A progressive wave of frequency $400 \text{ Hz}$ is travelling with velocity $336 \text{ m/s}$. How far apart are the two points on a wave which are $60^\circ$ out of phase?
Show Hint
$360^\circ$ corresponds to one full wavelength ($\lambda$). $60^\circ$ is $1/6$th of a wavelength.
Step 1: Understanding the Concept:
A progressive wave has a repeating spatial pattern characterized by its wavelength.
The phase difference between two points on the wave relates directly to the physical distance (path difference) between them.
A full cycle corresponds to a phase difference of $360^\circ$ (or $2\pi$ radians) and a path difference of one wavelength ($\lambda$). Step 2: Key Formula or Approach:
First, find the wavelength using the wave equation: $v = f \lambda \implies \lambda = \frac{v}{f}$.
The relation between phase difference $\Delta \phi$ and path difference $\Delta x$ is: $\Delta x = \frac{\lambda}{2\pi} \Delta \phi$ (if $\Delta \phi$ is in radians) or $\Delta x = \frac{\lambda}{360^\circ} \Delta \phi$ (if $\Delta \phi$ is in degrees). Step 3: Detailed Explanation:
Given frequency $f = 400 \text{ Hz}$.
Given wave velocity $v = 336 \text{ m/s}$.
Calculate the wavelength $\lambda$:
\[ \lambda = \frac{v}{f} = \frac{336}{400} = 0.84 \text{ m} \]
The given phase difference is $\Delta \phi = 60^\circ$.
Using the phase-path difference relationship in degrees:
\[ \Delta x = \frac{\lambda}{360^\circ} \times \Delta \phi \]
Substitute the values into the formula:
\[ \Delta x = \frac{0.84 \text{ m}}{360^\circ} \times 60^\circ \]
Simplify the fraction:
\[ \Delta x = 0.84 \times \left(\frac{1}{6}\right) \]
Perform the division:
\[ \Delta x = 0.14 \text{ m} \]
Step 4: Final Answer:
The distance between the two points is $0.14 \text{ m}$.