The equation of a progressive wave is given b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Comparing the given equation $y = 0.2 \cos \pi (0.04t + 0.02x - \frac{\pi}{6})$ with the standard wave equation $y = a \cos (\omega t + kx - \phi)

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(C) Comparing the given equation $y = 0.2 \cos \pi (0.04t + 0.02x - \frac{\pi}{6})$ with the standard wave equation $y = a \cos (\omega t + kx - \phi)$,we distribute $\pi$ inside the bracket:$y = 0.2 \cos (0.04\pi t + 0.02\pi x - \frac{\pi^2}{6})$.Here,the wave number $k = 0.02\pi$.We know that $k = \frac{2\pi}{\lambda}$,so $0.02\pi = \frac{2\pi}{\lambda}$.Solving for wavelength $\lambda$: $\lambda = \frac{2\pi}{0.02\pi} = 100 \ cm$.The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.Given $\Delta \phi = \frac{\pi}{2}$,we have $\frac{\pi}{2} = \frac{2\pi}{100} \Delta x$.$\Delta x = \frac{\pi}{2} 	imes \frac{100}{2\pi} = 25 \ cm$.

The equation of a progressive wave is given by $y = 0.2 \cos \pi (0.04t + 0.02x - \frac{\pi}{6})$. The distance is expressed in $cm$ and time in seconds. What will be the minimum distance between two particles having a phase difference of $\pi / 2$ in $cm$?

Similar Questions

The equation of a progressive wave is given by $y = a \sin \pi [\frac{t}{2} - \frac{x}{4}]$,where $t$ is in seconds and $x$ is in meters. The distance through which the wave moves in $8 \ s$ is .... $(m)$

What is a progressive wave?

The equation of a progressive wave is given by $x = 1.2 \sin(314t + 12.56y)$. What are the wavelength and the direction of the wave?

At a moment in a progressive wave,the phase of a particle executing $S.H.M.$ is $\frac{\pi}{3}$. What will be the phase of a particle $15 \ cm$ ahead at a time $\frac{T}{2}$ later,if the wavelength is $60 \ cm$?

$A$ sinusoidal progressive wave is generated in a string. Its equation is given by $y = (2 \text{ mm}) \sin(2\pi x - 100\pi t + \pi/3)$. The time when the particle at $x = 4 \text{ m}$ first passes through the mean position will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module