Question

A progressive wave is given by $Y = 12 \sin(5t - 4x)$. On this wave, how far apart are two points having a phase difference of $90^\circ$?

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{8}$
• C. $\frac{\pi}{16}$
• D. $\frac{\pi}{32}$

Hint:

Physics Tip: A phase difference of $360^{\circ}$ corresponds to a full wavelength ($\lambda$), $180^{\circ}$ corresponds to half a wavelength ($\lambda/2$), and $90^{\circ}$ corresponds to a quarter wavelength ($\lambda/4$). Here, $(\pi/2) / 4 = \pi/8$.

Answer 1

The Correct Option is B

Concept: Physics (Waves) - Phase Difference and Path Difference.

Step 1: Identify the wave parameters. The given wave equation is $Y = 12 \sin(5t - 4x)$. Comparing this to the standard form $Y = A \sin(\omega t - kx)$, we find the propagation constant $k = 4$.

Step 2: Relate $k$ to wavelength. We know that $k = \frac{2\pi}{\lambda}$. Therefore, $4 = \frac{2\pi}{\lambda} \implies \lambda = \frac{2\pi}{4} = \frac{\pi}{2}$.

Step 3: Use the phase-path relationship. The relation between phase difference ($\Delta\phi$) and path difference ($\Delta x$) is: $$\Delta\phi = \frac{2\pi}{\lambda} \Delta x \text{ or } \Delta\phi = k \cdot \Delta x$$

Step 4: Substitute and solve for $\Delta x$. Given $\Delta\phi = 90^{\circ} = \frac{\pi}{2}$ radians: $$\frac{\pi}{2} = 4 \cdot \Delta x$$ $$\Delta x = \frac{\pi}{8}$$ $$ \therefore \text{The two points are separated by a distance of } \frac{\pi}{8}. $$