course Phy 202 w~???|·n????????assignment #019
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10:31:59 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 --> Given the number of wavelength segments passing/ sec and the length of each segment, we can multiply the two to get the velocity they pass with.
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10:32:03 ** 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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10:34:23 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 --> If we know the wavelength and the velocity the waves pass with, we can divide the wavelength by the velocity to find the amount of time that passes between peaks at a given point. The time is the period. period = wavelength / velocity.
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10:34:27 ** 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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10:39:02 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 is the distance down the wave and x / v is the time the wave takes to travel that distance. What's at time t at position x is the same at time t - x/v at position x=0. The expression of that is: y = sin(`omega * (t - x / v)).
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10:39:04 ** 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 --> ok
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10:43:10 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 --> As wavelength decreases, more half-waves can fit onto the string. So, 1 half-wavelength = string length wavelength = 2 * string length This is 1 * 1/2 `lambda = L `lambda = 2 L. 2 wavelengths, 2 * 1/2 `lambda = L `lambda = L. 3 wavelengths, 3 * 1/2 `lambda = L `lambda = 2/3 L and so on for each following harmonic.... So the wavelengths are 2L, L, 2/3 L, 1/2 L, etc..
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10:43:15 ** 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 --> ok
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10:49:14 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 frequency is equal to the number of crests passing per time unit. If we have a 1-second portion of a wave divided into segments of the wavelength, the number of peaks equals the length of the entire portion divided by the length of a 1-wavelength segment. Which is the number of peaks passing per second. So, frequency is equal to wave velocity divided by wavelength.
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10:49:17 ** 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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10:50:40 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 --> We can divide the tension by the mass per unit length and then take the square root of what we get. velocity = sqrt ( tension / (mass/length) )
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10:50:43 ** 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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11:02:25 gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> The principle of superposition says that when two pulses on a string are travelling towards eachother, in the region where they overlap, the resultant displacement is the sum of the two seperate waves.
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11:02:31 ** 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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11:04:19 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 of incidence with a surface is the angle of the perpendicular to that surface. So, at a given angle of incidence, a ray reflects at an equal angle on the other side of the perpendicular.
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11:04:25 ** 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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course Phy 202 w~???|·n????????assignment #019
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10:31:59 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
......!!!!!!!!...................................
RESPONSE --> Given the number of wavelength segments passing/ sec and the length of each segment, we can multiply the two to get the velocity they pass with.
.................................................
......!!!!!!!!...................................
10:32:03 ** 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 **
......!!!!!!!!...................................
RESPONSE --> ok
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10:34:23 explain how we can reason out that the period of a periodic wave is equal to its wavelength divided by its velocity
......!!!!!!!!...................................
RESPONSE --> If we know the wavelength and the velocity the waves pass with, we can divide the wavelength by the velocity to find the amount of time that passes between peaks at a given point. The time is the period. period = wavelength / velocity.
.................................................
......!!!!!!!!...................................
10:34:27 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
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10:39:02 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 is the distance down the wave and x / v is the time the wave takes to travel that distance. What's at time t at position x is the same at time t - x/v at position x=0. The expression of that is: y = sin(`omega * (t - x / v)).
.................................................
......!!!!!!!!...................................
10:39:04 ** 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 --> ok
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10:43:10 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 --> As wavelength decreases, more half-waves can fit onto the string. So, 1 half-wavelength = string length wavelength = 2 * string length This is 1 * 1/2 `lambda = L `lambda = 2 L. 2 wavelengths, 2 * 1/2 `lambda = L `lambda = L. 3 wavelengths, 3 * 1/2 `lambda = L `lambda = 2/3 L and so on for each following harmonic.... So the wavelengths are 2L, L, 2/3 L, 1/2 L, etc..
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10:43:15 ** 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 --> ok
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10:49:14 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 frequency is equal to the number of crests passing per time unit. If we have a 1-second portion of a wave divided into segments of the wavelength, the number of peaks equals the length of the entire portion divided by the length of a 1-wavelength segment. Which is the number of peaks passing per second. So, frequency is equal to wave velocity divided by wavelength.
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10:49:17 ** 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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10:50:40 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 --> We can divide the tension by the mass per unit length and then take the square root of what we get. velocity = sqrt ( tension / (mass/length) )
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10:50:43 ** 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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11:02:25 gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> The principle of superposition says that when two pulses on a string are travelling towards eachother, in the region where they overlap, the resultant displacement is the sum of the two seperate waves.
.................................................
......!!!!!!!!...................................
11:02:31 ** 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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11:04:19 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 of incidence with a surface is the angle of the perpendicular to that surface. So, at a given angle of incidence, a ray reflects at an equal angle on the other side of the perpendicular.
.................................................
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11:04:25 ** 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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