Air is vibrating in a tube (closed at both ends) at its fundamental frequency of 160 Hz. If the tube is filled with helium instead, what is its fundamental frequency? Use 1020 m/s for the speed of sound in helium.
To find the fundamental frequency of the tube filled with helium, we need to consider the relationship between the fundamental frequency and the speed of sound.
The fundamental frequency of a vibrating tube is determined by the length of the tube and the speed of sound in the medium it contains. The fundamental frequency corresponds to the first harmonic, in which the tube has exactly one-half of a wave oscillating between the two ends.
In a closed tube, such as the one described, there are nodes at both ends, which means the length of the tube (L) is equal to one-fourth of the wavelength (λ) of the fundamental frequency.
Mathematically, we can express this relationship as:
λ = 4L
The speed of sound (v) is related to the frequency (f) and wavelength (λ) by the formula:
v = fλ
Rearranging this equation, we can solve for the frequency:
f = v / λ
Now, let's calculate the fundamental frequency (f₁) of the tube filled with helium. Given that the speed of sound in helium is 1020 m/s and the original fundamental frequency is 160 Hz.
First, we need to find the original wavelength (λ₁) using the formula:
λ₁ = 4L₁
Since the tube is closed at both ends, the length of the tube (L₁) is equal to one-fourth of the original wavelength (λ₁). Thus:
L₁ = λ₁ / 4
Substituting the value of L₁ back into the wavelength formula:
λ₁ = 4(λ₁ / 4)
Simplifying, we get:
λ₁ = λ₁
This means that the original wavelength (λ₁) remains the same.
Next, we can use the formula for frequency to find the new fundamental frequency (f₂) in helium:
f₂ = v / λ₂
Since the wavelength (λ₂) in helium is the same as the original wavelength (λ₁):
f₂ = v / λ₁
Substituting the given values:
f₂ = 1020 / λ₁
Finally, we can substitute the original fundamental frequency (160 Hz) to get the frequency in helium:
f₂ = 1020 / λ₁ = 1020 / (160 Hz)
Simplifying this, we find:
f₂ = 6.375 Hz
Therefore, the fundamental frequency of the tube filled with helium is 6.375 Hz.