1) For an open-open tube, the wavelengths are: 2 m, 1 m, 0.67 m, 0.50 m
2) For a closed-end tube, the wavelengths are: 4 m, 1.33 m, 0.80 m, 0.57 m
Explanation:
1)
For a tube open at both ends, both ends of the tube act as they are consecutive antinodes (in the fundamental mode of vibration). Since the distance between two consecutive antinodes is equal to half the wavelength, this means that the wavelength of the fundamental mode (first harmonics) is equal to twice the length of the tube:
[tex]\lambda_1 = 2L = 2(1 m) = 2 m[/tex]
The frequency of the 2nd harmonics is twice the frequency of the fundamental mode, this means that the wavelength of the 2nd harmonics is half the wavelength of the fundamental mode (remember that wavelength is inversely proportional to the frequency):
[tex]\lambda_2 = \frac{\lambda_1}{2}=1 m[/tex]
And therefore, the wavelengths of the 3rd and 4th harmonics are:
[tex]\lambda_3 = \frac{\lambda_1}{3}=\frac{2}{3}=0.67 m[/tex]
[tex]\lambda_4 = \frac{\lambda_1}{4}=\frac{2}{4}=0.50 m[/tex]
2)
For a tube closed at one end, the closed end acts as a node, while the open end acts as an antinode. The distance between a node and the next antinode is equal to 1/4 of the wavelength: therefore, the wavelength of the fundamental mode here is equal to 4 times the length of the tube,
[tex]\lambda_1 = 4 L = 4(1 m) = 4m[/tex]
In a closed end tube, only the odd modes of vibration are allowed. This means that the second harmonic has a frequency of
[tex]f_3 = 3 f_1[/tex]
And therefore, the corresponding wavelength of the second mode of vibration is
[tex]\lambda_3 = \frac{\lambda_1}{3}=\frac{4}{3}=1.33 m[/tex]
And continuing the sequence, the wavelengths of the 3rd and 4th mode of vibration are:
[tex]\lambda_5 = \frac{\lambda_1}{5}=\frac{4}{5}=0.80 m[/tex]
[tex]\lambda_7 = \frac{\lambda_1}{7}=\frac{4}{7}=0.57 m[/tex]
Learn more about waves and wavelength:
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