# 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

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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