if `x = a sin (omegat + pi//6)` and `x' = a cos omega`, t, then what is the phase difference between the two waves

Options

A. `pi//3`
B. `pi//6`
C. `pi//2`
D. `pi`

Explanation

To find the phase difference between the two waves given by the equations \( x = a \sin(\omega t + \frac{\pi}{6}) \) and \( x' = a \cos(\omega t) \), we can follow these steps:

### Step 1: Identify the phase of each wave
The first wave is given as:
\[ x = a \sin(\omega t + \frac{\pi}{6}) \]
Here, the phase \( \phi_1 \) is:
\[ \phi_1 = \frac{\pi}{6} \]

The second wave is given as:
\[ x' = a \cos(\omega t) \]
To find the phase of this wave, we can convert the cosine function to a sine function. We know that:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Thus, we can rewrite \( x' \) as:
\[ x' = a \sin\left(\omega t + \frac{\pi}{2}\right) \]
Here, the phase \( \phi_2 \) is:
\[ \phi_2 = \frac{\pi}{2} \]

### Step 2: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two waves is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the values we found:
\[ \Delta \phi = \frac{\pi}{2} - \frac{\pi}{6} \]

### Step 3: Simplify the phase difference
To perform the subtraction, we need a common denominator. The least common multiple of 2 and 6 is 6. Therefore, we can express \( \frac{\pi}{2} \) as:
\[ \frac{\pi}{2} = \frac{3\pi}{6} \]
Now substituting back into the equation:
\[ \Delta \phi = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Conclusion
The phase difference between the two waves is:
\[ \Delta \phi = \frac{\pi}{3} \]