The displacement of a particle performing S.H.M. is x = 3 sin (314 t) + 4 cos (314 t). Then the initial phase angle is,

Options

A. `tan^(-1)(7)`
B. `tan^(-1)(4)`
C. `tan^(-1)(3)`
D. `tan^(-1)(4//3)`

Explanation

To find the initial phase angle of the particle performing Simple Harmonic Motion (S.H.M.) given by the equation \( x = 3 \sin(314t) + 4 \cos(314t) \), we can follow these steps:

### Step 1: Identify the components of the displacement
The displacement \( x \) can be expressed as the sum of two components:
- \( y_1 = 3 \sin(314t) \)
- \( y_2 = 4 \cos(314t) \)

### Step 2: Represent the components as phasors
In phasor representation:
- The \( y_1 \) component (sine) can be represented as a vector of length 3 along the sine direction.
- The \( y_2 \) component (cosine) can be represented as a vector of length 4 along the cosine direction.

### Step 3: Calculate the resultant amplitude
The resultant amplitude \( R \) of the two components can be calculated using the Pythagorean theorem:
\[ R = \sqrt{(3^2 + 4^2)} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 4: Determine the phase angle
The phase angle \( \theta \) can be found using the tangent function:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{y_2}{y_1} = \frac{4}{3} \]

### Step 5: Calculate the phase angle
To find \( \theta \), we take the inverse tangent:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Final Result
Thus, the initial phase angle is:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]