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) x = 0.098
b) x = 2.72 m
Explanation:
(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.
When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.
Then, you have:
(1)
m: mass of the person = 62kg
k: spring constant = ?
v: velocity of the person just when he touches the fire net = ?
x: elongation of the fire net = 1.4 m
Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:
vo: initial velocity = 0 m/s
g: gravitational acceleration = 9.8 m/s^2
y: height from the person jumps = 20.0m
With this value you can find the spring constant k from the equation (1):
When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:
you solve the last expression for x:
When the person is lying on the fire net the elongation of the fire net is 0.098m
b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:
Next, you calculate x from the equation (1):
The net fire is stretched 2.72 m