Five sinusoidal waves have the same frequency `500 Hz` but their amplitudes are in the ratio `2 : 1//2 : 1//2 :1:1` and their phase angles `0 , pi//6 , pi//3 , pi//2 and pi`, respectively . The phase angle of resultant wave obtained by the superposition of these five waves is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To find the phase angle of the resultant wave obtained by the superposition of the five sinusoidal waves, we will follow these steps:

### Step 1: Identify the amplitudes and phase angles
The amplitudes are given in the ratio \(2 : \frac{1}{2} : \frac{1}{2} : 1 : 1\). We can express these amplitudes as:
- \(A_1 = 2A\)
- \(A_2 = \frac{1}{2}A\)
- \(A_3 = \frac{1}{2}A\)
- \(A_4 = A\)
- \(A_5 = A\)

The phase angles are:
- \(\phi_1 = 0\)
- \(\phi_2 = \frac{\pi}{6}\)
- \(\phi_3 = \frac{\pi}{3}\)
- \(\phi_4 = \frac{\pi}{2}\)
- \(\phi_5 = \pi\)

### Step 2: Write the equations for each wave
The equations for the five waves can be expressed in the form:
- \(y_1 = 2A \sin(\omega t + 0)\)
- \(y_2 = \frac{1}{2}A \sin(\omega t + \frac{\pi}{6})\)
- \(y_3 = \frac{1}{2}A \sin(\omega t + \frac{\pi}{3})\)
- \(y_4 = A \sin(\omega t + \frac{\pi}{2})\)
- \(y_5 = A \sin(\omega t + \pi)\)

### Step 3: Convert to phasor representation
We can represent each wave as a phasor in the complex plane:
- \(Y_1 = 2A\)
- \(Y_2 = \frac{1}{2}A e^{j\frac{\pi}{6}}\)
- \(Y_3 = \frac{1}{2}A e^{j\frac{\pi}{3}}\)
- \(Y_4 = A e^{j\frac{\pi}{2}}\)
- \(Y_5 = A e^{j\pi}\)

### Step 4: Calculate the components of each phasor
Now, we will calculate the x (real) and y (imaginary) components of each phasor:
1. \(Y_1: (2A, 0)\)
2. \(Y_2: \left(\frac{1}{2}A \cos\frac{\pi}{6}, \frac{1}{2}A \sin\frac{\pi}{6}\right) = \left(\frac{1}{2}A \cdot \frac{\sqrt{3}}{2}, \frac{1}{2}A \cdot \frac{1}{2}\right)\)
3. \(Y_3: \left(\frac{1}{2}A \cos\frac{\pi}{3}, \frac{1}{2}A \sin\frac{\pi}{3}\right) = \left(\frac{1}{2}A \cdot \frac{1}{2}, \frac{1}{2}A \cdot \frac{\sqrt{3}}{2}\right)\)
4. \(Y_4: (0, A)\)
5. \(Y_5: (-A, 0)\)

### Step 5: Sum the components
Now, we will sum the x and y components:
- **Total x-component**:
\[ X = 2A + \frac{1}{2}A \cdot \frac{\sqrt{3}}{2} + \frac{1}{2}A \cdot \frac{1}{2} - A \]
- **Total y-component**:
\[ Y = \frac{1}{2}A \cdot \frac{1}{2} + \frac{1}{2}A \cdot \frac{\sqrt{3}}{2} + A \]

### Step 6: Calculate the resultant phase angle
The phase angle \(\psi\) of the resultant wave can be calculated using:
\[ \tan \psi = \frac{Y}{X} \]
Thus, we can find \(\psi\) using:
\[ \psi = \tan^{-1}\left(\frac{Y}{X}\right) \]

### Step 7: Solve for the phase angle
After performing the calculations for \(X\) and \(Y\), we find that the resultant phase angle \(\psi\) is \(45^\circ\).

### Final Answer
The phase angle of the resultant wave is \(45^\circ\).

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