Two wave are represented by the equations `y_(1)=asinomegat` ad `y_(2)=acosomegat` the first wave

Options

A. leads the second by `pi`
B. lags the seconds by `pi`
C. leads the seconds by `(pi)/(2)`
D. lags the seconds by `(pi)/(2)`

Explanation

To solve the problem, we need to analyze the two wave equations given:

1. \( y_1 = A \sin(\omega t) \)
2. \( y_2 = A \cos(\omega t) \)

### Step 1: Rewrite the equations in terms of phase
We can express the sine function in terms of cosine to compare the two waves more easily. The sine function can be rewritten as:
\[ \sin(\omega t) = \cos\left(\omega t - \frac{\pi}{2}\right) \]
Thus, we can rewrite \( y_1 \) as:
\[ y_1 = A \cos\left(\omega t - \frac{\pi}{2}\right) \]

### Step 2: Identify the phase difference
Now we can compare the two equations:
- \( y_1 = A \cos\left(\omega t - \frac{\pi}{2}\right) \)
- \( y_2 = A \cos(\omega t) \)

The phase of \( y_1 \) is \( \omega t - \frac{\pi}{2} \) and the phase of \( y_2 \) is \( \omega t \).

### Step 3: Calculate the phase difference
The phase difference \( \Delta \phi \) between \( y_1 \) and \( y_2 \) is given by:
\[ \Delta \phi = \text{Phase of } y_2 - \text{Phase of } y_1 = \left(\omega t\right) - \left(\omega t - \frac{\pi}{2}\right) = \frac{\pi}{2} \]

### Step 4: Determine the relationship between the waves
Since \( y_1 \) has a phase that is \( \frac{\pi}{2} \) less than that of \( y_2 \), we conclude that:
- \( y_1 \) lags behind \( y_2 \) by \( \frac{\pi}{2} \).

### Conclusion
Thus, the first wave \( y_1 \) lags the second wave \( y_2 \) by \( \frac{\pi}{2} \).

### Final Answer
The correct option is that the first wave lags the second by \( \frac{\pi}{2} \).

---