n(1) is the frequency of the pipe closed...

`n_(1)` is the frequency of the pipe closed at one and `n_(2)` is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe so formed

`(n_(1)n_(2))/(2n_(2) + n_(1))`

`(2n_(2) n_(1))/(2n_(2) + n_(1))`

`(2n _(2) n_(1))/(n_(1) + n_(2))`

`(n_(2) + 2n_(1))/(n_(1) n_(2))`

Text Solution

Verified by Experts

The correct Answer is:

For a pipe open at both ends,
`n_(1) = v/(2L_(1)) :. L_(1) = v/(2n_(1))`
and for a pipe closed at one end,
`n_(2) = v/(4L_(2)) :. L_(2) = v/(4n_(2))`
When they are joined end to end, the new length
`L = L_(1) + L_(2) = v/2 [1/n_(1) + 1/(2n_(2))]`
` :. L_(1) + L_(2) = v/4 [(2n_(2) + n_(1))/(n_(1)n_(2))]`
` :. ` The new fundamental frequency of the pipe closed at one end will be
`n = v/(4(L_(1)+L_(2))) = (v*4 * n_(1)n_(2))/(4v(2n_(2) + n_(1))) = (n_(1)n_(2))/(2n_(2) + n_(1))`

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The fundamental frequency of a pipe closed at one end is 100Hz. If close end is open the fundamental frequency of same pipe wil be

100 Hz

200 Hz

50 Hz

400 Hz

n_1 " and " n_2 are the fundamental frequency of two open pipes of same diameter . If these two are joined end to end to form as single pipe, the fundamental frequency of the new pipe formed is

`n_1 +n_2`

`(n_1+n_2)/(2)`

`(n_1n_2)/(n_1+n_2)`

`(n_1+n_2)/(n_1n_2)`

What is the shortest length of a pipe open at one end and closed to other end resonates with the fundamental frequency n ?

v/2n

v/4n