A progressive wave is represented by `y = 12 sin (5t - 4x)` cm. On this wave, how far away are the two points having phase difference of `90^circ` ?

Options

A. `pi/2` cm
B. `pi/4` cm
C. `pi/8` cm
D. `pi/16` cm

Answer

To solve the problem, we need to find the distance between two points on a progressive wave that have a phase difference of \(90^\circ\). The wave is represented by the equation:

\[ y = 12 \sin(5t - 4x) \text{ cm} \]

### Step-by-Step Solution:

1. **Identify the wave parameters**:
- The given wave equation is in the form \(y = A \sin(\omega t - kx)\), where:
- Amplitude \(A = 12\) cm
- Angular frequency \(\omega = 5\) rad/s
- Wave number \(k = 4\) rad/m

2. **Determine the wavelength (\(\lambda\))**:
- The relationship between the wave number \(k\) and the wavelength \(\lambda\) is given by:
\[ k = \frac{2\pi}{\lambda} \]
- Rearranging gives:
\[ \lambda = \frac{2\pi}{k} \]
- Substituting \(k = 4\):
\[ \lambda = \frac{2\pi}{4} = \frac{\pi}{2} \text{ m} \]

3. **Phase difference and path difference relationship**:
- The phase difference \(\Delta \phi\) is related to the path difference \(\Delta x\) by the equation:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
- For a phase difference of \(90^\circ\) (which is \(\frac{\pi}{2}\) radians):
\[ \frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x \]

4. **Solve for the path difference (\(\Delta x\))**:
- Rearranging the equation gives:
\[ \Delta x = \frac{\lambda}{4} \]
- Substituting \(\lambda = \frac{\pi}{2}\):
\[ \Delta x = \frac{\frac{\pi}{2}}{4} = \frac{\pi}{8} \text{ m} \]

5. **Conclusion**:
- The distance between the two points having a phase difference of \(90^\circ\) is \(\frac{\pi}{8}\) cm.

### Final Answer:
\[ \Delta x = \frac{\pi}{8} \text{ cm} \]