course phy 202 There were questions on here from Problem sey 6 #11-14, which hasn't been assigned yet, so I didn't answer those. ??D??l????Y???????assignment #019019. `query 9
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19:34:53 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 --> Algebraic manipulation of the formula for wavelength: lambda = v / f multiply both sides by f lambda * f = velocity
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19:35:00 ** 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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19:43:16 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 --> in order to find the time between peaks you would divide wavelength by velocity. You want the units to cancel out to leave time. So wavelength (meter) / velocity (m/s) = seconds.
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19:43:26 ** 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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19:49:58 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 is the time lag. if x=0, x/v =0 and therefore can be excluded from the equation. If x equals any other number, it has a value and cannot be excluded from the equation.
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19:50:16 ** 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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19:51:35 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 --> These problems arent assigned until the next section
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