation `y=asinomegat+bcosomegat`.

Options

A. The motion is oscillatory but not SHM.
B. The motion is SHM with amplitude `a+b`.
C. The motion is SHM with amplitude `a^(2)+b^(2)`
D. The motion is SHM with amplitude `sqrt(a^(2)+b^(2))`.

Explanation

To solve the problem, we need to analyze the given equation of motion and determine whether it represents simple harmonic motion (SHM) and, if so, find the amplitude.

### Step-by-Step Solution:

1. **Identify the given equation**:
The displacement of the particle is given by:
\[ y = a \sin(\omega t) + b \cos(\omega t) \]

2. **Rearranging the equation**:
We can express the equation in a form that resembles the standard equation of SHM. To do this, we will use the trigonometric identity for the sine of a sum:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

3. **Using the Pythagorean theorem**:
We can visualize the coefficients \(a\) and \(b\) as the sides of a right triangle. The hypotenuse \(R\) can be calculated as:
\[ R = \sqrt{a^2 + b^2} \]

4. **Finding the angles**:
We can define angles using the definitions of sine and cosine:
\[ \cos \theta = \frac{a}{R} \quad \text{and} \quad \sin \theta = \frac{b}{R} \]

5. **Substituting back into the equation**:
Now, we can rewrite the original equation:
\[ y = R \left( \sin(\omega t) \cos \theta + \cos(\omega t) \sin \theta \right) \]
This simplifies to:
\[ y = R \sin(\omega t + \theta) \]

6. **Identifying SHM**:
The equation \(y = R \sin(\omega t + \theta)\) is in the standard form of SHM, where:
- \(R\) is the amplitude, and
- \(\theta\) is the phase constant.

7. **Conclusion**:
From the analysis, we find that the motion is indeed SHM with an amplitude given by:
\[ \text{Amplitude} = \sqrt{a^2 + b^2} \]

### Final Answer:
The motion is SHM with an amplitude of \(\sqrt{a^2 + b^2}\).

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