What are the three longest wavelengths for standing sound waves in a 124-cm-long tube that is (a) open at both ends and (b) open at one end, closed at the other?
You have to do some thinking. An open end gives a reflection of zero degrees, add that to two half wavelengths of 1.24m.
A closed end gives 180 degrees phase shift, so 1.24m is now only 90 degrees to give a full wavelength for the round trip.
I will be happy to critique your thinking.
how do i go from degree to meter...answer is required in m.
360 degrees is one wavelength. So count the reflection (either 0 deg,or 180 deg ..1/2 wavelength), then the two paths of 1.24m which provide the rest of the shift.
part A i used the equation n*(lambda/2)=length, for n- 1,2, and 3. this is coming back as incorrect
I would put 1.24 in for length and solve for the three wavelengths, that is what it asked.
now i am confused ...dont know where to begin
To find the three longest wavelengths for standing sound waves in a tube with different end conditions, we need to consider the fundamental frequency and its harmonics. The fundamental frequency is the lowest resonant frequency of the tube, while the harmonics are integer multiples of the fundamental frequency.
(a) Tube open at both ends:
For a tube open at both ends, the length of the tube corresponds to exactly half a wavelength (n/2). The fundamental frequency (n = 1) has one antinode in the center and gives half a wavelength. The second harmonic (n = 2) has two antinodes and gives a full wavelength, and so on.
To find the wavelength for the fundamental frequency in this case, we can use the formula:
λ = 2L/n
where λ is the wavelength, L is the length of the tube, and n is the harmonic number.
Given that L (length of the tube) is 124 cm, we can calculate the wavelength for the fundamental frequency:
λ₁ = 2L/n = 2(124 cm)/1 = 248 cm
Similarly, we can calculate the wavelengths for the second (n = 2) and third (n = 3) harmonics:
λ₂ = 2L/n = 2(124 cm)/2 = 124 cm
λ₃ = 2L/n = 2(124 cm)/3 ≈ 82.67 cm
Therefore, the three longest wavelengths for a tube open at both ends are 248 cm, 124 cm, and 82.67 cm.
(b) Tube open at one end, closed at the other:
For a tube open at one end and closed at the other, the length of the tube corresponds to a quarter wavelength (n/4). The fundamental frequency (n = 1) has one quarter-wavelength and one antinode at the open end. The second harmonic (n = 3) has three quarter-wavelengths and two antinodes, and so on.
To find the wavelength for the fundamental frequency in this case, we can use the formula:
λ = 4L/n
Using the given length of the tube (124 cm), we can calculate the wavelengths for the fundamental (n = 1), second (n = 3), and third (n = 5) harmonics:
λ₁ = 4L/n = 4(124 cm)/1 = 496 cm
λ₂ = 4L/n = 4(124 cm)/3 ≈ 165.33 cm
λ₃ = 4L/n = 4(124 cm)/5 ≈ 99.2 cm
Therefore, the three longest wavelengths for a tube open at one end and closed at the other are 496 cm, 165.33 cm, and 99.2 cm.