course Phy 202 Question: Query introductory set 6, problems 1-10
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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: ok 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: By definition, the period of a wave is the time it takes for one complete cycle, or the time between peaks. Since the wavelength is the length of each cycle and the velocity is the speed at which that cycle goes by, we need to divide the wavelength by the velocity to get time. 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: ok 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) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: We are told that the position is x. The period of the wave is x (position)/v (velocity). Although I don’t really understand what is meant by a delay in the time, I know that we subtract x from omega t because of this time delay. Confidence Rating:1
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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. ** STUDENT COMMENT (University Physics): According to the Y&F book (p.553) we get the expression for a sinusoidal wave moving the the +x-direction with the equation: Y(x,t) = A*cos[omega*(t-x/v)] I am not sure where the sine came from in the equation in the question. The book uses the cosine function to represent the waves motion.
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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: ok Your Self-Critique Rating: ********************************************* Question: 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The frequency of a wave by definition is the wave velocity divided by the wave length. Thus is we take a look at the number of peaks found in the length of one segment and divide this by the length of the entire wavelength, we will be able to find the frequency. Confidence Rating:2
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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: ok 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: We can find the velocity of the wave in the string by using the equation: V= (sqrt) [tension/(mass/length)] 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: ok Your Self-Critique Rating: 3 ********************************************* Question: gen phy explain in your own words the meaning of the principal of superposition YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: When two waves meet, the wave created is the sum of the size of each wave individually. Confidence Rating:
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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: ok 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: This is the law of reflection. If a ray’s angle of incidence is equal to its angle of reflection, this means that the ray is striking a surface and being reflected back at the exact same angle. 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: ok Your Self-Critique Rating: