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)
D Divya Saini

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

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