Two equations of two SHM `y = a sin (omegat-alpha)` and `y = a cos (omegat-alpha)`. The phase difference between the two is

Options

A. `0^(0)`
B. `alpha^(0)`
C. `90^(0)`
D. `180^(0)`

Answer

To find the phase difference between the two simple harmonic motions (SHM) given by the equations \( y = a \sin(\omega t - \alpha) \) and \( y = a \cos(\omega t - \alpha) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Equations**:
We have two equations:
- Equation 1: \( y_1 = a \sin(\omega t - \alpha) \)
- Equation 2: \( y_2 = a \cos(\omega t - \alpha) \)

2. **Recognize the Relationship Between Sine and Cosine**:
We know that the cosine function can be expressed in terms of the sine function:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
This means we can rewrite the second equation in terms of sine:
\[ y_2 = a \cos(\omega t - \alpha) = a \sin\left(\left(\omega t - \alpha\right) + \frac{\pi}{2}\right) \]

3. **Express Both Equations**:
Now we can express both equations in terms of sine:
- Equation 1: \( y_1 = a \sin(\omega t - \alpha) \)
- Equation 2: \( y_2 = a \sin\left(\omega t - \alpha + \frac{\pi}{2}\right) \)

4. **Determine the Phase Difference**:
The phase difference \( \Delta \phi \) between the two SHMs can be found by comparing the arguments of the sine functions:
\[ \Delta \phi = \left(\omega t - \alpha + \frac{\pi}{2}\right) - (\omega t - \alpha) = \frac{\pi}{2} \]

5. **Conclusion**:
Therefore, the phase difference between the two SHMs is:
\[ \Delta \phi = \frac{\pi}{2} \]

### Final Answer:
The phase difference between the two SHMs is \( \frac{\pi}{2} \).

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