The distance between two consecutive points with phase difference of 60^ in a wave of frequency 500 Hz is 6.0 m. The velocity with which the wave is traveling is km/s:

Given: Phase difference = 3, distance between the points x = 6 m, and frequency f = 500 Hz. Find: Wave velocity. The phase difference is given by = k x where k = 2 is the wave number. Substituting = 3 and x = 6 m, 3 = 2 6 So, = 36 m The wave velocity is v = f = 36 500 = 18000 m/s = 18 km/s Therefore, the velocity is 18 km/s. Given: Two consecutive points have phase difference 60^ over a distance of 6 m. Find: The speed of the wave. For a progressive wave, phase difference over distance is proportional to wavelength: 2 = x Using = 60^ = 3 we get /32 = 6 16 = 6 = 36 m Now use v = f v = 500 36 = 18000 m/s v = 18 km/s Thus, the required numerical value is 18.

The distance between two consecutive points with phase difference… — Solution | JEE Physics

NVAEasyJEE 2023Wave Motion Basics

The distance between two consecutive points with phase difference of $60^\circ$ in a wave of frequency $500 \, \text{Hz}$ is $6.0 \, \text{m}$ . The velocity with which the wave is traveling is _____ $\text{km/s}$ :

Answer

Correct answer:18

Step-by-step solution

Standard Method

Given: Phase difference $\Delta \phi = \frac{\pi}{3}$ , distance between the points $\Delta x = 6 \, \text{m}$ , and frequency $f = 500 \, \text{Hz}$ .

Find: Wave velocity.

The phase difference is given by

\Delta \phi = k \Delta x

where

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

is the wave number.

Substituting $\Delta \phi = \frac{\pi}{3}$ and $\Delta x = 6 \, \text{m}$ ,

\frac{\pi}{3} = \frac{2\pi}{\lambda} \cdot 6

So,

\lambda = 36 \, \text{m}

The wave velocity is

v = \lambda f = 36 \cdot 500 = 18000 \, \text{m/s} = 18 \, \text{km/s}

Therefore, the velocity is $18 \, \text{km/s}$ .

Using phase difference and wavelength relation

Given: Two consecutive points have phase difference $60^\circ$ over a distance of $6 \, \text{m}$ .

Find: The speed of the wave.

For a progressive wave, phase difference over distance is proportional to wavelength:

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

Using

\Delta \phi = 60^\circ = \frac{\pi}{3}

we get

\frac{\pi/3}{2\pi} = \frac{6}{\lambda}

\frac{1}{6} = \frac{6}{\lambda}

\lambda = 36 \, \text{m}

Now use

v = f\lambda

v = 500 \times 36 = 18000 \, \text{m/s}

v = 18 \, \text{km/s}

Thus, the required numerical value is 18.

Common mistakes

Using $60$ directly instead of converting $60^\circ$ to radians or using the fraction of a full cycle is incorrect. Phase formulas require angular phase consistency. Convert $60^\circ$ to $\frac{\pi}{3}$ or use that it is $\frac{1}{6}$ of $360^\circ$ .
Confusing the given $6 \, \text{m}$ with the wavelength is incorrect. The question gives the distance corresponding to a phase difference of only $60^\circ$ , not a full cycle. Use the phase-distance relation first to find $\lambda$ .
Forgetting to convert $18000 \, \text{m/s}$ into $\text{km/s}$ leads to the wrong final numerical answer. Since the blank asks for $\text{km/s}$ , divide by $1000$ and report only 18 in the answer field.

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