Four simple harmonic vibrations `x_(1) = 8sinepsilont`, `x_(2) = 6sin(epsilont+pi/2)`, `x_(3)=4sin(epsilont+pi)` and `x_(4) = 2sin(epsilont+(3pi)/2)` are superimposed on each other. The resulting amplitude and its phase difference with `x_(1)` are respectively.

Options

A. `20, tan^(-)(1/2)`
B. `4sqrt(2),(pi/2)`
C. `20, tan^(-1)(2)`
D. `4sqrt(2),(pi/4)`

Answer

To solve the problem of superimposing the four simple harmonic motions (SHMs), we will follow these steps:

### Step 1: Identify the amplitudes and phase angles of each SHM
The given SHMs are:
- \( x_1 = 8 \sin(\epsilon t) \) → Amplitude = 8, Phase = 0
- \( x_2 = 6 \sin(\epsilon t + \frac{\pi}{2}) \) → Amplitude = 6, Phase = \( \frac{\pi}{2} \)
- \( x_3 = 4 \sin(\epsilon t + \pi) \) → Amplitude = 4, Phase = \( \pi \)
- \( x_4 = 2 \sin(\epsilon t + \frac{3\pi}{2}) \) → Amplitude = 2, Phase = \( \frac{3\pi}{2} \)

### Step 2: Convert each SHM into phasor representation
- \( x_1 \) is represented as a vector of length 8 along the positive x-axis.
- \( x_2 \) is represented as a vector of length 6 along the positive y-axis.
- \( x_3 \) is represented as a vector of length 4 along the negative x-axis.
- \( x_4 \) is represented as a vector of length 2 along the negative y-axis.

### Step 3: Calculate the resultant vector in the x and y directions
- In the x-direction:
- Contribution from \( x_1 \): +8
- Contribution from \( x_3 \): -4
- Total in x-direction: \( 8 - 4 = 4 \)

- In the y-direction:
- Contribution from \( x_2 \): +6
- Contribution from \( x_4 \): -2
- Total in y-direction: \( 6 - 2 = 4 \)

### Step 4: Determine the resultant amplitude using the Pythagorean theorem
The resultant amplitude \( A_R \) can be calculated as:
\[ A_R = \sqrt{(4)^2 + (4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \]

### Step 5: Calculate the phase difference with respect to \( x_1 \)
To find the phase difference \( \theta \):
\[ \tan \theta = \frac{\text{y-component}}{\text{x-component}} = \frac{4}{4} = 1 \]
Thus, \( \theta = \tan^{-1}(1) = \frac{\pi}{4} \).

### Final Result
The resulting amplitude is \( 4\sqrt{2} \) and the phase difference with \( x_1 \) is \( \frac{\pi}{4} \).

### Summary
The final answer is:
- Resulting Amplitude: \( 4\sqrt{2} \)
- Phase Difference with \( x_1 \): \( \frac{\pi}{4} \)

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