Two waves are given by `y_(1)=asin(omegat-kx)` and `y_(2)=a cos(omegat-kx)`. The phase difference between the two waves is -

Options

A. `(lambda//2pi)phi`
B. `((phi+(pi//2))/(2pi))lambda`
C. `(2pi)/(lambda)(phi-(pi)/(2))`
D. `(pi)/2`

Answer

To find the phase difference between the two waves given by \( y_1 = A \sin(\omega t - kx) \) and \( y_2 = A \cos(\omega t - kx) \), we can follow these steps:

### Step 1: Identify the forms of the waves
The first wave is given as:
\[ y_1 = A \sin(\omega t - kx) \]
The second wave is given as:
\[ y_2 = A \cos(\omega t - kx) \]

### Step 2: Rewrite the cosine function in terms of sine
We can express the cosine function in terms of sine using the identity:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Applying this to \( y_2 \):
\[ y_2 = A \cos(\omega t - kx) = A \sin\left(\left(\omega t - kx\right) + \frac{\pi}{2}\right) \]

### Step 3: Identify the phase of each wave
From the rewritten form:
- The phase of \( y_1 \) is \( \phi_1 = \omega t - kx \)
- The phase of \( y_2 \) is \( \phi_2 = \left(\omega t - kx\right) + \frac{\pi}{2} \)

### Step 4: Calculate the phase difference
The phase difference \( \Delta \phi \) between the two waves is given by:
\[ \Delta \phi = \phi_2 - \phi_1 \]
Substituting the phases we found:
\[ \Delta \phi = \left(\omega t - kx + \frac{\pi}{2}\right) - (\omega t - kx) = \frac{\pi}{2} \]

### Conclusion
Thus, the phase difference between the two waves is:
\[ \Delta \phi = \frac{\pi}{2} \]

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

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