Two simple harmonic motions A and B are given respectively by the following equations.
`y_(1)=asin[omegat+(pi)/(6)]`
`y_(2)=asin[omegat+(3pi)/(6)]`
Phase difference between the waves is

Options

A. `(pi)/(2)`
B. `(pi)/(6)`
C. `(pi)/(3)`
D. Zero

Explanation

To find the phase difference between the two simple harmonic motions given by the equations:

1. \( y_1 = A \sin\left(\omega t + \frac{\pi}{6}\right) \)
2. \( y_2 = A \sin\left(\omega t + \frac{3\pi}{6}\right) \)

we will follow these steps:

### Step 1: Identify the phases of the two equations
The phase of the first equation \( y_1 \) is:
\[ \Phi_1 = \omega t + \frac{\pi}{6} \]
The phase of the second equation \( y_2 \) is:
\[ \Phi_2 = \omega t + \frac{3\pi}{6} \]

### Step 2: Calculate the phase difference
The phase difference \( \Delta \Phi \) is given by:
\[ \Delta \Phi = \Phi_2 - \Phi_1 \]
Substituting the values of \( \Phi_1 \) and \( \Phi_2 \):
\[ \Delta \Phi = \left(\omega t + \frac{3\pi}{6}\right) - \left(\omega t + \frac{\pi}{6}\right) \]

### Step 3: Simplify the expression
When we simplify the expression:
\[ \Delta \Phi = \omega t + \frac{3\pi}{6} - \omega t - \frac{\pi}{6} \]
The \( \omega t \) terms cancel each other out:
\[ \Delta \Phi = \frac{3\pi}{6} - \frac{\pi}{6} \]
This simplifies to:
\[ \Delta \Phi = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Conclusion
The phase difference between the two simple harmonic motions is:
\[ \Delta \Phi = \frac{\pi}{3} \]