A spaceship lands on a new planet. the crew decides to measure the speed of sound in the planets atmosphere. taking a tube with a length of 50 cm, they find that one resonant frequency occurs at 2,520 Hz, and the next resonant frequency at 2,940 hz. what is the speed of the sound? what is the number of each harmonic? what type of pipe is the crew using (open at both ends, closed at both ends, or one end open and the other closed)?
show steps please!
To find the speed of sound in the planet's atmosphere, we can use the formula:
v = λ * f
Where:
v is the speed of sound
λ (lambda) is the wavelength of sound waves
f is the frequency of the sound waves
First, let's find the wavelength (λ) using the given information.
The length of the tube (L) is 50 cm, and we know that one resonant frequency (f₁) occurs at 2,520 Hz, and the next resonant frequency (f₂) occurs at 2,940 Hz.
For an open pipe, the length of the pipe (L) is equal to one-fourth (1/4) of the wavelength (λ), and for a closed pipe, the length of the pipe (L) is equal to one-half (1/2) of the wavelength (λ).
In this case, the pipe is open at both ends. Therefore, the equation for the first harmonic (n = 1) is:
λ₁ = 4L
Substituting the given length (50 cm):
λ₁ = 4 * 50 cm = 200 cm
Now, let's use the equation for the second harmonic (n = 2):
λ₂ = 4L / 2 = 2L
We know the resonant frequency (f₂) for the second harmonic:
λ₂ = v / f₂
Rearranging the equation, we can solve for the wavelength (λ₂):
λ₂ = v / f₂
Now we can equate these two expressions for wavelength (λ₁ and λ₂):
λ₁ = λ₂
4L = 2L
2L = 200 cm
L = 100 cm
Now that we know the length (L) of the pipe, we can find the wavelength (λ) for the second harmonic:
λ₂ = 2L = 2 * 100 cm = 200 cm
To find the speed of sound (v), we can use the equation:
v = λf
Substituting the values we have:
v = λ₁ * f₁ = 200 cm * 2,520 Hz
v ≈ 504,000 cm/s
Therefore, the speed of sound in the planet's atmosphere is approximately 504,000 cm/s.
To determine the number of each harmonic, we can use the relationship between the frequency (f) and the harmonic number (n):
f = n * (v / λ)
Substituting the values we have:
For the first harmonic (n = 1):
f₁ = 1 * (504,000 cm/s / 200 cm) = 2,520 Hz
For the second harmonic (n = 2):
f₂ = 2 * (504,000 cm/s / 200 cm) = 2,940 Hz
Therefore, the first harmonic occurs at 2,520 Hz, and the second harmonic occurs at 2,940 Hz.
Since the pipe is open at both ends, we can conclude that the crew is using an open-ended pipe.