A vibratory motion is represented by `x=2Acosomegat+Acos(omegat+(pi)/(2))+Acos(omegat+pi)+(A)/(2)cos(omegat+(3pi)/(2))`.
the resultant amplitude of the motion is

Options

A. `(9A)/(2)`
B. `(sqrt(5)A)/(2)`
C. `(5A)/(2)`
D. 2A

Explanation

To find the resultant amplitude of the given vibratory motion represented by

\[ x = 2A \cos(\omega t) + A \cos\left(\omega t + \frac{\pi}{2}\right) + A \cos(\omega t + \pi) + \frac{A}{2} \cos\left(\omega t + \frac{3\pi}{2}\right), \]

we can break down the expression step by step.

### Step 1: Rewrite the cosine terms using trigonometric identities
1. The term \( \cos\left(\omega t + \frac{\pi}{2}\right) \) can be rewritten as \( -\sin(\omega t) \).
2. The term \( \cos(\omega t + \pi) \) can be rewritten as \( -\cos(\omega t) \).
3. The term \( \cos\left(\omega t + \frac{3\pi}{2}\right) \) can be rewritten as \( \sin(\omega t) \).

Substituting these identities into the equation gives:

\[ x = 2A \cos(\omega t) + A(-\sin(\omega t)) + A(-\cos(\omega t)) + \frac{A}{2}\sin(\omega t) \]

### Step 2: Combine like terms
Now, combine the terms involving \( \cos(\omega t) \) and \( \sin(\omega t) \):

\[ x = (2A - A) \cos(\omega t) + \left(-A + \frac{A}{2}\right) \sin(\omega t) \]

This simplifies to:

\[ x = A \cos(\omega t) - \frac{A}{2} \sin(\omega t) \]

### Step 3: Express in terms of a single cosine function
To express this in a single cosine function, we can use the phasor method. We can represent the coefficients of \( \cos(\omega t) \) and \( \sin(\omega t) \) as vectors.

Let:
- \( A_1 = A \) (the coefficient of \( \cos(\omega t) \))
- \( A_2 = -\frac{A}{2} \) (the coefficient of \( \sin(\omega t) \))

### Step 4: Calculate the resultant amplitude
The resultant amplitude \( R \) can be calculated using the Pythagorean theorem:

\[ R = \sqrt{A_1^2 + A_2^2} \]

Substituting the values:

\[ R = \sqrt{A^2 + \left(-\frac{A}{2}\right)^2} \]

Calculating \( R \):

\[ R = \sqrt{A^2 + \frac{A^2}{4}} = \sqrt{A^2\left(1 + \frac{1}{4}\right)} = \sqrt{A^2 \cdot \frac{5}{4}} = A \sqrt{\frac{5}{4}} = \frac{A \sqrt{5}}{2} \]

### Final Result
Thus, the resultant amplitude of the motion is:

\[ \frac{A \sqrt{5}}{2} \]