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 \( x_1 = A \sin(\omega t + \frac{\pi}{6}) \) and \( x_2 = A \cos(\omega t) \), we can follow these steps:

### Step 1: Identify the phase of each motion
- For \( x_1 = A \sin(\omega t + \frac{\pi}{6}) \), the phase \( \phi_1 \) is \( \frac{\pi}{6} \).
- For \( x_2 = A \cos(\omega t) \), we can express the cosine function in terms of sine:
\[ x_2 = A \cos(\omega t) = A \sin\left(\omega t + \frac{\pi}{2}\right) \]
Thus, the phase \( \phi_2 \) is \( \frac{\pi}{2} \).

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

### Step 3: Find a common denominator
To perform the subtraction, we need a common denominator. The least common multiple of 2 and 6 is 6:
\[ \Delta \phi = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} \]

### Step 4: Simplify the result
Now, simplify \( \frac{2\pi}{6} \):
\[ \Delta \phi = \frac{\pi}{3} \]

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

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