An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of
1 answer:
Length of the pipe = 0.39 m
Third harmonic frequency = 1400 Hz
For the third harmonic:
Wavelength =
The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength
Distance from center = 0.25 × wavelength
Distance =
Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm
Distance =
Distance from the center to the nearest anti - node = 6.5 cm
Hence, the nearest distance to the anti - node from the center = 6.5 cm
So, option C is correct.
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Answer:
(a) 89 m/s
(b) 11000 N
Explanation:
Note that answers are given to 2 significant figures which is what we have in the values in the question.
(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.
(b) The tension, , is given by
where is the speed,
is the tension and
is the mass per unit length.
Hence,
To determine , we need to know the mass of the cable. We use the density formula:
where is the mass and
is the volume.
If the length is denoted by , then
The density of steel = 8050 kg/m3
The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is