Question

the first overtone of an open organ pipe beats with the first overtone of a closed organn pipe with a beat frequency 2.2Hz . the fundamental frequency of the closed organ pipe is 110Hz find the lengths of the pipes , take velocity of sound = > 330 m /s

Answer 1

Hello pari 

according to our question 

let L1 and L2 be the lengths of open organ pipe and closed organ pipe respectively 

first overtone of open organ pipe = > 2n1 = > 2 ×ν/2L2  = > v/L1 

first overtone of closed organ pipe = > 3n2 = > 3 ×v/4L2  = > +,-3.3

sd n2 is the fundamental frequency of closed organ pipe 

n2 => v/4L2= > 330/4×110 = > 0.75 m 

from equation 1 

v/L1 = > 3v/4L2 = > +,-2.2

taking positive sign v/L1 = > 3×110+2.2 = > 332.2 

L1 = > v/332.2 = >330m/332.2  = > 0.993 m = > 99.3 c, 

taking negative sign 

v/L1 = > 3n2 -2.2 = > 3×110 -2.2 = 327.8 

L1 = > v/327.8  = > 330/327.8 => 1.006 m = > 100.6 cm is our required

answer pari hope this helps you !

@ engineer gopal 

Answer 2

$\mathbb{ANSWER:-}$

1.006 m

$\mathfrak{STEP \:BY \:STEP \: SOLUTION:- \: }$

let L1 and L2 be the lengths of open organ pipe and closed organ pipe respectively

first overtone of open organ pipe = > 2n1 = > 2 ×ν/2L2 = > v/L1

first overtone of closed organ pipe = > 3n2 = > 3 ×v/4L2 = > +,-3.3

sd n2 is the fundamental frequency of closed organ pipe

n2 => v/4L2= > 330/4×110 = > 0.75 m

from equation 1

v/L1 = > 3v/4L2 = > +,-2.2

taking positive sign v/L1 = > 3×110+2.2 = > 332.2

L1 = > v/332.2 = >330m/332.2 = > 0.993 m = > 99.3 c,

taking negative sign

v/L1 = > 3n2 -2.2 = > 3×110 -2.2 = 327.8

L1 = > v/327.8 = > 330/327.8 => 1.006 m = > 100.6 cm is our required

Hope it may help you...

Please follow me ....

And mark this answer as Brainiest answer....