Answer:
The person's ear canal length is 2.91 cm.
Explanation:
Given that,
Frequency = 3000 Hz
Temperature = 30°C
We need to calculate the speed
Using equation of sound wave with temperature
[texs]v(t)=331+0.6\times T[/tex]
Put the value of T
[tex]v(t)=331+0.6\times30^{\circ}\ C[/tex]
[tex]v(t)=349\ m/s[/tex]
We need to calculate the length
Using formula of closed organ pipe
[tex]f=\dfrac{v}{4L}[/tex]
[tex]L=\dfrac{v}{4f}[/tex]
Where, L = length
v = velocity
f = frequency
Put the value into the formula
[tex]L=\dfrac{349}{4\times3000}[/tex]
[tex]L=0.0291\ m[/tex]
[tex]L=2.91\times10^{-2}\ m[/tex]
[tex]L=2.91\ cm[/tex]
Hence, The person's ear canal length is 2.91 cm.
