What would happen the the wavelength of an organ pipe if you fill it with helium instead of air? What would happen to the fundamental frequency? Explain
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To determine what would happen to the wavelength and fundamental frequency of an organ pipe if it is filled with helium instead of air, we need to understand the concept of standing waves in a closed pipe.
In a closed pipe, such as an organ pipe, the fundamental frequency and wavelength are determined by the length of the pipe. The fundamental frequency, represented by f₁, is the lowest frequency at which the pipe can resonate or vibrate. The wavelength, represented by λ₁, corresponds to the distance between two adjacent nodes (points of zero amplitude) in the standing wave pattern.
The formula that relates the fundamental frequency, wavelength, and the speed of sound in a medium is:
v = fλ,
where:
- v is the speed of sound in the medium,
- f is the fundamental frequency, and
- λ is the wavelength.
Now let's consider the effect of replacing air with helium in the organ pipe. The speed of sound in a medium is determined by the properties of the medium, such as its density and elasticity. Since helium is less dense than air, it has a different speed of sound.
When calculating the fundamental frequency, we can assume that the speed of sound in the new medium is given by the equation:
v₂ = v₁√(γ₁/γ₂),
where:
- v₂ is the speed of sound in helium,
- v₁ is the speed of sound in air (approximately 343 m/s at room temperature),
- γ₁ is the specific heat ratio (ratio of specific heat at constant pressure to specific heat at constant volume) for air (approximately 1.4), and
- γ₂ is the specific heat ratio for helium (approximately 1.67).
Now, we can use the relationship between the fundamental frequency, wavelength, and speed of sound to determine the change in wavelength when the medium changes, assuming the length of the pipe remains constant.
Since the length of the pipe stays the same, the wavelength will also remain the same. Therefore, λ₂ = λ₁.
Using the equation v = fλ, we can rewrite it as:
λ₂ = v₂/f₂,
where f₂ is the new fundamental frequency in helium.
Now we can substitute the expressions for v and λ:
λ₂ = (v₁√(γ₁/γ₂))/f₂.
Since λ₂ = λ₁, we can write:
λ₁ = (v₁√(γ₁/γ₂))/f₁.
Now, we can express the relationship between f₁ and f₂:
f₂ = (v₁√(γ₁/γ₂))/λ₁.
We can see that if the speed of sound in helium is lower than the speed of sound in air, the new fundamental frequency (f₂) would be higher for the same length of the pipe. Conversely, if the speed of sound in helium is higher, the new fundamental frequency would be lower.
In summary, when an organ pipe is filled with helium instead of air, the wavelength would remain the same, assuming the length of the pipe remains constant. However, the fundamental frequency would change depending on the speed of sound in helium compared to air.
To calculate the exact change, you would need to know the specific length of the organ pipe and the speed of sound in helium.