The displacement of a particle executing a S.H.M. is given by x= A "sin" `omega t +A "cos" omega t`. What is the amplitude of motion ?

Options

A. `2A`
B. `sqrt(2) A`
C. `sqrt(3) A`
D. A

Explanation

To find the amplitude of the motion given the displacement of a particle executing simple harmonic motion (S.H.M.), we start with the expression for displacement:

\[ x = A \sin(\omega t) + A \cos(\omega t) \]

### Step 1: Rewrite the Displacement
We can rewrite the displacement in a more manageable form. We know that:

\[ \sin(\omega t) \text{ and } \cos(\omega t) \]

can be combined using the Pythagorean identity. To do this, we can express the sine and cosine terms as components of a vector.

### Step 2: Identify the Components
The displacement can be viewed as the sum of two perpendicular components:

- One component along the sine direction (y-axis)
- One component along the cosine direction (x-axis)

### Step 3: Use the Pythagorean Theorem
The resultant amplitude \( A_r \) can be found using the Pythagorean theorem:

\[ A_r = \sqrt{(A \cos(\omega t))^2 + (A \sin(\omega t))^2} \]

### Step 4: Substitute the Values
Substituting the values into the equation gives:

\[ A_r = \sqrt{A^2 + A^2} \]

### Step 5: Simplify the Expression
This simplifies to:

\[ A_r = \sqrt{2A^2} \]

### Step 6: Final Calculation
Taking the square root:

\[ A_r = A \sqrt{2} \]

### Conclusion
Thus, the amplitude of the motion is:

\[ A_r = A \sqrt{2} \]