The displacement of the wave given by equation `y(x, t)=a sin (kx - omega t +phi), ` where`phi=0` at point x and t = 0 is same as that at point

Options

A. `x+2npi`
B. `x+(2npi)/k`
C. `kx + 2npi`
D. Both (a) and (b)

Explanation

To solve the problem, we start with the wave equation given:

\[ y(x, t) = a \sin(kx - \omega t + \phi) \]

where \(\phi = 0\) at \(x\) and \(t = 0\). We need to find the displacement at this point and determine how many points exist where the displacement is the same.

### Step 1: Substitute \(\phi\) and \(t\)

Since \(\phi = 0\), we can simplify the equation:

\[ y(x, t) = a \sin(kx - \omega t) \]

Now, we need to evaluate this at \(t = 0\):

\[ y(x, 0) = a \sin(kx) \]

### Step 2: Understand the condition for displacement

We are looking for points \(x\) where the displacement \(y(x, t)\) is the same. This means we want to find all \(x\) such that:

\[ a \sin(kx) = a \sin(kx') \]

for some other point \(x'\).

### Step 3: Use the properties of the sine function

The sine function has the property that:

\[ \sin(\theta) = \sin(\theta + 2n\pi) \]

for any integer \(n\). This means that:

\[ kx = kx' + 2n\pi \]

### Step 4: Solve for \(x'\)

Rearranging the equation gives:

\[ kx - kx' = 2n\pi \]

Dividing through by \(k\):

\[ x - x' = \frac{2n\pi}{k} \]

This implies:

\[ x' = x - \frac{2n\pi}{k} \]

### Step 5: Identify the number of solutions

The integer \(n\) can take any integer value (positive, negative, or zero). Therefore, for each integer \(n\), there is a corresponding point \(x'\) where the displacement is the same as at point \(x\).

### Conclusion

Since \(n\) can take infinitely many integer values, there are infinitely many points \(x'\) where the displacement is the same as at point \(x\).

Thus, the answer to the question is that there are infinitely many points where the displacement is the same.

---