In a stationary wave, the distance between a node and the next antinode is $10cm$. What is the wavelength?
Answer
298.5k+ views
Hint: Node is a point of minimum displacement of the standing wave and antinode is the point of maximum displacement of the standing wave. To solve this question, we must know the distance between a node and an antinode, and then, how to relate the distance between them to the wavelength of the wave.
Complete step by step solution:
While observing a standing wave, we can see that the distance between a node and an antinode is one-half the distance between a crest and a trough.
So, next comes the relation between wavelength and the distance between a node and an antinode.
We know the distance between a crest and a trough is one-half the wavelength. Therefore, the distance between a node and an antinode is one-fourth the wavelength.
Hence, mathematically, we can write, assuming $l$ to be the distance between a node and an antinode:
$l=\dfrac{\lambda }{4}$
According to the question:
$l=10cm$
$\Rightarrow \dfrac{\lambda }{4}=10cm$
$\Rightarrow \lambda =40cm$
Therefore, the wavelength of the standing wave is $40cm$.
Note: We must be very careful in writing the mathematical relation of the distance between a node and an antinode and the wavelength, as while writing this relation is a very common silly mistake. We must not confuse between a standing wave and a stationary wave; they are the same wave.
Sometimes, there will be a problem like two waves traveling in opposite directions coincide, the distance of the node and antinode of the resultant wave is, say $xcm$. And we are supposed to the wavelength of the resultant wave. We must know that a standing wave is formed by the interference of two waves. Therefore, we have to find the same thing as we did in this question.
Complete step by step solution:
While observing a standing wave, we can see that the distance between a node and an antinode is one-half the distance between a crest and a trough.
So, next comes the relation between wavelength and the distance between a node and an antinode.
We know the distance between a crest and a trough is one-half the wavelength. Therefore, the distance between a node and an antinode is one-fourth the wavelength.
Hence, mathematically, we can write, assuming $l$ to be the distance between a node and an antinode:
$l=\dfrac{\lambda }{4}$
According to the question:
$l=10cm$
$\Rightarrow \dfrac{\lambda }{4}=10cm$
$\Rightarrow \lambda =40cm$
Therefore, the wavelength of the standing wave is $40cm$.
Note: We must be very careful in writing the mathematical relation of the distance between a node and an antinode and the wavelength, as while writing this relation is a very common silly mistake. We must not confuse between a standing wave and a stationary wave; they are the same wave.
Sometimes, there will be a problem like two waves traveling in opposite directions coincide, the distance of the node and antinode of the resultant wave is, say $xcm$. And we are supposed to the wavelength of the resultant wave. We must know that a standing wave is formed by the interference of two waves. Therefore, we have to find the same thing as we did in this question.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

