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A pipe is closed at one end produces a fundamental note of 412 Hz. It is cut into 2 parts of equal length. The fundamental notes produced by the two peices are

  • Option 1)

    (824, 1648)Hz

  • Option 2)

    (412, 824)Hz

  • Option 3)

    (206, 412)Hz

  • Option 4)

    (206, 824)Hz

 

Answers (1)

best_answer

f= \frac{v}{4l }= 412Hz

f_{1}= \frac{v}{l }=4\times 412Hz =1648

f_{2}= \frac{v}{2l }=2\times 412Hz =824

 

String fixed at one end and free from other end -

f_{n}= n\frac{V}{4L}

n= 1,3,5.......

 

- wherein

f_{1}= \frac{V}{4L} is fundamental frequency

 

 


Option 1)

(824, 1648)Hz

This is incorrect

Option 2)

(412, 824)Hz

This is incorrect

Option 3)

(206, 412)Hz

This is incorrect

Option 4)

(206, 824)Hz

This is incorrect

Posted by

divya.saini

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