What is the minimum phase difference between two simple harmonic oscillations given by `y_(1)=(1)/(2) "sin" omega t+(sqrt(3))/(2) "cos" omega t`
`y_(2)= "sin" omega t+ "cos" omega t` ?

Options

A. `(pi)/(12)`
B. `(7pi)/(12)`
C. `(pi)/(6)`
D. `-(pi)/(6)`

Explanation

To find the minimum phase difference between the two simple harmonic oscillations given by:

1. \( y_1 = \frac{1}{2} \sin(\omega t) + \frac{\sqrt{3}}{2} \cos(\omega t) \)
2. \( y_2 = \sin(\omega t) + \cos(\omega t) \)

we will follow these steps:

### Step 1: Rewrite \( y_1 \) in the form \( A \sin(\omega t + \phi) \)

To rewrite \( y_1 \) in the form \( A \sin(\omega t + \phi) \), we can use the following trigonometric identity:

\[ A \sin(\omega t + \phi) = A \left( \sin(\omega t) \cos(\phi) + \cos(\omega t) \sin(\phi) \right) \]

Comparing coefficients, we have:

\[ A \cos(\phi) = \frac{1}{2} \quad \text{and} \quad A \sin(\phi) = \frac{\sqrt{3}}{2} \]

### Step 2: Find the amplitude \( A \)

To find \( A \), we can use the Pythagorean identity:

\[ A = \sqrt{(A \cos(\phi))^2 + (A \sin(\phi))^2} \]

Substituting the values we have:

\[ A = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

### Step 3: Find \( \phi \)

Now we can find \( \phi \) using:

\[ \cos(\phi) = \frac{1/2}{1} = \frac{1}{2} \quad \text{and} \quad \sin(\phi) = \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2} \]

From this, we can determine:

\[ \phi = \frac{\pi}{3} \quad (\text{since } \cos(\phi) = \frac{1}{2} \text{ and } \sin(\phi) = \frac{\sqrt{3}}{2}) \]

Thus, we can rewrite \( y_1 \) as:

\[ y_1 = \sin\left(\omega t + \frac{\pi}{3}\right) \]

### Step 4: Rewrite \( y_2 \) in the form \( A \sin(\omega t + \phi) \)

Now we will rewrite \( y_2 \):

\[ y_2 = \sin(\omega t) + \cos(\omega t) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin(\omega t) + \frac{1}{\sqrt{2}} \cos(\omega t) \right) \]

Using the identity again, we can see that:

\[ \phi = \frac{\pi}{4} \quad (\text{since } \cos(\phi) = \sin(\phi) = \frac{1}{\sqrt{2}}) \]

Thus, we can rewrite \( y_2 \) as:

\[ y_2 = \sqrt{2} \sin\left(\omega t + \frac{\pi}{4}\right) \]

### Step 5: Calculate the phase difference

Now we can find the phase difference \( \Delta \phi \) between \( y_1 \) and \( y_2 \):

\[ \Delta \phi = \left| \phi_1 - \phi_2 \right| = \left| \frac{\pi}{3} - \frac{\pi}{4} \right| \]

Finding a common denominator (12):

\[ \Delta \phi = \left| \frac{4\pi}{12} - \frac{3\pi}{12} \right| = \left| \frac{\pi}{12} \right| \]

### Conclusion

Thus, the minimum phase difference between the two simple harmonic oscillations is:

\[ \Delta \phi = \frac{\pi}{12} \]