course Phy 232 y?l??????????·???c?·assignment #019019. `query 9
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21:17:09 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 --> knowing how many waves pass per second (frequency) and how much of a wave has to go by in that time (wavelength) you can multiply the two and get dist/time => velocity
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21:17:11 ** 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 --> sounds great
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21:17:15 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 --> dist/velocity yeilds time between peaks
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21:17:19 ** 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 --> good
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21:17:21 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 --> position is related to time which we get by the x/v so multiplyng that by the sine wave and decreasing that from the x=0 point will give us the motion at that position
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21:17:22 ** 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 -->
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21:17:26 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 --> first harmonic is when half the wave fits on the string so it will take 2 times the length for one full wave second harmonic is when 2 halves the wave fit on the string, so it takes 1 length for one wave third harmonic is when 3 halves fit on the length which means it takes 2/3 the length for one wave
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21:17:28 ** 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 --> sounds right on
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21:17:29 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 --> freq = velocity/lambda - so we divide the wavelength of each harmonic into the velocity
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21:17:30 ** 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 --> looks right
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21:17:32 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 --> equation 15.13 v = sqrt(F/mu) = sqrt(tension/mass density)
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21:17:33 ** We divide tension by mass per unit length and take the square root: v = sqrt ( tension / (mass/length) ). **
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RESPONSE -->
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21:17:35 gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> adding the amplitude of each wave together to get the resulting wave
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21:17:36 ** 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 --> same thing
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21:17:37 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 --> looking at the reflection will give you the same angle from the reflective surface as if you could see the incidence.
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21:17:39 ** 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 --> mine makes sense to me
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