course PHY 122 June 18 4:26 pmI also submit these work forms in addition to the open query. I don't know if I should or not. Please let me know.
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RESPONSE --> velocity is equal to the wavelength (the distance between peaks) times the frequency. frequency is the number of peaks in a certain interval of time. if we know the wavelength is a certain length then we can multiply the frequency times the wavelength to say tht the wave will move X meters in X unit of time (seconds). this is the velocity.
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16:00:44 ** we know how many wavelength segments will pass every second, and we know the length of each, so that multiplying the two gives us the velocity with which they must be passing **
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RESPONSE --> okay
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16:04:58 explain how we can reason out that the period of a periodic wave is equal to its wavelength divided by its velocity
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RESPONSE --> period is also frequency which represents the number of peaks passing per unit of time (peaks/ second) or Hertz. a wave passing at a velocity with a wavelength: the frequency is the number of peaks divided by time or the velocity divided by the wavelength.
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16:06:46 ** If we know how far it is between peaks (wavelength) and how fast the wavetrain is passing (velocity) we can divide the distance between peaks by the velocity to see how much time passes between peaks at a given point. That is, period is wavelength / velocity. **
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RESPONSE --> yes. I understand this concept.
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16:09:18 explain why the equation of motion at a position x along a sinusoidal wave is A sin( `omega t - x / v) if the equation of motion at the x = 0 position is A sin(`omega t)
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RESPONSE --> because the motion at the position x along a sinusoidal wave will have a different position that when x = 0.
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16:12:43 ** the key is the time delay. Time for the disturbance to get from x = 0 to position x is x / v. What happens at the new position is delayed by time x/v, so what happens there at clock time t happened at x=0 when clock time was t = x/v. In more detail: If x is the distance down the wave then x / v is the time it takes the wave to travel that distance. What happens at time t at position x is what happened at time t - x/v at position x=0. That expression should be y = sin(`omega * (t - x / v)). } The sine function goes from -1 to 0 to 1 to 0 to -1 to 0 to 1 to 0 ..., one cycle after another. In harmonic waves the motion of a point on the wave (think of the motion of a black mark on a white rope with vertical pulses traveling down the rope) will go thru this sort of motion (down, middle, up, middle, down, etc.) as repeated pulses pass. If I'm creating the pulses at my end, and that black mark is some distance x down in rope, then what you see at the black mark is what I did at time x/v earlier. **
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RESPONSE --> okay
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16:16:48 Query introductory set six, problems 11-14 given the length of a string how do we determine the wavelengths of the first few harmonics?
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RESPONSE --> you send out into your original string; depending on the number of nodes it has, the length of the string must correspond to a certain peak-to-peak distance, or to 1/2 of 1/3 or 1/4 etc. of a wavelength. The wavelength must therefore be twice, three times, four times, etc. the length of the string.
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16:18:16 ** As wavelength decreases you can fit more half-waves onto the string. You can fit one half-wave, or 2 half-waves, or 3, etc.. So you get 1 half-wavelength = string length, or wavelength = 2 * string length; using `lambda to stand for wavelength and L for string length this would be 1 * 1/2 `lambda = L so `lambda = 2 L. For 2 wavelengths fit into the string you get 2 * 1/2 `lambda = L so `lambda = L. For 3 wavelengths you get 3 * 1/2 `lambda = L so `lambda = 2/3 L; etc. } Your wavelengths are therefore 2L, L, 2/3 L, 1/2 L, etc.. **
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RESPONSE --> okay
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16:21:49 Given the wavelengths of the first few harmonics and the velocity of a wave disturbance in the string, how do we determine the frequencies of the first few harmonics?
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RESPONSE --> we can find the frequencies of the first few harmonics if we know the velocity and the wavelength. v = sqrt of tension* mass per unit length
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16:22:54 ** The frequency is the number of crests passing per unit of time. We can imagine a 1-second chunk of the wave divided into segments each equal to the wavelength. The number of peaks is equal to the length of the entire chunk divided by the length of a 1-wavelength segment. This is the number of peaks passing per second. So frequency is equal to the wave velocity divided by the wavelength. **
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RESPONSE --> Correct. The solution I propsed you would have to know tension and mass. Here we can use only velocity and wavelength to find frequency.
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16:23:35 Given the tension and mass density of a string how do we determine the velocity of the wave in the string?
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RESPONSE --> As I stated incorrectly before, the velocity can be determined by taking the square root of tension divided by mass per unit length.
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16:23:41 ** We divide tension by mass per unit length and take the square root: v = sqrt ( tension / (mass/length) ). **
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RESPONSE --> correct.
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16:23:48 gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> different section.
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16:23:52 ** the principle of superposition tells us that when two different waveforms meet, or are present in a medium, the displacements of the two waveforms are added at each point to create the waveform that will be seen. **
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RESPONSE --> okay
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16:23:58 gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?
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RESPONSE --> skipped.
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16:24:03 ** angle of incidence with a surface is the angle with the perpendicular to that surface; when a ray comes in at a given angle of incidence it reflects at an equal angle on the other side of that perpendicular **
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RESPONSE --> okay.
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ЩWհ assignment #010 010. `query 9 Physics II 06-18-2009"