The equation of a particle executing a linear S.H.M. is given by `x=4 "cos" omega t + "sin" omega t`. The "tan"gent of its initial phase angle is given by

Options

A. `"tan" alpha =(3)/(4)`
B. `"tan" alpha =(4)/(3)`
C. `"tan" alpha =1`
D. `"tan" alpha =sqrt(3)`

Explanation

To find the tangent of the initial phase angle for the given equation of a particle executing linear simple harmonic motion (S.H.M.), we start with the equation:

\[ x = 4 \cos(\omega t) + 3 \sin(\omega t) \]

### Step 1: Identify the coefficients
From the equation, we can identify the coefficients of \(\cos(\omega t)\) and \(\sin(\omega t)\):
- Coefficient of \(\cos(\omega t)\) is \(4\).
- Coefficient of \(\sin(\omega t)\) is \(3\).

### Step 2: Use the standard form of S.H.M.
The general form of the equation for S.H.M. can be expressed as:
\[ x = A \cos(\omega t + \phi) \]
where \(A\) is the amplitude and \(\phi\) is the phase angle.

### Step 3: Relate the coefficients to the amplitude and phase angle
We can express the coefficients in terms of the amplitude \(A\) and phase angle \(\phi\):
- \(A \cos(\phi) = 4\)
- \(A \sin(\phi) = 3\)

### Step 4: Find the tangent of the phase angle
To find the tangent of the phase angle \(\phi\), we can use the following relationship:
\[ \tan(\phi) = \frac{\sin(\phi)}{\cos(\phi)} \]
Substituting the values from the coefficients:
\[ \tan(\phi) = \frac{A \sin(\phi)}{A \cos(\phi)} = \frac{3}{4} \]

### Step 5: Conclusion
Thus, the tangent of the initial phase angle \(\phi\) is:
\[ \tan(\phi) = \frac{3}{4} \]

### Final Answer
The tangent of the initial phase angle is \(\frac{3}{4}\).

---