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  • 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

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