Respuesta :
Answer:
The lowest frequency is 45.01 Hz.
The second lowest frequency is 135.03 hz.
The highest frequency is 19993.4 Hz.
Explanation:
Given that,
Length = 1.85 m
Range of frequency = 20 hz - 20000 Hz
Temperature = 18.0°C
Suppose, the tube is closed at one end lowest frequency , second lowest frequency and highest frequency
We need to calculate the velocity of sound
Using formula of sound velocity
[tex]v=v_{0}+0.61\times T[/tex]
Put the value into the formula
[tex]v=332+0.061\times18[/tex]
[tex]v=333.09\ m/s[/tex]
(a). For closed end,
We need to calculate the lowest and second lowest frequency
Using formula of frequency
[tyex]f_{n}=(2n+1)\times\dfrac{v}{4l}[/tex]
If n =0
[tex]f_{0}=\dfrac{v}{4l}[/tex]
Put the value into the formula
[tex]f_{0}=\dfrac{333.09}{4\times1.85}[/tex]
[tex]f_{0}=45.01\ Hz[/tex]
If n = 1
[tex]f_{1}=\dfrac{3v}{4l}[/tex]
Put the value into the formula
[tex]f_{1}=\dfrac{3\times333.09}{4\times1.85}[/tex]
[tex]f_{1}=135.03\ Hz[/tex]
Now, The maximum audible range is 20000 Hz.
We need to calculate the value of n
Using formula of frequency
[tex]f_{n}=(2n+1)\dfrac{v}{4l}[/tex]
Put the value into the formula
[tex]20000=(2n+1)\times45.01[/tex]
[tex]20000=2n\times45.01+45.01[/tex]
[tex]n=\dfrac{20000-45.01}{2\times45.01}[/tex]
[tex]n=221.6[/tex]
We need to calculate the maximum frequency
Using formula of frequency
[tex]f_{n}=(2n+1)\dfrac{v}{4l}[/tex]
Put the value into the formula
[tex]f_{221.6}=(2\times221.6+1)\times45.01[/tex]
[tex]f_{221.6}=19993.4\ Hz[/tex]
Hence, The lowest frequency is 45.01 Hz.
The second lowest frequency is 135.03 hz.
The highest frequency is 19993.4 Hz.
