course Phy 202
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19:56:29 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 --> Length of WV * # of WV = velocity
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19:56:54 ** 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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19:58:32 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 --> The distance between peaks / velocity is the amount of time between peaks.
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19:58:47 ** 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 --> WL / V = period
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20:00:43 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 * sin(omega t - x/v) equation of motion at x = 0 is A *sin(omega t). We can use the equation of motion and incorporate its clocktime at x/v.
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20:01:49 ** 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 --> x distance down wave x/v is time is takes the wave to travel that distance time t at x is what happened at t-x/v at position x=0. I have never understood sine or any of that, but I am grasping this without diving too far in the sine function.
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20:03:18 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 --> The length of the string tells us how many waves can fit into the cycle. When you know how man waves you can separate them into lengths and use that to determine the wavelengths.
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20:04:03 ** 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 --> 1*1/2 Lambda = 2 L 2* 1/2 Lamba = L 3*1/2 Lambda = 2/3 L. okay.
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20:05:24 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 --> Given Harmonics Velocity of wave disturbance in string Frequency = Velocity/ Wavelength We know velocity and we can take the wave and find its segments.
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20:05:32 ** 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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20:06:03 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 --> Velocity of the wave = tension / mass/length
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20:06:26 ** We divide tension by mass per unit length and take the square root: v = sqrt ( tension / (mass/length) ). **
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RESPONSE --> Oh and take the square root v = sqrt ( tension / mass / length)
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20:07:07 gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> The visible wave form is the two wavelengths that encounter one another added together.
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20:07:35 ** 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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20:08:34 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 reflection is made when the ray hits the reflecting surface that is perpendicular to the ray. The perpendicular angle is the angle of incidence.
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20:09:04 ** 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 --> When a ray comes in at a given angle of incidence, ti reflects the other side of the perpendicular. Close.
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