What frequencies will a 150 m long tube produce in the audib

What frequencies will a 1.50 m long tube produce in the audible range (20 Hz - 20,000 Hz) at 15.0°C for the following cases? (a) the tube is closed at one end lowest frequency Hz second lowest frequency Hz highest frequency (rounded to the nearest Hz) (b) the tube is open at both ends lowest frequency Hz Hz second lowest frequency Hz highest frequency (rounded to the nearest Hz)

Solution

a) for closed at one end :

L = n lambda /2    + lambda/4

4L = (2n + 1) lambda

lambda = 4L / (2n + 1)

and frequency, f = v / lambda

at 15 deg C, v = 340.4 m/s

f = 340.4 / (4 L / (2n + 1))

f = (2n + 1) (340.4) / (4 L )

and L = 1.50 m

f = (2n + 1) ( 56.73)

where n = 0, 1 , 2 ..

for lowest, n = 1

f = 56.73 Hz ............lowest

for highest :

for n = 176

f = (2 x176 + 1) (56.73) =20025.69

hence at 175

f = (2x175 + 1) (56.73) = 19912.23 Hz .........Highest

b) When tube is open to the both side:

(n +1) * lambda / 2 = L

lambda = 2 L / (n + 1)

where L = 0. 1 , 2 , ....

f = v / (2 L / (n +1) )

f = (n +1 ) (v/2L) = (n + 1) (340.4 / (2 x 1.5))

f = ( n + 1) 113.47

lowest :

n = 0

f = 113.47 hz

highest:

n = 176

f = (175 + 1) 113.47 = 19970.72 Hz ......Ans