course Phy 232 010. `query 9 *********************************************
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Given Solution: ** 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 ** Your Self-Critique: Okay. Your Self-Critique Rating:3 ********************************************* Question: explain how we can reason out that the period of a periodic wave is equal to its wavelength divided by its velocity ********************************************* Your Solution: The wavelength is a distance, the velocity is a distance per second. If you divide wavelength by velocity, you get a time or the period. Confidence Rating: 3
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Given Solution: ** 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. ** Your Self-Critique: Okay. Your Self-Critique Rating: 3 ********************************************* Question: 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) ********************************************* Your Solution: I am not sure, but one thing I remember from when I learned LaPlace transforms is that when there is a time delay there is normally some number being subtracted from t. It looks as if a disturbance is occurring at x = 0, there will be a time delay of x/v before that disturbance is seen at x. Confidence Rating:2
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Given Solution: ** 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. ** Your Self-Critique: Okay. Your Self-Critique Rating:3 ********************************************* Question: Query introductory set six, problems 11-14 given the length of a string how do we determine the wavelengths of the first few harmonics? ********************************************* Your Solution: If the two ends of the string are fixed, the string will be divided into halfs. What I mean by this is the wavelengths will be L/2 if there is one full cycle on the string. L/4 for 2 cycles, L/8 and so on. Confidence Rating:2
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Given Solution: ** 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.. ** Your Self-Critique: Okay
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Given Solution: ** 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. ** Your Self-Critique: Okay Your Self-Critique Rating:3 ********************************************* Question: Given the tension and mass density of a string how do we determine the velocity of the wave in the string? ********************************************* Your Solution: The velocity of a wave on a string is v = (F/u)^0.5. Confidence Rating:3
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Given Solution: ** We divide tension by mass per unit length and take the square root: v = sqrt ( tension / (mass/length) ). ** Your Self-Critique: Okay. Your Self-Critique Rating:3 ********************************************* Question: gen phy explain in your own words the meaning of the principal of superposition ********************************************* Your Solution: I am not in this class. Confidence Rating:3
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Given Solution: ** 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. ** Your Self-Critique: Okay Your Self-Critique Rating:3 ********************************************* Question: gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence? ********************************************* Your Solution: I am not in this class Confidence Rating:3
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Given Solution: ** 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 ** Your Self-Critique: Okay. Your Self-Critique Rating:3 "