q_a_assignment9

course Phy 202

assignment #009???y??b??????Physics II

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07-12-2006

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13:40:47

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 -->

If a wave is passing a point at the same time, ittherefore has a constant frequency. The lenth of the wave tells us how far it moves. By multiplyig these 2 numbers, you get v=f*lamda

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13:41:08

** 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 -->

That is what I had said.

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13:47:55

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 -->

a period is the time it takes the wave to move one full cycle. In order to find that, you have to determine how long the wave is ( wavelength) and how fast it is moving (velocity). so, you take the wavelength and divide it by the velocity to show how long it will take.

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13:48:03

** 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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14:15:55

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 should equal A and doesn't so therefore, you have to calculate that into it.

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14:19:53

** 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)).

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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 -->

I hadn't thought of it that way. I am pretty sure I understand it.

It's a very good idea at this point to articulate your understanding in your own words. That way, if there's something you don't understand it will probably show itself and I can help you address it.

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13:27:45

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 length of the string lets us know how much half a wavelength is because there has to be two nodes, one on each end.

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13:29:57

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 -->

I got the same results but din't go on to the 2nd and 3rd wavelength. I understand t though.

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13:32:45

** 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 -->

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? the frequency is the number of the number of peaks per unit time.

I wasn't given a chance to answer this before the answer came up but I understand the question and the answer.

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14:44:36

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 velocity is equal to the square root of the thension divided by the mass divided by the Length.

v-sqrt(FT/(m/L))

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14:44:51

** We divide tension by mass per unit length and take the square root:

v = sqrt ( tension / (mass/length) ). **

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RESPONSE -->

I got the same answer.

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15:02:21

gen phy explain in your own words the meaning of the principal of superposition

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RESPONSE -->

superposition is when two waves pass each other and overlap and it tresults in displacement .

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15:02:36

** 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 -->

I understnad.

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15:04:58

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 incicence is the angle coming into a surface and the angle of reflection is the reflected ray.

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15:05:12

** 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 -->

i understand

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Good work. See my notes and let me know if you have questions.