✔ 最佳答案
Fundemental frequency (fo) is the lowest frequency at which a pipe resonates with sound, when a Node at the bottom and an antinode at the top, with no additional Node or antinode in betweem.
Question 1)
wavelength of tuning fork = 340ms/400Hz = 0.85m
first fundemental wavelength at which resonates = 0.85/4 = 0.2125m
second fundemental wavelength = 0.85 x 3/4 = 0.6375m
Question 2)
Don't understand the diagram, don't too sure whether i get the question either but here is my guess:
L = 1/4 λo
Length of the pipe, L = 20m
λo = 4L = 4 x 20m = 80m... (this is the wavelength of the wave, a bit to great.... so i think i got the question wrong, but anyway, lets carry on.)
speed of wave, v = frequency (f) x wavelength (λ)
v = fλ
sub λo = 80m
hence,
340m/s = f x 80m
fundemental frequency (fo) = 340m/s /80m = 4.25 Hz
Overtones occurs at 3fo, 5fo, 7fo etc.
Hence,
1st overtone:
f = 3v/4L *explanation for this formula edited below
f = 12.75 Hz
The easier way => 3 x fundemental frequency (fo) = 3 x 4.25Hz (from above)
2nd overtone at 5fo:
5 x 4.25Hz = 21.25Hz
End of answer^^
*extra part, explanation of f = 3v/4L
back to the first step:
L = 1/4λo (why 1/4? because this is the fundemental mode, i.e. distance between one node and antinode)
but for the over tone, we use L = 3/4λo (Why? because this is the next f. mode i.e. Node/ Antinode/ Node/ Antinode. Draw it out, it helps)
ok. so L = 3/4λo
λo = 4/3L
next we refer to the formula: v = fλ, where λ = 4/3L, sub it into v= fλ to find:
v = f (4/3L)
BUT we are looking for the fundemental frequency, hence rearrange formula to find:
f = 3v/4L
Hope it helps^^ my follow physicist
參考: self try n error... LOL
