In a stationary wave, the distance between a node and the next antinode is $10cm$. What is the wavelength?
Answer
253.2k+ 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
States of Matter Chapter For JEE Main Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Circuit Switching vs Packet Switching: Key Differences Explained

Mass vs Weight: Key Differences Explained for Students

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

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

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Other Pages
JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

