course Mth 271
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14:34:51 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 --> frequency of a wave (measured in Hz) = f wavelenght = 'lambda velocity = distance / time = f * 'dt * 'lambda / 'dt = f * 'lambda In words, this means: The number of cycles of the wave (f * 'dt) of a certain length ('lambda) per time interval ('dt) must pass. So, velocity of a wave = frequency * wavelength
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14:35:01 ** 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 --> Exactly.
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14:38:58 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 --> Take Problem 5 for example. It is asking for the period of a wave with 4 meters between peaks, if the wave passes at 16 meters/sec. If you take a 1-second section ot the wave, it will be 16 meters long, and it will have peaks that are 4 meters apart. Each peak is 4/16ths of the 1-second section. So the time before peaks will be 4/16ths of a second. The time between peaks if the period of the cycle.
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14:39:01 ** 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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14:45:39 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 particle at position x lags the x = 0 particle by the time required for the wave to propogate over the distance x. So you have to factor in the time it takes for the wave to go the distance x which is (x/v).
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14:46:23 ** 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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14:54:23 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 --> To determine the harmonics you have to know there are nodes at the ends of the string seperated by the length of the string. Harmonic 1 has 2 nodes. Harmonics 2 has 3 nodes. Harmonic 3 has 4 nodes. So these harmonics have 1, 2, and 3, half wavelengths in the same distance. So, to get the wavelengths of the harmonics you just multiply the length of the string by the inverse of the fraction of the wavelength. For example: String length = 23 meters Harmonic 1 : 1/2 wavelenght = 46 meters
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14:55:42 ** 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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14:59:40 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 --> v = 'sqrt (tension) / (mass / unit length) Once we find the velocity, you divide that number by the wavelength to get the frequency in Hz.
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14:59:47 ** 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 --> Okay
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15:01:18 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 --> v = 'sqrt (tension) / (mass/unit length) You take the square root of the quotient of the tension divided by the mass per unit of length.
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15:01:23 ** We divide tension by mass per unit length and take the square root: v = sqrt ( tension / (mass/length) ). **
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RESPONSE --> Okay
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15:04:04 gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> In a region where two waves overlap, the displacement equals the sum of their seperate displacements. A crest is positive and a trough is negative.
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15:04:11 ** 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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15:06:55 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 angle that the incoming wave (incident) makes with the reflecting surface is equal to the angle made by the reflected wave.
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15:07:00 ** 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 -->
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