What is the phase difference between two simple harmonic motions represented by `x_(1)=A"sin"(omegat+(pi)/(6))` and `x_(2)=A "cos"omegat`?

Options

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

Explanation

To find the phase difference between the two simple harmonic motions represented by the equations \( x_1 = A \sin(\omega t + \frac{\pi}{6}) \) and \( x_2 = A \cos(\omega t) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the phases of both equations**:
- For the first equation \( x_1 = A \sin(\omega t + \frac{\pi}{6}) \), the phase is \( \phi_1 = \omega t + \frac{\pi}{6} \).
- For the second equation \( x_2 = A \cos(\omega t) \), we can rewrite it in terms of sine:
\[ x_2 = A \cos(\omega t) = A \sin\left(\omega t + \frac{\pi}{2}\right) \]
Thus, the phase for the second equation is \( \phi_2 = \omega t + \frac{\pi}{2} \).

2. **Calculate the phase difference**:
- The phase difference \( \Delta \phi \) is given by:
\[ \Delta \phi = \phi_1 - \phi_2 \]
- Substituting the phases we found:
\[ \Delta \phi = \left(\omega t + \frac{\pi}{6}\right) - \left(\omega t + \frac{\pi}{2}\right) \]
- The \( \omega t \) terms cancel out:
\[ \Delta \phi = \frac{\pi}{6} - \frac{\pi}{2} \]

3. **Simplify the expression**:
- To simplify \( \frac{\pi}{6} - \frac{\pi}{2} \), we need a common denominator. The common denominator for 6 and 2 is 6:
\[ \Delta \phi = \frac{\pi}{6} - \frac{3\pi}{6} = \frac{\pi - 3\pi}{6} = \frac{-2\pi}{6} = -\frac{\pi}{3} \]

4. **Take the absolute value**:
- Since phase difference is typically expressed as a positive value, we take the absolute value:
\[ |\Delta \phi| = \frac{\pi}{3} \]

5. **Conclusion**:
- The phase difference between the two simple harmonic motions is \( \frac{\pi}{3} \).

### Final Answer:
The phase difference between the two simple harmonic motions is \( \frac{\pi}{3} \).