The minimum phase difference between the two simple harmonic oscillations `y_1=1/2sinomegat+(sqrt3/2) cos omegat` and `y_2=sinomegat+cosomegat` is

Options

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

Explanation

To find the minimum phase difference between the two simple harmonic oscillations given by \( y_1 = \frac{1}{2} \sin(\omega t) + \frac{\sqrt{3}}{2} \cos(\omega t) \) and \( y_2 = \sin(\omega t) + \cos(\omega t) \), we can follow these steps:

### Step 1: Rewrite \( y_1 \) in the form of \( R \sin(\omega t + \phi_1) \)

We can express \( y_1 \) in a single sine function using the formula:
\[ R \sin(\omega t + \phi) = R \sin(\omega t) \cos(\phi) + R \cos(\omega t) \sin(\phi) \]
Here, we identify:
- \( R \cos(\phi) = \frac{1}{2} \)
- \( R \sin(\phi) = \frac{\sqrt{3}}{2} \)

To find \( R \) and \( \phi \):
1. Calculate \( R \):
\[ R = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]
2. Calculate \( \phi \):
\[ \tan(\phi) = \frac{\sin(\phi)}{\cos(\phi)} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \implies \phi = \frac{\pi}{3} \]

Thus, we can rewrite \( y_1 \) as:
\[ y_1 = \sin(\omega t + \frac{\pi}{3}) \]

### Step 2: Rewrite \( y_2 \) in the form of \( R \sin(\omega t + \phi_2) \)

For \( y_2 = \sin(\omega t) + \cos(\omega t) \):
1. We can express it as:
\[ y_2 = \sqrt{1^2 + 1^2} \sin\left(\omega t + \frac{\pi}{4}\right) = \sqrt{2} \sin\left(\omega t + \frac{\pi}{4}\right) \]
where \( R = \sqrt{2} \) and \( \phi_2 = \frac{\pi}{4} \).

### Step 3: Calculate the phase difference

Now, we have:
- \( \phi_1 = \frac{\pi}{3} \)
- \( \phi_2 = \frac{\pi}{4} \)

The phase difference \( \Delta \phi \) is given by:
\[ \Delta \phi = |\phi_1 - \phi_2| = \left| \frac{\pi}{3} - \frac{\pi}{4} \right| \]

### Step 4: Find a common denominator and calculate

To subtract these fractions, we find a common denominator:
\[ \frac{\pi}{3} = \frac{4\pi}{12}, \quad \frac{\pi}{4} = \frac{3\pi}{12} \]
Thus,
\[ \Delta \phi = \left| \frac{4\pi}{12} - \frac{3\pi}{12} \right| = \left| \frac{\pi}{12} \right| \]

### Final Answer

The minimum phase difference between the two simple harmonic oscillations is:
\[ \Delta \phi = \frac{\pi}{12} \]