Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The equation of a stationary wave is- <img alt="\mathrm{ y=0.8 \cos \left(\frac{\pi x}{20}\right) \sin 200 \pi t \text {, where } x }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

The equation of a stationary wave is- \mathrm{ y=0.8 \cos \left(\frac{\pi x}{20}\right) \sin 200 \pi t \text {, where } x } is in \mathrm{ \mathrm{cm} \: \: and\: \: t } is in second. The separation between consecutive nodes will be-
 

Option: 1

20 \mathrm{~cm}
 


Option: 2

10 \mathrm{~cm}
 


Option: 3

40 \mathrm{~cm}
 


Option: 4

30 \mathrm{~cm}


Answers (1)

best_answer

\mathrm{\frac{2 \pi}{\lambda}=\frac{\pi}{20} \Rightarrow \lambda=40 \mathrm{~cm}}

separation Between two consecutive nodes.

\mathrm{ =\frac{\lambda}{2}=\frac{40}{2}=20 \mathrm{~cm} \text {. } }

Hence option 1 is correct.

 

Posted by

mansi

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech Admissions 2025 LIVE: JEE Main, MET registration underway; how to apply
anam-khan

JEE Main 2025 Registration: NTA issues advisory for Aadhaar authentication, name mismatch issues
anam-khan

BTech Admissions 2025: JEE Main registration last date, direct link

Latest Question