The displacement of a particle varies with time according to the relation `y=a sin omega t +b cos omega t `.

Options

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

Explanation

To solve the problem, we need to analyze the given displacement equation of the particle:

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

**Step 2: Recognize the form of the equation**
This equation can be recognized as a combination of sine and cosine functions, which suggests that it may represent Simple Harmonic Motion (SHM).

**Step 3: Rewrite the equation in a single sine or cosine form**
To express \( y \) in a single trigonometric function, we can use the following trigonometric identity:
\[ R \sin(\omega t + \phi) = R \sin(\omega t) \cos(\phi) + R \cos(\omega t) \sin(\phi) \]
where \( R \) is the amplitude and \( \phi \) is the phase angle.

**Step 4: Identify the coefficients**
From the equation:
- Let \( A = b \) (coefficient of \( \cos(\omega t) \))
- Let \( B = a \) (coefficient of \( \sin(\omega t) \))

**Step 5: Calculate the amplitude**
The amplitude \( R \) can be calculated using the formula:
\[ R = \sqrt{A^2 + B^2} \]
Substituting \( A \) and \( B \):
\[ R = \sqrt{b^2 + a^2} \]

**Step 6: Conclusion**
Thus, the motion is confirmed to be Simple Harmonic Motion (SHM), and the amplitude of the motion is:
\[ R = \sqrt{a^2 + b^2} \]

**Final Answer:**
The amplitude of the motion is \( \sqrt{a^2 + b^2} \).

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