The equation of a wave is y = 60 sin (1200 t | vedclass

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

Question

(A) The given wave equation is $y = A \sin (\omega t - k x)$,where $A = 60 \ \mu m = 60 	imes 10^{-6} \ m$,$\omega = 1200 \ rad/s$,and $k = 6 \ m^{-1

Vedclass · Accepted Answer

$3.6 	imes 10^{-4}$

Vedclass · Answer

(A) The given wave equation is $y = A \sin (\omega t - k x)$,where $A = 60 \ \mu m = 60 	imes 10^{-6} \ m$,$\omega = 1200 \ rad/s$,and $k = 6 \ m^{-1}$.The maximum particle velocity $(v_p)$ is given by $v_p = A \omega$.$v_p = (60 	imes 10^{-6} \ m) 	imes (1200 \ rad/s) = 72000 	imes 10^{-6} \ m/s = 7.2 	imes 10^{-2} \ m/s$.The wave velocity $(v_w)$ is given by $v_w = \frac{\omega}{k}$.$v_w = \frac{1200}{6} = 200 \ m/s$.The ratio of maximum particle velocity to wave velocity is $\frac{v_p}{v_w} = \frac{A \omega}{\omega/k} = A k$.$\frac{v_p}{v_w} = (60 	imes 10^{-6} \ m) 	imes (6 \ m^{-1}) = 360 	imes 10^{-6} = 3.6 	imes 10^{-4}$.

The equation of a wave is $y = 60 \sin (1200 t - 6 x)$,where '$y$' is in micron,'$t$' is in second,and '$x$' is in metre. The ratio of the maximum particle velocity to the wave velocity of wave propagation is:

Similar Questions

Two waves of frequencies $20 Hz$ and $30 Hz$ travel out from a common point. The phase difference between them after $0.6 s$ is:

The equation $y = A \cos^2 \left( 2\pi nt - 2\pi \frac{x}{\lambda} \right)$ represents a wave with

If a given wave has a propagation constant of $\frac{5 \pi}{7} \, rad/m$,then the phase difference between two particles having a distance difference of $\frac{49}{22} \, m$ is ..... $rad.$ (Take $\pi = \frac{22}{7}$)

Write the definition of the angular frequency of a wave and its $SI$ unit.

What is the phase difference between two successive crests in a wave?

Vedclass Test Series

Exam Paper Generator

Online Exam Module