The distance between the successive node and | vedclass

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

Question

(C) In a stationary wave,nodes $(N)$ are points of minimum displacement (zero amplitude),and anti-nodes $(A)$ are points of maximum displacement (maxi

Vedclass · Accepted Answer

$\frac{\lambda}{4}$

Vedclass · Answer

(C) In a stationary wave,nodes $(N)$ are points of minimum displacement (zero amplitude),and anti-nodes $(A)$ are points of maximum displacement (maximum amplitude).The distance between two consecutive nodes is $\frac{\lambda}{2}$.The distance between two consecutive anti-nodes is $\frac{\lambda}{2}$.The distance between a node and the next consecutive anti-node is exactly half the distance between two consecutive nodes,which is $\frac{1}{2} 	imes \frac{\lambda}{2} = \frac{\lambda}{4}$.Therefore,the distance between the successive node and anti-node is $\frac{\lambda}{4}$.

The distance between the successive node and anti-node is

Similar Questions

In a stationary wave,which of the following statements is correct regarding strain?

Two identical progressive waves moving in opposite directions superimpose to produce a stationary wave. The wavelength of each progressive wave is $\lambda$. The wavelength of the stationary wave 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:

$A$ stretched string fixed at both ends has $m$ nodes,then the length of the string will be

The distance between the nearest node and antinode in a stationary wave is

Vedclass Test Series

Exam Paper Generator

Online Exam Module