For simple harmonic vibrations
`y_(1)=8cos omegat`
`y_(2)=4 cos (omegat+(pi)/(2))`
`y_(3)=2cos (omegat+pi)`
`y_(4)=cos(omegat+(3pi)/(2))` are superimposed on one another. The resulting amplitude and phase are respectively

Options

A. `sqrt(45) and tan^(-1)((1)/(2))`
B. `sqrt(45) and tan^(-1)((1)/(3))`
C. `sqrt(75) and tan^(-1)(2)`
D. `sqrt(75) and tan^(-1)((1)/(3))`

Answer

To solve the problem of finding the resulting amplitude and phase of the superimposed simple harmonic vibrations given by:

1. \( y_1 = 8 \cos(\omega t) \)
2. \( y_2 = 4 \cos\left(\omega t + \frac{\pi}{2}\right) \)
3. \( y_3 = 2 \cos\left(\omega t + \pi\right) \)
4. \( y_4 = \cos\left(\omega t + \frac{3\pi}{2}\right) \)

we will follow these steps:

### Step 1: Identify the Amplitudes and Phases
- For \( y_1 \): Amplitude \( A_1 = 8 \), Phase \( \phi_1 = 0 \)
- For \( y_2 \): Amplitude \( A_2 = 4 \), Phase \( \phi_2 = \frac{\pi}{2} \)
- For \( y_3 \): Amplitude \( A_3 = 2 \), Phase \( \phi_3 = \pi \)
- For \( y_4 \): Amplitude \( A_4 = 1 \), Phase \( \phi_4 = \frac{3\pi}{2} \)

### Step 2: Convert to Phasor Representation
We can represent these amplitudes and phases as vectors (phasors) in the complex plane:
- \( y_1 \) corresponds to the vector \( 8 \) at angle \( 0 \)
- \( y_2 \) corresponds to the vector \( 4 \) at angle \( \frac{\pi}{2} \)
- \( y_3 \) corresponds to the vector \( -2 \) at angle \( \pi \)
- \( y_4 \) corresponds to the vector \( -1 \) at angle \( \frac{3\pi}{2} \)

### Step 3: Calculate the Resultant Amplitude
To find the resultant amplitude, we need to add these vectors. We will first calculate the x and y components of each vector:

- For \( y_1 \):
- \( x_1 = 8 \)
- \( y_1 = 0 \)

- For \( y_2 \):
- \( x_2 = 0 \)
- \( y_2 = 4 \)

- For \( y_3 \):
- \( x_3 = -2 \)
- \( y_3 = 0 \)

- For \( y_4 \):
- \( x_4 = 0 \)
- \( y_4 = -1 \)

Now, summing the components:
- Total \( x \) component:
\[ X = x_1 + x_2 + x_3 + x_4 = 8 + 0 - 2 + 0 = 6 \]

- Total \( y \) component:
\[ Y = y_1 + y_2 + y_3 + y_4 = 0 + 4 + 0 - 1 = 3 \]

### Step 4: Calculate the Resultant Amplitude
The resultant amplitude \( A \) can be calculated using the Pythagorean theorem:
\[ A = \sqrt{X^2 + Y^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \]

### Step 5: Calculate the Phase Angle
The phase angle \( \phi \) can be calculated using:
\[ \tan(\phi) = \frac{Y}{X} = \frac{3}{6} = \frac{1}{2} \]
Thus,
\[ \phi = \tan^{-1}\left(\frac{1}{2}\right) \]

### Final Result
The resulting amplitude and phase are:
- Amplitude: \( 3\sqrt{5} \)
- Phase: \( \tan^{-1}\left(\frac{1}{2}\right) \)