Answer:
The frequencies are [tex](f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)[/tex]
Explanation:
From the question we are told that
The length of the ear canal is [tex]l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m[/tex]
The speed of sound is assumed to be [tex]v_s = 344 \ m/s[/tex]
Now taking look at a typical ear canal we see that we assume it is a closed pipe
Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as
[tex]f = \frac{v_s}{4 * l }[/tex]
substituting values
[tex]f = \frac{344}{4 * 0.013 }[/tex]
[tex]f = 6615.4 \ Hz[/tex]
Also the the second harmonic for the pipe (ear canal) is mathematically represented as
[tex]f_1 = \frac{3v_s}{4 * l}[/tex]
substituting values
[tex]f_1 = \frac{3 * 344}{4 * 0.013}[/tex]
[tex]f_1 = 19846.2 \ Hz[/tex]
Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies
