Equation of a progressive wave is given by
`y=0.2cospi(0.04t+0.02x-(pi)/(6))`
The distance is expressed in cm and time in second. The minimum distance between two particles having the phase difference of `pi//2` is................

Answer

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

\[ y = 0.2 \cos \left( \pi (0.04t + 0.02x) - \frac{\pi}{6} \right) \]

### Step 1: Identify the wave number (k) and wavelength (λ)

The general form of a progressive wave can be expressed as:

\[ y = A \cos(\omega t + kx + \phi) \]

From the given equation, we can identify the wave number \( k \) from the term \( 0.02x \).

The wave number \( k \) is given by:

\[ k = 0.02 \, \text{cm}^{-1} \]

To find the wavelength \( \lambda \), we use the relationship:

\[ k = \frac{2\pi}{\lambda} \]

Rearranging gives:

\[ \lambda = \frac{2\pi}{k} \]

Substituting \( k = 0.02 \):

\[ \lambda = \frac{2\pi}{0.02} = 100\pi \, \text{cm} \]

### Step 2: Relate phase difference to path difference

The phase difference \( \Delta \phi \) is related to the path difference \( \Delta x \) by the formula:

\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]

We know that \( \Delta \phi = \frac{\pi}{2} \). We can rearrange the formula to solve for \( \Delta x \):

\[ \Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi} \]

### Step 3: Substitute the values

Now substituting \( \Delta \phi = \frac{\pi}{2} \) and \( \lambda = 100\pi \):

\[ \Delta x = \frac{\frac{\pi}{2} \cdot 100\pi}{2\pi} \]

The \( \pi \) cancels out:

\[ \Delta x = \frac{100}{4} = 25 \, \text{cm} \]

### Conclusion

The minimum distance between two particles having a phase difference of \( \frac{\pi}{2} \) is:

\[ \Delta x = 25 \, \text{cm} \]

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