Two waves are represented by the equations `y_1=a sin omega t` and `y_2=a cos omegat`. The first wave

Options

A. (a) Leads the second by `pi`
B. (b) Lags the second by `pi`
C. (c) Leads the second by `pi/2`
D. (d) Lags the second by `pi/2`

Explanation

To solve the problem of determining the phase relationship between the two waves represented by the equations \( y_1 = A \sin(\omega t) \) and \( y_2 = A \cos(\omega t) \), we can follow these steps:

### Step 1: Rewrite the second wave in terms of sine
We know that the cosine function can be expressed in terms of the sine function using the phase shift identity:
\[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \]
Using this identity, we can rewrite \( y_2 \):
\[ y_2 = A \cos(\omega t) = A \sin\left(\omega t + \frac{\pi}{2}\right) \]

### Step 2: Compare the two wave equations
Now we have:
- \( y_1 = A \sin(\omega t) \)
- \( y_2 = A \sin\left(\omega t + \frac{\pi}{2}\right) \)

### Step 3: Determine the phase difference
From the equations, we can see that \( y_2 \) is a sine wave that is shifted by \( \frac{\pi}{2} \) radians (or 90 degrees) ahead of \( y_1 \). This means that \( y_1 \) lags behind \( y_2 \).

### Step 4: Conclusion
Since \( y_2 \) leads \( y_1 \) by \( \frac{\pi}{2} \), we can conclude that:
- The first wave \( y_1 \) lags behind the second wave \( y_2 \) by \( \frac{\pi}{2} \).

Thus, the answer is that the first wave lags by \( \frac{\pi}{2} \).

### Final Answer
The first wave \( y_1 \) lags behind the second wave \( y_2 \) by \( \frac{\pi}{2} \).

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