Assignment 10

course Phy 232

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

010. `query 9

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Question: 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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Your Solution:

The velocity can be represented by the product of its wavelength and frequency because we know the distance from one trough to the next , wavelength, which travels a constant speed, v, and advances a distance of one wavelength in a certain time period, frequency. By looking at the units to we see that frequency is in Hz (s^-1) and wavelength is in meters, so by multiplying the two we get m/s.

Confidence Rating:

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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: 2

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Question: 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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Your Solution:

We need to know the distance between peaks (wavelength) and the velocity of the moving waves. If we divide the distance (meters) by the velocity (m/s), we get the period (seconds), or how much time passes between wave troughs.

Confidence Rating: 2

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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

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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)

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Your Solution:

At t=0, the particle at x = 0 is at its maximum possible displacement. The wave disturbance travels from this original point x = 0 to some point to the right. To find the time that it traveled away from the origin, we divide the distance that it traveled, x, by its velocity, v (x/v).

If time, t, is the time that it started at x=0, then by using t-x/v, we can find the displacement of point x at time t.

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.

The choice of the cosine function is arbitrary. Either function, or a combination of both, can come out of the solution to the wave equation (that's the partial differential equation which relates the second derivative with respect to position to the second derivative with respect to time).

The sine and cosine functions differ only by a phase difference of 90 degrees, and either can be used to describe simple harmonic motion or the motion of harmonic waves. The choice simply depends on the initial conditions of the system.

We don't want to get into solving the wave equation here, but the point can be illustrated by considering simple harmonic motion, which is characterized by F_net = - k x (leading to m x '' = - k x or x '' = -k/m * x, where derivatives are with respect to time).

The general solution to the equation x '' = - k / m * x is x = B sin(omega t) + C cos(omega t), where B and C are arbitrary constants and omega = sqrt(k/m).

B sin(omega t) + C cos(omega t) = A sin(omega t + phi), where A and phi are determined by B, C and the choice to use the sine function on the right-hand side.

B sin(omega t) + C cos(omega t) = A cos(omega t + phi), where A and phi are determined by B, C and the choice to use the cosine function on the right-hand side. The value of A will be the same as if we had used the sine function on the right, and the value of phi will differ by 90 degrees or pi/2 radians.

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. **

Self-Critique:

Your solution uses the same equation as me; however, I have a cosine function and you used a sine function. I don’t know why they are different from the book and your answer. Is there a certain reason you would use one or the other?

Good question; see my comments above.

Your Self-Critique Rating: 2

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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?

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Your Solution:

The length of the string must be some integer of a half-wavelength (so, lambda/2, 2*(lambda/2), etc.).

L = n*lambda/2 where n = 1,2,3,4, etc.

So solving for lambda we get:

Lambda = 2*L/n

Some examples of wavelengths are:

2*L, L,2/3*L, ½*L, 2/5*L, etc.

Confidence Rating: 3

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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

Your Self-Critique Rating: 3

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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?

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Your Solution:

Frequencyn = v/lambdan where n = 1,2,3,4, etc.

The general frequency equation is plugging in what we found above for lambda:

n*v/lambda*2 = frequency

Confidence Rating:

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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, I just plugged in what we had found for the wavelength to get my final equation.

Your Self-Critique Rating: 3

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Question: 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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Your Solution:

Tension, F

I am not sure what is meant by mass density. In order to find velocity, I need the mass per unit length, u

V = sqrt[F/(mass/length)]

Confidence Rating: 2

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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, I just assumed that mass density just meant the mass. I got the same equation though.

Your Self-Critique Rating: 2

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

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Your Solution:

I’m in university physics

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:

Your Self-Critique Rating:

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Question: gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?

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Your Solution:

I’m in university physics

Confidence Rating:

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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:

Your Self-Critique Rating:

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&#Good responses. See my notes and let me know if you have questions. &#