Two S.H.M.s along the same straight line in the same direction and of the same period are given by the equations, `x_1=3" "sin(4pit+pi/6)` and `x_2=4" " sin(4pit+pi/3)`. The initial phase of the resultant motion is

Options

A. `tan^(-1)((6+4sqrt3)/(sqrt3))`
B. `tan^(-1)((2sqrt3+3)/(2+2sqrt3))`
C. `tan^(-1)((3+4sqrt3)/(4+3sqrt3))`
D. `tan^(-1)((4+6sqrt3)/(sqrt6))`

Answer

To find the initial phase of the resultant motion of the two simple harmonic motions (SHMs) given by the equations:

1. \( x_1 = 3 \sin(4\pi t + \frac{\pi}{6}) \)
2. \( x_2 = 4 \sin(4\pi t + \frac{\pi}{3}) \)

we will follow these steps:

### Step 1: Identify the Amplitudes and Phases
From the equations, we can identify:
- Amplitude of \( x_1 \) (A1) = 3
- Phase of \( x_1 \) (φ1) = \( \frac{\pi}{6} \)
- Amplitude of \( x_2 \) (A2) = 4
- Phase of \( x_2 \) (φ2) = \( \frac{\pi}{3} \)

### Step 2: Calculate the Phase Difference
The phase difference (θ) between the two SHMs is given by:
\[ \theta = \phi_2 - \phi_1 = \frac{\pi}{3} - \frac{\pi}{6} \]
To compute this, we need a common denominator:
\[ \frac{\pi}{3} = \frac{2\pi}{6} \]
Thus,
\[ \theta = \frac{2\pi}{6} - \frac{\pi}{6} = \frac{\pi}{6} \]

### Step 3: Vector Addition of the Two SHMs
To find the resultant SHM, we can use vector addition. The magnitudes of the vectors are the amplitudes, and the angle between them is the phase difference we just calculated.

### Step 4: Use the Law of Cosines
The resultant amplitude \( R \) can be calculated using the formula:
\[ R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\theta)} \]
Substituting the values:
\[ R = \sqrt{3^2 + 4^2 + 2 \cdot 3 \cdot 4 \cdot \cos\left(\frac{\pi}{6}\right)} \]
We know that \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \):
\[ R = \sqrt{9 + 16 + 24 \cdot \frac{\sqrt{3}}{2}} = \sqrt{25 + 12\sqrt{3}} \]

### Step 5: Calculate the Initial Phase of the Resultant Motion
The initial phase \( \phi \) of the resultant motion can be calculated using:
\[ \tan(\phi) = \frac{A_1 \sin(\phi_1) + A_2 \sin(\phi_2)}{A_1 \cos(\phi_1) + A_2 \cos(\phi_2)} \]
Calculating the sine and cosine components:
- \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \)
- \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)
- \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)
- \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \)

Now substituting these into the equation:
\[ \tan(\phi) = \frac{3 \cdot \frac{1}{2} + 4 \cdot \frac{\sqrt{3}}{2}}{3 \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2}} = \frac{\frac{3}{2} + 2\sqrt{3}}{\frac{3\sqrt{3}}{2} + 2} \]

### Step 6: Solve for the Initial Phase
This will give us the value of \( \phi \).

### Final Result
The initial phase of the resultant motion can be determined from the above calculations.