#$&*
course Phy 122
3/14 10:27This is a practice for Test 2.
I appreciate your feedback. " "Problem Number 1
What is the velocity of a wave whose wavelength and frequency are 3 meters and 62 Hz, respectively?
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62 wavelengths/ sec * 3 meters/wavelength = 186 meters/sec
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Problem Number 2
What is the wavelength of a wave traveling at 70 meters/second, if 35 equally spaced peaks pass a given point every second.
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70 meters/second * 1 sec/35 wavelenths = 2 meters/wavelength
Each wavelength is 2 meters.
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Problem Number 3
what are the first three frequencies that you would expect to observe in a string which is stretched out between two points 11 meters apart under a constant tension of 39 Newtons. The string has a mass of 1.1 kg.
.First harmonic: 1/2 wavelength = 11 meters, so 1 wavelength = 22 meters
Second harmonic: 2/2 wavelength = 11 meters, so 1 wavelength = 11 meters
Third harmonic : 3/2 wavelength = 11 meters, so 1 wavelenth = 7.33 meters
v = `sqrt( Tension/mass/unit)
v= `sqrt( (39N/(1.1kg/11m)) = 19.7 m/s
First harmonic: 19.7m/s * wavelength/22 meters = .895wavelengths/sec = .895 Hz
Second harmonic: 19.7m/s * wavelength/11 meters = 1.79 wavelengths/sec = 1.79 Hz
Third harmonic: 19.7m/s * wavelength/7.3 meters = 2.69 wavelengths/sec = 2.69 Hz
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Problem Number 4
The mass density of a string is 16.7 grams/meter, its length is 35 meters and its tension is such that the velocity of a transverse wave in the string is 54 m/s. How much energy is there in the string if it carries a traveling wave with amplitude 1.95 meters and frequency 128 Hz?
College and University Physics students also answer the following:
How much power does it take to maintain this wave?
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vMAX = `omega * A
VMax = 128 cycles/sec * 2 `pi Radians/cycle * 1.95 meters = 1568m/s
Mass of string = .0167 kg/meter * 35 meters = .5845kg
Energy = .5 * mass * VMax^2
Energy = .5 * .5845 kg * 2459512 m^2/s/2
Energy = 718792 J or approximately 719000 Joules.
Attempting the College Physics part:
It travels 35 meters at 54m/s. 35m/ 54m/s = .648 secs
719000J / .648 sec = 1,110,000 J/s or 1,110,000 watts
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Problem Number 5
Ultraviolet waves are produced in phase by two sources separated by 2.27 * 10^-7 meters. The waves have wavelength .52 * 10^-7 meters are observed at a distance of several meters from the sources. A line from the sources to the point of observation makes an angle of 36.1 degrees with the perpendicular bisector of the line segment joining the sources.
By what fraction of a wavelength will the wave emitted from the source further from the observer lag the wave emitted from the closer source?
What are the first three angles at which the strength of the radiation reaching the observer would be minimized?
What are the first two angles at which the radiation reaching the observer would be maximized?
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path difference = sine(36.1) * 2.27 * 10^-7 = 1.34 * 10^-7 meters
1.34 * 10^-7 meters / (.52 *10^-7 m/ wavelenth) = 2.58 wavelengths
Strength of radiation will be minimized when difference is 1/2, 3/2 and 5/2 wavelenths
sin`theta *2.27 * 10^-7 = 1/2(.52 * 10^-7m)
sin `theta = 1/2(.52 * 10^-7m)/( 2.27 * 10^-7)
`theta = arcsin(1/2(.52 * 10^-7m)/ (2.27 * 10^-7))
`theta = 6.58 degrees
sin`theta *2.27 * 10^-7 = 3/2(.52 * 10^-7m)
sin `theta = 3/2(.52 * 10^-7m)/( 2.27 * 10^-7)
`theta = arcsin(3/2(.52 * 10^-7m)/ (2.27 * 10^-7))
`theta = 20.1 degrees
sin`theta *2.27 * 10^-7 = 5/2(.52 * 10^-7m)
sin `theta = 5/2(.52 * 10^-7m)/( 2.27 * 10^-7)
`theta = arcsin(5/2(.52 * 10^-7m)/ (2.27 * 10^-7))
`theta = 34.9 degrees
Strength of the radiation will be maximized when the difference is 2/2 and 4/2 wavelengths
sin`theta *2.27 * 10^-7 = 2/2(.52 * 10^-7m)
sin `theta = 2/2(.52 * 10^-7m)/( 2.27 * 10^-7)
`theta = arcsin(2/2(.52 * 10^-7m)/ (2.27 * 10^-7))
`theta = 13.24 degrees
sin`theta *2.27 * 10^-7 = 4/2(.52 * 10^-7m)
sin `theta = 4/2(.52 * 10^-7m)/( 2.27 * 10^-7)
`theta = arcsin(4/2(.52 * 10^-7m)/ (2.27 * 10^-7))
`theta = 27.3 degrees
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Problem Number 6
Suppose that you are holding one end of a Slinky and the other end is attached to a wall. The Slinky is held at some constant tension. If you tweak the Slinky by displacing some coils perpendicular to a direction parallel to the direction in which waves propagate through it, the disturbance requires 1.1 seconds to travel its length. You want to create a standing wave in the Slinky by moving your end in a low-amplitude simple harmonic motion perpendicular to the direction of wave propagation (the SHM has a small enough amplitude that you can regard your end, for practical purposes, as a node).
As you answer the following, give a full description of what happens and why, as well as a correct mathematical analysis.
If you wish to create a wave with nodes at its two ends and only one antinode, then what must be the frequency and period of the SHM with which you drive the wave?
What would be the frequency of your SHM if the wave, with nodes at both ends, contained exactly two antinodes?
What if the number of antinodes was three?
What if the number of antinodes was four?
What is the wavelength of each mode of vibration, in terms of the length of the Slinky (e.g., 1/3 of the length, 5 times the length, etc.)?
What is the ratio sequence corresponding to the sequence of frequencies?
.For two nodes with one antinode:
The pulse will travel down to the wall and back to you. This will take 1.1 seconds * 2 or 2.2 seconds.
This is the period (2.2 sec) so frequency will be 1cycle / 2.2 seconds = .455Hz
Since the wavelength is down and back the wavelength is 2 times the length of the slinky
This represents the first harmonic
For two antinodes you will have three nodes: one at each end and one in the middle.
The pulse will travel down to the wall and back - which will constitute two wavelengths
This period will be 1.1 seconds and the frequency will be 2 cycle/2.2 seconds = .910 Hz
This is the second harmonic
For three antinodes, there will be four nodes.
There will be three cycles in a round trip. 2.2 seconds/3 cycles = a period of .733 seconds
Frequency is 1 cycle/.733 seconds =1.36 Hz
Wavelength will be 2/3 the length of the slinky.
For four antinodes there will be five nodes
Four cycles in a round trip of 2.2 seconds. 2.2 seconds/4 cycles = a period of .55 seconds
Frequency is 1 cycle/.55 seconds = 1.82 Hz
Four cycles to 2 lengths of the slinky means each wavelength is 2 times the length of the slinky.
Ratios:
.910/.455 = 2
1.36/.910 = 3/2
1.82/1.36 = 4/3
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Problem Number 7
This was blank
I skipped #8 - the extra credit."
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Everything looks very good. I expect that you'll do very well on the test.
Let me know if you have additional questions.
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