A pipe of length 6.8 m is closed at one end and sustains a standing wave at its second overtone. determine the distance between
a node and an adjacent antinode.
1 answer:
For a pipe closed at one end, the second overtone is the 3rd natural frequency as shown in the diagram.
For the fundamental frequency,
λ = 4L
where
λ = wavelength
L = the length of the tube
For the second overtone (3rd frequency)
λ = (4/3)L
Because L = 6.8 m, therefore
λ = (4/3)*(6.8 m) = 9.0667 m
The distance between a node and an adjacent antinode is λ/2, which is
9.0667/2 = 4.533 m
Answer: 4.5 m
For the fundamental frequency,
λ = 4L
where
λ = wavelength
L = the length of the tube
For the second overtone (3rd frequency)
λ = (4/3)L
Because L = 6.8 m, therefore
λ = (4/3)*(6.8 m) = 9.0667 m
The distance between a node and an adjacent antinode is λ/2, which is
9.0667/2 = 4.533 m
Answer: 4.5 m
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<h2>The temperature of the air is 66.8° C</h2>
Explanation:
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= 798 x 0.48 = 383 m/sec
Suppose the temperature at this time = T K
Thus 383 ∝ I
The velocity of sound is 329 m/s at 273 K ( given )
Thus 329 ∝ II
Dividing I by II , we have
=
or = 1.25
and T = 339.8 K = 66.8° C