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In an organ pipe ( may be closed or open...

In an organ pipe ( may be closed or open of `99 cm` length standing wave is setup , whose equation is given by longitudinal displacement `xi = (0.1 mm) cos ( 2pi)/( 0.8) ( y + 1 cm) cos 2 pi (400) t`
where `y` is measured from the top of the tube in metres and `t` in second. Here `1 cm` is th end correction.

Assume end correction approximately equals to `(0.3) xx`(diameter of tube) , estimate the moles of air pressure inside the tube (Assume tube is at `NTP` , and at `NTP , 22.4 litre` contain 1 mole)

A

First overtone

B

Second overtone

C

Third harmonic

D

Fundamental mode

Text Solution

Verified by Experts

The correct Answer is:
B
`(2pi)/(80) = (2pi)/(lambda)` So `lambda = 80`
and effective length of air column `= 99 + 1 = 100 cm`
So `(l)/(lambda) = (5)/(4) rArr 1 = 5(lambda)/(4)`, so five half loops will be formed
`l = 5 ((lambda)/(4))` so second overtone
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