The instantaneous displacement `x` of a particle executing simple harmonic motion is given by `x=a_1sinomegat+a_2cos(omegat+(pi)/(6))`. The amplitude `A` of oscillation is given by

Options

A. `sqrt(a_1^2+a_2^2+2a_1a_2cos((pi)/(6)))`
B. `sqrt(a_1^2+a_2^2+2a_1a_2cos((pi)/(3)))`
C. `sqrt(a_1^2+a_2^2-2a_1a_2cos((pi)/(6)))`
D. `sqrt(a_1^2+a_2^2-2a_1a_2cos((pi)/(3)))`

Explanation

To find the amplitude \( A \) of the oscillation given the instantaneous displacement \( x \) of a particle executing simple harmonic motion as

\[ x = A_1 \sin(\omega t) + A_2 \cos\left(\omega t + \frac{\pi}{6}\right), \]

we can follow these steps:

### Step 1: Rewrite the cosine term
First, we can rewrite the cosine term using the angle addition formula:

\[ \cos\left(\omega t + \frac{\pi}{6}\right) = \cos(\omega t)\cos\left(\frac{\pi}{6}\right) - \sin(\omega t)\sin\left(\frac{\pi}{6}\right). \]

Substituting the values of \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \) and \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \), we get:

\[ \cos\left(\omega t + \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \cos(\omega t) - \frac{1}{2} \sin(\omega t). \]

### Step 2: Substitute back into the equation
Now substituting this back into the original equation for \( x \):

\[ x = A_1 \sin(\omega t) + A_2\left(\frac{\sqrt{3}}{2} \cos(\omega t) - \frac{1}{2} \sin(\omega t)\right). \]

This simplifies to:

\[ x = \left(A_1 - \frac{A_2}{2}\right) \sin(\omega t) + \frac{\sqrt{3}}{2} A_2 \cos(\omega t). \]

### Step 3: Identify coefficients
Let:

\[ B = A_1 - \frac{A_2}{2} \quad \text{(coefficient of } \sin(\omega t)\text{)} \]
\[ C = \frac{\sqrt{3}}{2} A_2 \quad \text{(coefficient of } \cos(\omega t)\text{)}. \]

### Step 4: Calculate the amplitude
The amplitude \( A \) of the oscillation can be found using the formula:

\[ A = \sqrt{B^2 + C^2}. \]

Substituting the values of \( B \) and \( C \):

\[ A = \sqrt{\left(A_1 - \frac{A_2}{2}\right)^2 + \left(\frac{\sqrt{3}}{2} A_2\right)^2}. \]

### Step 5: Simplify the expression
Now we can simplify this expression:

\[ A = \sqrt{\left(A_1 - \frac{A_2}{2}\right)^2 + \frac{3}{4} A_2^2}. \]

### Final Result
Thus, the amplitude \( A \) of the oscillation is given by:

\[ A = \sqrt{\left(A_1 - \frac{A_2}{2}\right)^2 + \frac{3}{4} A_2^2}. \]