Here are some of the equations provided, plus the questions: The modal frequenci

Question

Here are some of the equations provided, plus the questions:

The modal frequencies of a pipe with two open ends are given by

fn = (n * v)/(2L)                                                                          (7)  

which is the same as for a string except that the speed of the wave is just the speed of sound in air (you can assume 343.5 m/s, but this will vary depending on the temperature of the air).

Similarly, if one end of the pipe is closed off, the node that is created at the end of the pipe causes the relationship between the length of the pipe and frequency to be

fp = (n * v)/(4L)

2a) Using the pipe equations, calculate the length of tube needed to create a fundamental frequency of 330 Hz for a tube with:

-both sides open

-one side is closed off

2b) Calculate the length of tube needed to create a 330 Hz frequency while accounting for end effects for both cases in part a. For this analysis, use a radius of the pipe of 0.5 in (0.0127 m).

Explanation / Answer

Solution:

2a)

I) From the given formula, for two open ends,

fn = (n * v)/(2L)

L = (n * v)/(2fn)

here n=1, v=343.5 m/s and fn= 330 Hz

then L = 1 * 343.5/(2*330)

= 0.5204 meter

II) from the formula for both open ends,

fp = (n * v)/(4L)

L =   (n * v)/(4fp)

= 1 * 343.5/(4*330)

= 0.2602 meter

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