What is the phase difference between two simple harmonic motions represented by `X_(1)=Asin[omegat+(pi)/(3)]andX_(2)=Acosomegat`?

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 \( X_1 = A \sin(\omega t + \frac{\pi}{3}) \) and \( X_2 = A \cos(\omega t) \), we can follow these steps:

### Step 1: Identify the equations
We have:
- \( X_1 = A \sin(\omega t + \frac{\pi}{3}) \)
- \( X_2 = A \cos(\omega t) \)

### Step 2: Convert one equation to a common trigonometric function
To find the phase difference, we need to express both equations in terms of the same trigonometric function. Here, we can convert \( X_2 \) (which is in cosine) to sine.

Using the identity:
\[ \cos(\theta) = \sin\left(\frac{\pi}{2} + \theta\right) \]
we can rewrite \( X_2 \):
\[ X_2 = A \cos(\omega t) = A \sin\left(\frac{\pi}{2} + \omega t\right) \]

### Step 3: Write the equations with their phases
Now we can express both equations with their respective phases:
- For \( X_1 \): The phase is \( \phi_1 = \omega t + \frac{\pi}{3} \)
- For \( X_2 \): The phase is \( \phi_2 = \frac{\pi}{2} + \omega t \)

### Step 4: Calculate the phase difference
The phase difference \( \Delta \phi \) is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the phases:
\[ \Delta \phi = \left(\frac{\pi}{2} + \omega t\right) - \left(\omega t + \frac{\pi}{3}\right) \]

### Step 5: Simplify the expression
Now, simplifying the expression:
\[ \Delta \phi = \frac{\pi}{2} - \frac{\pi}{3} \]
To subtract these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6:
\[ \Delta \phi = \frac{3\pi}{6} - \frac{2\pi}{6} = \frac{3\pi - 2\pi}{6} = \frac{\pi}{6} \]

### Conclusion
Thus, the phase difference between the two simple harmonic motions is:
\[ \Delta \phi = \frac{\pi}{6} \]

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