A wave represented by the equation y_1 = a co | vedclass

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

Question

(B) For a stationary wave to have a node at $x = 0$,the resultant displacement $y = y_1 + y_2$ must be zero at $x = 0$ for all $t$.Given $y_1 = a \cos

Vedclass · Accepted Answer

$a \cos(kx + \omega t + \pi)$

Vedclass · Answer

(B) For a stationary wave to have a node at $x = 0$,the resultant displacement $y = y_1 + y_2$ must be zero at $x = 0$ for all $t$.Given $y_1 = a \cos(kx - \omega t)$.At $x = 0$,$y_1 = a \cos(-\omega t) = a \cos(\omega t)$.For a node at $x = 0$,the second wave $y_2$ must satisfy $y_1 + y_2 = 0$ at $x = 0$,so $y_2 = -a \cos(\omega t)$.Since the wave is traveling in the opposite direction ($-x$ direction),the wave function must be of the form $y_2 = a \cos(kx + \omega t + \phi)$.At $x = 0$,$y_2 = a \cos(\omega t + \phi) = -a \cos(\omega t) = a \cos(\omega t + \pi)$.Comparing the arguments,we get $\phi = \pi$.Thus,the equation for the other wave is $y_2 = a \cos(kx + \omega t + \pi)$.

$A$ wave represented by the equation $y_1 = a \cos(kx - \omega t)$ is superimposed with another wave to form a stationary wave such that the point $x = 0$ is a node. The equation for the other wave is

Similar Questions

Two waves of same frequency $(n)$ are approaching each other with same velocity of $12 \ m/s$ along the same linear path and interfere. The distance between two consecutive nodes is

Standing stationary waves can be obtained in an air column even if the interfering waves are

What kind of waves can survive as standing waves?

$A$ travelling wave represented by $y = A \sin (\omega t - kx)$ is superimposed on another wave represented by $y = A \sin (\omega t + kx)$. The resultant is

$A$ wave represented by the equation $y = a \cos (kx - \omega t)$ is superposed with another wave to form a stationary wave such that the point $x = 0$ is a node. The equation for the other wave is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module