Two waves are represented by `y_(1)= a sin (omega t + ( pi)/(6)) and y_(2) = a cos omega t `. What will be their resultant amplitude

Options

A. a
B. `sqrt(2) a `
C. `sqrt(3) a`
D. 2a

Answer

To find the resultant amplitude of the two waves represented by \( y_1 = a \sin(\omega t + \frac{\pi}{6}) \) and \( y_2 = a \cos(\omega t) \), we can follow these steps:

### Step 1: Rewrite the second wave in sine form
We know that \( \cos(\omega t) \) can be expressed in terms of sine:
\[ y_2 = a \cos(\omega t) = a \sin\left(\omega t + \frac{\pi}{2}\right) \]

### Step 2: Identify the amplitudes and phases
From the equations:
- For \( y_1 \):
- Amplitude \( a_1 = a \)
- Phase \( \phi_1 = \frac{\pi}{6} \)

- For \( y_2 \):
- Amplitude \( a_2 = a \)
- Phase \( \phi_2 = \frac{\pi}{2} \)

### Step 3: Calculate the phase difference
The phase difference \( \phi \) between the two waves is given by:
\[ \phi = \phi_2 - \phi_1 = \frac{\pi}{2} - \frac{\pi}{6} \]
To compute this:
\[ \phi = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Step 4: Use the formula for resultant amplitude
The formula for the resultant amplitude \( A \) of two waves with equal amplitudes \( a_1 \) and \( a_2 \) and a phase difference \( \phi \) is:
\[ A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos(\phi)} \]
Substituting \( a_1 = a \), \( a_2 = a \), and \( \phi = \frac{\pi}{3} \):
\[ A = \sqrt{a^2 + a^2 + 2a \cdot a \cdot \cos\left(\frac{\pi}{3}\right)} \]

### Step 5: Calculate \( \cos\left(\frac{\pi}{3}\right) \)
We know that:
\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]
Thus, substituting this value into the equation:
\[ A = \sqrt{a^2 + a^2 + 2a^2 \cdot \frac{1}{2}} \]
This simplifies to:
\[ A = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} \]

### Step 6: Final Result
Therefore, the resultant amplitude is:
\[ A = \sqrt{3}a \]

### Conclusion
The resultant amplitude of the two waves is \( \sqrt{3}a \).

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