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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Answer:
distance = 6.1022 x10^16[m]
Explanation:
To solve this problem we must use the formula of the average speed which relates distance to time, so we have
v = distance / time
where:
v = velocity = 3 x 10^8 [m/s]
distance = x [meters]
time = 6.45 [light years]
Now we have to convert from light-years to seconds in order to get the distance in meters.
Now using the formula:
distance = v * time
distance = (3*10^8)*203407200
distance = 6.1022 x10^16[m]