course Phy 202 The physics marathon continues. ¾Ð¾™¢Š•ªèšz€¬þ×£løwí_assignment #019
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07:38:20 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 --> We know this to be true because frequency is, in very general terms, the speed of the waves still in terms of the wave. This value multiplied by the wavelength will provide a result that is in meters/second, indicating that it has to be velocity.
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07:38:29 ** 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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07:43:15 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 --> Wavelength is the distance between peaks in meters, and the velocity is the speed of the wave in meters/second, so dividing the wavelength by the velocity will give a result in the form of seconds as would any d / v. In this case it is the time per wavelength.
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07:43:26 ** 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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08:04:15 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 --> x/v represents a time lag in this equation. Time lag is the time difference between the current position x and x = 0 given the velocity. It is calculated by dividing the distance by the velocity, as described by x/v.
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08:07:07 ** 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 --> The time delay between x and x = 0 is the time required for the disturbance to reach position x on the wave.
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08:37:06 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 two ends of the string can be considered nodes, or points of destructive interference. Therefore, the first wave will have 2 nodes, increasing by 1 for each harmonic. harmonic 1: 1/2(wavelength) = d, so wavelength = 2d harmonic 2: 2/2(wavelength) = d, so wavelength = d harmonic 3: 3/2(wavelength) = d, so wavelength = 2/3d harmonic 4: 4/2(wavelength) = d, so wavelength = 2d et cetera.
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08:41:11 ** 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 --> By 2d for the last one I meant 1/2d
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08:47:12 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 --> The given wavelengths and velocity can be used in the following equation to find the frequencies of the first few harmonics: f = v / lambda
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08:47:24 ** 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 --> OK
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08:56:11 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 --> The following formula gives the velocity of the wave based on these numbers: v = sqrt(tension / mass per unit length)
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08:56:30 ** 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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09:01:07 gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> When interference occurs and two waves pass over one another, the principle of superposition states that the sum of their separate displacements is the calculated resultant displacement, regardless of the state of the waves (trough vs. trough, peak vs. trough ( = 0 ), peak vs. peak).
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09:01:15 ** 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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09:08:12 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 the incoming wave makes with the reflection surface is equal to the angle the reflected wave will make with the reflection surface.
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09:08:21 ** 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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