Question

An organ pipe open at both ends has a length L. The resonant frequency is 283.3 Hertz and here are two half wavelengths and two quarter wavelengths established in the resonating open tube. The velocity of sound is 340 meter/second. Determine the length L of the tube.

a.) 1.40 meter

b.) 4.80 meter

c.) 1.80 meter

d.) 6.20 meter

Answer 1

we know that in organ pipes at open end always an antinode will form and the distance from a an antiinode to the next node is equal to the (1/4) of a wavelength and
the distance between any two antinodes and nodes is equal to half of the wavelength

so from the description given the resonant frequency is of third harmonic

we know that the relation between v,lambda,f; v = lambda*f ==> f = v/lambda

here L = 3*lambda/2 ==>lambda = 2*L/3

we know that in organ pipes the harmonic are given by

   fn = n*v/2L
where n values are n =1,2,3,4,5.......

   so the third harmonic is f3 = 3*v/2L

the length of the pipe in terms of wavelength is L = lambda/4+ lambda/2+lambda/2 + lambda/4 = 3*lambda/2

given resonant (3rd harmonic) frequency is f3 = 283.3 Hz ,
velocity of sound is v = 340 m/s

   f3 = 3*v/2L

   L = 3*v/(2*f3) m
  
   L = 3*340/(2*283.3) m

   L = 1.800 m

so the length of the organ pipe is L = 1.80 m

the answer is option (c) <<<<<-------------ANSWER