Two particle are executing simple harmonic motion. At an instant of time t their displacement are `y_(1)=a "cos"(omegat)` and `y_(2)=a "sin" (omegat)`
Then the phase difference between `y_(1)` and `y_(2)` is

Options

A. `120^(@)`
B. `90^(@)`
C. `180^(@)`
D. zero

Explanation

To find the phase difference between the two particles executing simple harmonic motion, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Displacements**:
We have two displacements given:
- \( y_1 = A \cos(\omega t) \)
- \( y_2 = A \sin(\omega t) \)

2. **Convert \( y_1 \) to Sine Form**:
We can express \( y_1 \) in terms of sine:
\[ y_1 = A \cos(\omega t) = A \sin\left(\omega t + \frac{\pi}{2}\right) \]
This is because \( \cos(x) = \sin\left(x + \frac{\pi}{2}\right) \).

3. **Identify the Phases**:
Now we can identify the phases of both displacements:
- Phase of \( y_1 \) is \( \phi_1 = \omega t + \frac{\pi}{2} \)
- Phase of \( y_2 \) is \( \phi_2 = \omega t \)

4. **Calculate the Phase Difference**:
The phase difference \( \Delta \phi \) between \( y_1 \) and \( y_2 \) is given by:
\[ \Delta \phi = \phi_1 - \phi_2 \]
Substituting the values:
\[ \Delta \phi = \left(\omega t + \frac{\pi}{2}\right) - \omega t = \frac{\pi}{2} \]

5. **Convert to Degrees**:
To express the phase difference in degrees, we convert \( \frac{\pi}{2} \) radians to degrees:
\[ \Delta \phi = \frac{\pi}{2} \times \frac{180}{\pi} = 90^\circ \]

### Final Answer:
The phase difference between \( y_1 \) and \( y_2 \) is \( 90^\circ \).

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