Three component sinusoidal waves progressing in the same directions along the same path have the same period byt their amplitudes are `A, A/2 and A/3.` The phases of the variation at any position x on their path at time `t = 0 are 0, -pi/2 and -pi` respectively. Find the amplitude and phase of the resultant wave.

Explanation

To find the amplitude and phase of the resultant wave from the three component waves, we can follow these steps:

### Step 1: Write the equations for each wave
The three component waves can be expressed as:
1. Wave 1: \( y_1 = A \sin(\omega t) \) (Amplitude = A, Phase = 0)
2. Wave 2: \( y_2 = \frac{A}{2} \sin(\omega t - \frac{\pi}{2}) \) (Amplitude = \( \frac{A}{2} \), Phase = \( -\frac{\pi}{2} \))
3. Wave 3: \( y_3 = \frac{A}{3} \sin(\omega t - \pi) \) (Amplitude = \( \frac{A}{3} \), Phase = \( -\pi \))

### Step 2: Convert the second and third waves to cosine form
Using the identity \( \sin(x - \frac{\pi}{2}) = \cos(x) \) and \( \sin(x - \pi) = -\sin(x) \):
- Wave 2: \( y_2 = \frac{A}{2} \cos(\omega t) \)
- Wave 3: \( y_3 = -\frac{A}{3} \sin(\omega t) \)

### Step 3: Express all waves in terms of sine
To combine all three waves, we can express the cosine wave in terms of sine:
- \( \cos(\omega t) = \sin(\omega t + \frac{\pi}{2}) \)
Thus, Wave 2 becomes:
- Wave 2: \( y_2 = \frac{A}{2} \sin(\omega t + \frac{\pi}{2}) \)

### Step 4: Combine the waves
Now we can express all three waves in sine form:
1. Wave 1: \( A \sin(\omega t) \)
2. Wave 2: \( \frac{A}{2} \sin(\omega t + \frac{\pi}{2}) \)
3. Wave 3: \( -\frac{A}{3} \sin(\omega t) \)

Combining these:
- Resultant: \( y_R = A \sin(\omega t) + \frac{A}{2} \cos(\omega t) - \frac{A}{3} \sin(\omega t) \)

### Step 5: Combine like terms
Combine the sine terms:
- \( y_R = \left(A - \frac{A}{3}\right) \sin(\omega t) + \frac{A}{2} \cos(\omega t) \)
- \( y_R = \frac{2A}{3} \sin(\omega t) + \frac{A}{2} \cos(\omega t) \)

### Step 6: Find the resultant amplitude and phase
To find the resultant amplitude \( R \) and phase \( \phi \):
- \( R = \sqrt{\left(\frac{2A}{3}\right)^2 + \left(\frac{A}{2}\right)^2} \)
- \( R = \sqrt{\frac{4A^2}{9} + \frac{A^2}{4}} \)
- Finding a common denominator (36):
- \( R = \sqrt{\frac{16A^2}{36} + \frac{9A^2}{36}} = \sqrt{\frac{25A^2}{36}} = \frac{5A}{6} \)

### Step 7: Find the phase angle \( \phi \)
Using the tangent function:
- \( \tan(\phi) = \frac{\text{Vertical Component}}{\text{Horizontal Component}} = \frac{\frac{A}{2}}{\frac{2A}{3}} = \frac{3}{4} \)
- \( \phi = \tan^{-1}\left(\frac{3}{4}\right) \)

### Final Result
The amplitude of the resultant wave is \( \frac{5A}{6} \) and the phase is \( \tan^{-1}\left(\frac{3}{4}\right) \).

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