`x=x_(1)+x_(2)` (where `x_(1)=4 cos omega t ` and `x_(2)=3 sin omega t`) is the equation of motion of a particle along x-axis. The phase different between `x_(1)` and x is

Options

A. `37^(@)`
B. `53^(@)`
C. `90^(@)`
D. none of these

Explanation

To find the phase difference between \( x_1 \) and \( x \) in the equation \( x = x_1 + x_2 \), where \( x_1 = 4 \cos(\omega t) \) and \( x_2 = 3 \sin(\omega t) \), we can follow these steps:

### Step 1: Write the equation of motion
We have:
\[ x = x_1 + x_2 = 4 \cos(\omega t) + 3 \sin(\omega t) \]

### Step 2: Use the sine addition formula
We can express \( x \) in the form of a single sine function using the sine addition formula:
\[ x = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude and \( \phi \) is the phase angle.

### Step 3: Identify coefficients
From the equation \( x = 4 \cos(\omega t) + 3 \sin(\omega t) \), we can identify:
- Coefficient of \( \cos(\omega t) \): \( 4 \)
- Coefficient of \( \sin(\omega t) \): \( 3 \)

### Step 4: Calculate the amplitude \( A \)
The amplitude \( A \) can be calculated using the formula:
\[ A = \sqrt{(4^2 + 3^2)} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 5: Calculate the phase angle \( \phi \)
The phase angle \( \phi \) can be found using:
\[ \tan(\phi) = \frac{\text{Coefficient of } \sin(\omega t)}{\text{Coefficient of } \cos(\omega t)} = \frac{3}{4} \]
Thus,
\[ \phi = \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 6: Find the phase difference between \( x_1 \) and \( x \)
We know that:
- \( x_1 = 4 \cos(\omega t) = 4 \sin\left(\omega t + \frac{\pi}{2}\right) \)
- The phase of \( x_1 \) is \( \frac{\pi}{2} \) (or 90 degrees).

Now, the phase difference \( \Delta \theta \) between \( x_1 \) and \( x \) is given by:
\[ \Delta \theta = \text{Phase of } x_1 - \text{Phase of } x = \frac{\pi}{2} - \phi \]

### Step 7: Substitute \( \phi \)
Substituting \( \phi \):
\[ \Delta \theta = 90^\circ - \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 8: Calculate the numerical value
Using a calculator:
\[ \tan^{-1}\left(\frac{3}{4}\right) \approx 36.87^\circ \]
Thus,
\[ \Delta \theta \approx 90^\circ - 36.87^\circ \approx 53.13^\circ \]

### Final Answer
The phase difference between \( x_1 \) and \( x \) is approximately \( 53^\circ \).

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