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`)

Options

A. Upper end of the tube and lower end of the tube are respectively open and closed
B. Air column is vibrating in fundamental mode
C. Air column is vibrating in second overtone
D. Equation of the standing wave in terms of excess pressure is `DeltaP=(125piN//m^(2))sin((2pi)/(0.8))(y+0.01m)cos(800pit)`

Answer

B

Explanation

`epsi=0.1mm" "cos((2pi))/(0.8))(y+1cm)cos2pi(400)t`
correction is 1 cm so at y=-1 cm
Lower end y=99cm
`epsi=0 implies`Node so it is closed.
`(2pi)/(0.8)=(2pi)/(lamda)implies lamda=0.8` effective length of air column `0.99+0.01=1m`
`(l)/(lamda)=(5)/(4) implies l=(5lamda)/(4) implies`second overtone
`DeltaP=-B(depsi)/(dx)=(125pi(N)/(m^(2)))sin((2pi)/(0.8))(y+1cm)cos(800pit)`.