The phase difference between the following tw | vedclass

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

Question

(B) Given equations are:$y_1 = a \sin(\omega t - kx)$$y_2 = b \cos(\omega t - kx + \frac{\pi}{3})$To find the phase difference,we convert the cosine f

Vedclass · Accepted Answer

$\frac{5\pi}{6}$

Vedclass · Answer

(B) Given equations are:$y_1 = a \sin(\omega t - kx)$$y_2 = b \cos(\omega t - kx + \frac{\pi}{3})$To find the phase difference,we convert the cosine function into a sine function using the identity $\cos(	heta) = \sin(	heta + \frac{\pi}{2})$.$y_2 = b \sin(\omega t - kx + \frac{\pi}{3} + \frac{\pi}{2})$$y_2 = b \sin(\omega t - kx + \frac{2\pi + 3\pi}{6})$$y_2 = b \sin(\omega t - kx + \frac{5\pi}{6})$The phase of $y_1$ is $\phi_1 = \omega t - kx$.The phase of $y_2$ is $\phi_2 = \omega t - kx + \frac{5\pi}{6}$.The phase difference $\Delta\phi = \phi_2 - \phi_1 = \frac{5\pi}{6}$.

The phase difference between the following two waves $y_1$ and $y_2$ is:
$y_1 = a \sin(\omega t - kx)$
$y_2 = b \cos(\omega t - kx + \frac{\pi}{3})$

Similar Questions

The displacement of two sinusoidal waves is given by the equations:
$y_1 = 8 \sin(20x - 30t)$
$y_2 = 8 \sin(25x - 40t)$
Then the phase difference between the waves at time $t = 2 \ s$ and distance $x = 5 \ cm$ will be:

$A$ plane progressive wave is given by $y = 2 \cos 6.284(330t - x)$. What is the period of the wave?

The figure shows the shape of a part of a long string in which transverse waves are produced by attaching one end of the string to a tuning fork of frequency $250 \ Hz$. What is the velocity of the waves in $ms^{-1}$?

The displacement $y$ of a wave travelling in the $x$-direction is given by $y = 10^{-4} \sin(600t - 2x + \frac{\pi}{3})$ metres,where $x$ is expressed in metres and $t$ in seconds. The speed of the wave motion,in $m \ s^{-1}$,is

What will be the change in phase of a wave due to reflection from a free support?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The phase difference between the following two waves $y_1$ and $y_2$ is:$y_1 = a \sin(\omega t - kx)$$y_2 = b \cos(\omega t - kx + \frac{\pi}{3})$