Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • How to solve this problem- - Oscillations and Waves - JEE Main-6

A simple harmonic progressive wave is represented by the equation y = 8\sin(0.1x -2t), where  x\; \&\; y are in cm and t is in second. At any instant the phase difference between two particles seperated by 2  cm is x- direction is 

  • Option 1)

    18\degree

  • Option 2)

    36\degree

  • Option 3)

    54\degree

  • Option 4)

    72\degree

 

Answers (1)

best_answer

y = a sin 2\pi \left [ \frac{t}{T} - \frac{x}{\lambda}\right ] \Rightarrow \lambda = 10 cm

\phi = \frac{2\pi}{\lambda} X path difference \Rightarrow \frac{2\pi}{10} X 2 = \frac{2}{5} X 180^{o} = 72^{o}

 

Relation between phase difference and path difference -

Phase difference \left ( \Delta \phi \right )

= \frac{2\pi }{\lambda }\times path \: dif\! \! ference\left ( \Delta x \right )\\\lambda =wave\; length

-

 

 

 


Option 1)

18\degree

This is incorrect.

Option 2)

36\degree

This is incorrect.

Option 3)

54\degree

This is incorrect.

Option 4)

72\degree

This is correct.

Posted by

prateek

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

JEE Mains 2025 session 1 registration deadline on November 22; apply on jeemain.nta.nic.in
anam-khan

BTech Admission 2025 Live: JEE Mains, SRMJEEE, MET registrations underway; MHT CET syllabus updates
anam-khan

Centre issues coaching guidelines, prohibits centres from using false claims like '100% selection'

Latest Question