Respuesta :
Explanation:
We know
the frequency of the tube with one end open and the other end closed follows the given relations as
[tex]f_{1}[/tex] : [tex]f_{2}[/tex] : [tex]f_{3}[/tex] : [tex]f_{4}[/tex] = 1 : 3 : 5 : 7
∴ the 4th allowed wave is [tex]f_{4}[/tex] = 7 [tex]f_{1}[/tex]
= [tex]\frac{7v}{4l}[/tex]
We know [tex]f_{4}[/tex] = 1975 Hz and v = 343 m/s ( as given in question )
∴[tex]l = \frac{7\times v}{4\times f_{4}}[/tex]
[tex]l = \frac{7\times 343}{4\times 1975}[/tex]
= 0.303 m
We know that v = [tex]f_{4}[/tex] x [tex]λ_{4}[/tex]
[tex]\lambda _{4}= \frac{v}{f_{4}}[/tex]
[tex]\lambda _{4}= \frac{343}{1975}[/tex]
= 0.17 m
Now when the warmer air is flowing, the speed gets doubled and the mean temperature increases. And as a result the wavelength increases but the amplitude and the frequency remains the same.
So we can write
v ∝ λ
or [tex]\frac{v_{1}}{v_{2}}= \frac{\lambda _{1}}{\lambda _{2}}[/tex]
Therefore, the wavelength becomes doubled = 0.17 x 2
= 0.34 m
Now the new length of the air column becomes doubled
∴ [tex]l^{'}[/tex] = 0.3 x 2
= 0.6 m
∴ New speed, [tex]v^{'}[/tex] = 2 x 343
= 686 m/s
∴ New frequency is [tex]f^{'}=\frac{v^{'}}{4\times l^{'}}[/tex]
[tex]f^{'}=\frac{686}{4\times 0.6}[/tex]
= 283 Hz
∴ The new frequency remains the same.
Now we know
[tex]v_{s}[/tex] = 12 m/s, [tex]v_{o}[/tex] = 4 m/s, [tex]f_{o}[/tex] = 1975 Hz
Therefore, apparent frequency is [tex]f^{'}=f^{o}\left ( \frac{v+v_{s}}{v+v_{o}} \right )[/tex]
[tex]f^{'}=1975\left ( \frac{343+12}{343+4} \right )[/tex]
= 2020.5 Hz
