Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Need clarity, kindly explain! - Oscillations and Waves - JEE Main-10

A 1 cm long string vibrates with fundamental frequency of 256 Hz. If the length is reduced to \frac{1}{}4 cm keeping the tension unaltered, new fundamental frequency will be

  • Option 1)

    64

  • Option 2)

    256

  • Option 3)

    512

  • Option 4)

    1024

 

Answers (1)

best_answer

n\propto \frac{1}{l} \Rightarrow \frac{n_{2}}{n_{1}}= \frac{l_{1}}{l_{2}}

n_{2}=n_{1}\frac{l_{1}}{l_{2}}\Rightarrow n_{2}= \frac{1\times 256}{1/4} = 1024

 

Vibration of composite string when joint is a node (N) -

\frac{n_{1}}{n_{2}}= \frac{l_{1}}{l_{2}}\sqrt{\frac{\mu _{1}}{\mu _{2}}}

 

- wherein

n_{1}\: and\: n_{2} are number of loops in two cases

l_{1}\: and\: l_{2} are length of two parts

\mu_{1}\: and\: \mu_{2} are linear density in two parts

 

 


Option 1)

64

This is incorrect

Option 2)

256

This is incorrect

Option 3)

512

This is incorrect

Option 4)

1024

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

Engineering aspirant dies by suicide in Kota, fourth case this year
anam-khan

What is non-Aadhaar declaration in JEE Main 2025? Explain mismatch between image, face
anam-khan

Forgot JEE Main password 2025 for downloading admit card? Here’s what to do

Latest Question