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?
I did 340/(4*1.24)=68.54
3*340/(4*1.24)=205.64
but i am confused
To find the three longest wavelengths for standing sound waves in a 124-cm-long tube that is open at both ends, you can use the formula:
λ = 2L/n
where λ represents the wavelength, L represents the length of the tube, and n is the harmonic number.
For an open tube, the harmonics are odd multiples of the fundamental frequency, which is the frequency of the first harmonic.
So, for an open tube, the three longest wavelengths would correspond to the third, fifth, and seventh harmonics. To find their values, you can calculate:
λ1 = 2(124 cm)/1 = 248 cm
λ3 = 2(124 cm)/3 = 82.67 cm
λ5 = 2(124 cm)/5 = 49.6 cm
Therefore, the three longest wavelengths corresponding to the open tube are 248 cm, 82.67 cm, and 49.6 cm.
For a tube that is open at one end and closed at the other, the harmonics are odd integer multiples of a quarter wavelength. In this case, the formula for the wavelength is:
λ = 4L/n
Using the same approach as before to find the three longest wavelengths, you obtain:
λ1 = 4(124 cm)/1 = 496 cm
λ3 = 4(124 cm)/3 = 165.33 cm
λ5 = 4(124 cm)/5 = 99.2 cm
Hence, the three longest wavelengths corresponding to the tube that is open at one end and closed at the other are 496 cm, 165.33 cm, and 99.2 cm.
For a tube that is open at both ends:
To find the three longest wavelengths, we can start by calculating the fundamental frequency, which corresponds to the first harmonic (n = 1).
The speed of sound in air is approximately 340 m/s.
(a) Open at both ends:
The fundamental frequency is given by f1 = v / λ1, where v is the speed of sound and λ1 is the wavelength of the first harmonic.
Since the tube is open at both ends, the antinodes (points of maximum displacement) will be at the ends. This means that the wavelength is equal to twice the length of the tube, λ1 = 2L = 2(124 cm) = 248 cm.
Using the formula f1 = v / λ1, we can find the fundamental frequency:
f1 = 340 m/s / 2.48 m = 137.10 Hz.
The second harmonic (n = 2) will have a wavelength that is half that of the fundamental frequency, λ2 = λ1 / 2 = 124 cm / 2 = 62 cm.
Using the formula f2 = v / λ2, we can find the frequency for the second harmonic:
f2 = 340 m/s / 0.62 m = 548.39 Hz.
Similarly, for the third harmonic (n = 3), the wavelength will be λ3 = λ1 / 3 = 124 cm / 3 = 41.33 cm.
Using the formula f3 = v / λ3, we can find the frequency for the third harmonic:
f3 = 340 m/s / 0.41 m = 825.97 Hz.
Therefore, the three longest wavelengths for a tube that is open at both ends are:
λ1 = 248 cm (corresponding to a frequency of 137.10 Hz),
λ2 = 62 cm (corresponding to a frequency of 548.39 Hz), and
λ3 = 41.33 cm (corresponding to a frequency of 825.97 Hz).
(b) Open at one end, closed at the other:
In this case, the antinode is at the open end and the node (point of no displacement) is at the closed end. This means that the tube length corresponds to a quarter-wavelength, L = λ1 / 4.
Using the formula f1 = v / λ1, we can find the fundamental frequency:
f1 = 340 m/s / 4.96 m = 68.55 Hz.
For the second harmonic (n = 2), the wavelength will be λ2 = λ1 / 2 = 2L = 2(4.96 m) = 9.92 m.
Using the formula f2 = v / λ2, we can find the frequency for the second harmonic:
f2 = 340 m/s / 9.92 m = 34.27 Hz.
Similarly, for the third harmonic (n = 3), the wavelength will be λ3 = λ1 / 3 = (4/3)L = (4/3)(4.96 m) = 6.61 m.
Using the formula f3 = v / λ3, we can find the frequency for the third harmonic:
f3 = 340 m/s / 6.61 m = 51.51 Hz.
Therefore, the three longest wavelengths for a tube that is open at one end and closed at the other are:
λ1 = 4.96 m (corresponding to a frequency of 68.55 Hz),
λ2 = 9.92 m (corresponding to a frequency of 34.27 Hz), and
λ3 = 6.61 m (corresponding to a frequency of 51.51 Hz).