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: - Oscillations and Waves - JEE Main-10

Two vibrating string of same material but length l and 2l have raddii 2r & r. They are streched under same tension. Both the strings vibrate in their fundamental modes. Find out the ratio of frequencies n1/n2.

  • Option 1)

    2

  • Option 2)

    4

  • Option 3)

    8

  • Option 4)

    1

 

Answers (1)

best_answer

n=\frac{v}{2l}= \frac{\sqrt\frac{T}{M}}{2l}

M= mass / length 

M= \rho s 

n=\frac{1}{2l}= {\sqrt\frac{T}{\rho s}}

S = \pi r^{2} area

n \propto \frac{1}{rl}\Rightarrow \frac{n_{1}}{n_{2}}= \frac{r\times 2l}{2r\times l}= 1

 

Standing wave in a string fixed at both ends -

 

- wherein

\nu _{n}= \frac{n}{2L}.\sqrt{\frac{T}{\mu }}

n= 1,2,3...........

\nu _{n} frequency of nth harmonic .

T= tension in string

\mu = mass / length

 

 


Option 1)

2

This is incorrect

Option 2)

4

This is incorrect

Option 3)

8

This is incorrect

Option 4)

1

This is correct

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