course PHY 202 fñåHŠ¥£ês¿×°ÇíêÊï~öðäüïºXá€IÛÃŽassignment #019
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21:11:07 Query introductory set 6, problems 1-10 explain how we know that the velocity of a periodic wave is equal to the product of its wavelength and frequency
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RESPONSE --> Velocity of a periodic wave is the the speed at which wave crests move, measured as displacement/unit of time. The wavelength represents displacement, or the distance between two successive crests, and the unit if time is measured by the period. Period is the time elapsed between two successive crests. You then must take the wavelength times period(1/f) to get the velocity.
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21:11:18 ** 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 --> ok
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21:17:24 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 defined by the time that elapses between two successive crests passing by the same point in space or 1/f. Since frequency is equal to velocity/wavelength, then it follows that 1/(velocity/wavelengh)=wavelength/frequency.
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21:17:34 ** 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 --> ok
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21:21:24 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 --> A=Amplitude of the motion Sine or cosine can be used in the equation of motion omega=2pi(frequency) t=time t lag=x/v I know how to calculate it, but don't necessarily know why the equation of motion at a position x along the wave is equal ot the equation of motion.
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21:23:50 ** 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 --> ok
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21:32:53 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 --> If you have a length of string, then the first few harmonics must have 2, 3, and 4 nodes in the length. So nodes represent half wavelenths, so they must have 1/2 wavelength, 2/2 wavelength and 3/2 wavelenths.
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21:35:45 ** 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 --> clear explanation of 2L, 2/3L and 1/2 L
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21:37:56 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 --> once you know the wavelengths of the first few harmonics and then the velocity, you must multiply the wavelengths by the velocity to arrive at the frequencies as the equation f=v(wavelength) suggests.
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21:39:30 ** 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 --> Yes, f=velocity/wavelength.
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21:41:05 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 --> given the formula, v=sqrt(Ft)/m/L Where V=velocity Ft=tension in the string, m=mass L=length of the string. Solve for your unknowns to arrive at the answer.
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21:41:12 ** We divide tension by mass per unit length and take the square root: v = sqrt ( tension / (mass/length) ). **
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RESPONSE --> ok
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21:47:13 gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> The principle of superpositon just describes the moment when two waves pass each other and overlap. To find the displacement of the wave at that time, you must add the seperate displacements of each of the waves.
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21:47:26 ** 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 --> ok
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21:56:20 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 --> The basic concept is that waves reflect off of objects in their path. When a wave strikes the boundary between two materials, part of the wave is reflected and part is transmitted.
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21:58:46 ** 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 --> ok
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Ò“{ÊÇì”E½ñ˜±â–ì½xã©Ṳ̂ assignment #019 019. `query 9 Physics II 07-16-2007"