Give the distance between consecutive nodes and antinodes in terms of wavelength $\lambda$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) In a stationary wave,the distance between two consecutive nodes is $\frac{\lambda}{2}$.
Similarly,the distance between two consecutive antinodes is $\frac{\lambda}{2}$.
The distance between a node and its consecutive antinode is half of the distance between two consecutive nodes.
Therefore,the distance between a consecutive node and antinode is $\frac{1}{2} \times \frac{\lambda}{2} = \frac{\lambda}{4}$.

Explore More

Similar Questions

In case of a stationary wave pattern,which of the following statements is $CORRECT$?

Stationary waves are produced in a $10\,m$ long stretched string. If the string vibrates in $5$ segments and the wave velocity is $20\,m/s$,the frequency is ..... $Hz$.

For a stationary wave,$Y = 10 \sin \left( \frac{\pi x}{15} \right) \cos (48 \pi t) \text{ cm}$,the distance between a node and the successive antinode is (in $\text{ cm}$)

The figure shows an incident pulse $P$ reflected from a rigid support. Which one of $A, B, C, D$ represents the reflected pulse correctly?

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

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo