Two `SHM's` are represented by `y = a sin (omegat - kx)` and `y = b cos (omegat - kx)`. The phase difference between the two is :

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(3pi)/(4)`

Answer

To find the phase difference between the two simple harmonic motions represented by the equations \( y_1 = a \sin(\omega t - kx) \) and \( y_2 = b \cos(\omega t - kx) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Phases**:
- The first equation is \( y_1 = a \sin(\omega t - kx) \).
- The phase of \( y_1 \) is \( \phi_1 = \omega t - kx \).
- The second equation is \( y_2 = b \cos(\omega t - kx) \).
- The phase of \( y_2 \) is \( \phi_2 = \omega t - kx \).

2. **Convert Cosine to Sine**:
- We can express the cosine function in terms of sine:
\[ \cos(\theta) = \sin\left(\frac{\pi}{2} + \theta\right) \]
- Therefore, we can rewrite \( y_2 \) as:
\[ y_2 = b \cos(\omega t - kx) = b \sin\left(\frac{\pi}{2} + (\omega t - kx)\right) \]
- This means the phase of \( y_2 \) can be expressed as:
\[ \phi_2 = \frac{\pi}{2} + (\omega t - kx) \]

3. **Calculate the Phase Difference**:
- The phase difference \( \Delta \phi \) is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
- Substituting the values of \( \phi_1 \) and \( \phi_2 \):
\[ \Delta \phi = \left(\frac{\pi}{2} + (\omega t - kx)\right) - (\omega t - kx) \]
- Simplifying this expression:
\[ \Delta \phi = \frac{\pi}{2} \]

4. **Conclusion**:
- The phase difference between the two simple harmonic motions is \( \frac{\pi}{2} \).

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