Question

given a 5.0m pipe measured end to end with an inner diameter of 15cm and a variable frequency mini speaker. assumed an air temperature of 30°C(hot).
a. one end of the pipe is closed by taping a circular wood cylinder over the hole at one end. the wood does not go inside the pipe,it covers its the hole. compute the lowest audible resonant frequency of the pipe.
b. sketch the node antinode pattern in the pipe for the resonant frequency you determined from part A.
c. what is the wavelength corresponding to the above frequency?
d. how could you experimentally determine the resonant frequency in part A using mini speaker?
e. repeat part a,,c with both pipe open. you will need to use a length correction on both ends.

Answer 1

When one of the ends is closed, standing waves are formed with the sound played inside the pipe such that nodes are formed at the closed end and antinodes are formed at the open end.

The smallest resonant frequency will thus be the fundamental mode whose wavelength will be:

=>

speed of propagation of sound is 349 m/s at 30 degree Celsius

so, the frequency for this is:

which is not in the audible range.

the next smallest wavelength will be:

so, the frequency for this will be:

this is the lowest audible resonant frequency of the pipe.

this will create a node at one of the closed ends, an antinode at the open end and another node at a distance of x = (2L/3) from the closed end.

the wavelength corresponding to this frequency is 6.667 Hz.