Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Need explanation for: - Equation of travelling wave on stretched string of linear density 5g/m is where distance and time a - Oscillations and Waves - JEE Main

Equation of travelling wave on stretched string of linear density 5g/m is y = 0.003\sin \left ( 450t - 9x \right ) where distance and time are measured in SI units. The tension in the string is:

  • Option 1)

    5N

  • Option 2)

    7.5N

  • Option 3)

    10N

  • Option 4)

    12.5 N

Answers (1)

best_answer

 

Relation between particle velocity and wave speed -

V_{P}= -V\: \frac{dy}{dx}

- wherein

V_{P}= particle velocity

V = wave velocity

\frac{dy}{dx}= slope of curve

 

 

Speed of wave on string -

v= \sqrt{\frac{T}{\mu }}
 

- wherein

T= Tension in the string

\mu = linear mass density

 

V=\frac{\omega }{K}=\frac{450}{9}=50m/s

V=\sqrt{\frac{T}{M}}

\Rightarrow \frac{T}{M}=2500

T=2500\times 5\times 10^{-3}

=12.5 N

 

 

 

 


Option 1)

5N

Option 2)

7.5N

Option 3)

10N

Option 4)

12.5 N
Posted by

admin

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 Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’
anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year
anam-khan

What after JEE Main results 2024? All you need to know about JEE Adv, CSAB, JoSAA counselling

Latest Question