Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Please,please help me - Oscillations and Waves - JEE Main-7

Equation y = 0.15\sin 5x\cos 300t describes a stationary wave. The wavelength of stationary wave is 

  • Option 1)

    0

  • Option 2)

    1.256 mtr

  • Option 3)

    2.512 mtr

  • Option 4)

    0.62 mtr

 

Answers (1)

best_answer

Compare with standard equation 

\frac{2\pi }{\lambda }= 5\Rightarrow \lambda = \frac{6.25}{5}= 1.256 mtr

 

Equation of stationary waves -

y_{1}= A\sin \left ( kx- \omega t \right )

y_{2}= A\sin \left ( kx+ \omega t \right )

 

- wherein

Resultant wave

y= \left ( 2A\sin kx.\cos \omega t \right )

 

 


Option 1)

0

This is incorrect

Option 2)

1.256 mtr

This is correct

Option 3)

2.512 mtr

This is incorrect

Option 4)

0.62 mtr

This is incorrect

Posted by

Aadil

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 aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question