A tube with a length L is closed at both ends and is filled with air. The wavelength of the fundamental mode of this tube is given by L*c, where c is a dimensionless constant. What is this constant, c, numerically? Also, the frequency of the first overture is how many times the fundamental frequency of this closed tube (numerically)?

start in the middle. the resonance to each end is 1/4 lambda, so the totalresonance is 1/2 lambda. C is 1/2

I tried it but the answer is still wrong =/. I'm not so sure if this has something to do with the other equation n = c/v but I doubt it.

To find the constant 'c' and the ratio of the frequency of the first overtone to the fundamental frequency, let's first understand the fundamental mode and the first overtone of a closed tube.

A closed tube has both ends sealed, which means it has a node (point of zero displacement) at both ends and an antinode (point of maximum displacement) in the middle. The fundamental mode has a single half-wavelength between the two ends of the tube.

The wavelength of the fundamental mode (λ₁) is equal to 2 times the length of the closed tube (L), so we have:
λ₁ = 2L

To find the constant 'c', we need to relate the wavelength (λ₁) to the fundamental frequency (f₁) using the speed of sound (v). The formula for the speed of sound is:
v = f₁ * λ₁

Rearranging the equation, we get:
λ₁ = v / f₁

Substituting the value of λ₁ from above, we have:
2L = v / f₁
f₁ = v / (2L)

Thus, the constant 'c' can be calculated as the inverse of 2:
c = 1 / 2 = 0.5

Now let's consider the ratio of the frequency of the first overtone (f₂) to the fundamental frequency (f₁) for a closed tube. The first overtone has two half-wavelengths between the two ends of the tube, which gives us:
λ₂ = L

Using the formula for the speed of sound (v) again, we have:
λ₂ = v / f₂

Substituting the value of λ₂ from above, we get:
L = v / f₂
f₂ = v / L

To find the ratio of the frequency of the first overtone to the fundamental frequency, we divide f₂ by f₁:
(f₂ / f₁) = (v / L) / (v / (2L)) = (v / L) * (2L / v) = 2

Therefore, the frequency of the first overtone is twice (numerically) the frequency of the fundamental frequency for this closed tube.

In summary:
- The constant 'c' is numerically 0.5.
- The frequency of the first overtone is twice the frequency of the fundamental mode.