Assignment 10

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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:

Thinking of the wave as a 1-second segment, the frequency tells us how many waves pass in 1 second. By knowing the length of each wave, all that I would need to do is multiply the length by the frequency to get the velocity. This can be verified looking at the units:

Length (m) * frequency (s^-1) = velocity (m/s)

confidence rating #$&*:

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confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

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3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

• 2 indicates that you believe you addressed most of the discrepancies between the given solution and your solution but are unsure of some aspects of the situation; you would at this point consider including a question or a statement of what you're not sure you understand

• 1 indicates that you believe you understand the overall idea of the solution but have not been able to address the specifics of the discrepancies between your solution and the given solution; in this case you would normally include a question or a statement of what you're not sure you understand

• 0 indicates that you don't understand the given solution, and/or can't make a reasonable judgement about whether or not your solution is correct; in this case you would be expected to address the given solution phrase-by-phrase and state what you do and do not understand about each phrase)

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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:

Well, I have previously shown that the velocity of the wave is equal to the frequency times the wavelength. The period of the wave is equal to the inverse of the frequency (T = 1/f or f = 1/T). If I substitute this relationship into the formula that I already know and rearrange it, then I can show that the period is equal to the wavelength divided by velocity:

v = f*’lambda

v = (1/T)*lambda

T*v = lambda

T = lambda/v

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 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:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

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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:

If the equation of motion at x=0 is A*sin(‘omega t), then there is a point at an arbitrary distance x that also follows this motion. This position x occurs at a “time lag”. The time lag is equal to x/v. Therefore, the equation for the y position at the arbitrary point x is now:

y = A sin(‘omega (t - “time lag”))

y = A sin(‘omega (t - x/v))

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

confidence rating #$&*: (Type in a number from 0 to 3, indicating your level of confidence in your solution.

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3 means you are at least 90% confident of your solution, or that you are confident you got at least 90% of the solution

2 means that you are more that 50% confident of your solution, or that you are confident you got at least 50% of the solution

1 means that you think you probably got at least some of the solution correct but don't think you got the whole thing

0 means that you're pretty sure you didn't get anything right)

Response: 3

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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.

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.

STUDENT QUESTION

I don’t understand why we would subtract x/v from the omega*t value. Why wouldn’t it be addition?

INSTRUCTOR RESPONSE

What happens at x at clock time t is what happened at the origin at a certain earlier clock time.

The time required to for a pulse to propagate from the origin to position x is x / v.

So What happens at x at clock time t is what happened at the origin at clock time t - x / v.

At clock time t - x / v the point at the origin was at y position y = A sin (omega (t - x / v)).

Your Self-Critique:

Self-critique (if necessary): (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise explain in your own words how your solution differs from the given solution, and demonstrate what you did not originally understand but now understand about the problem and its solution.)

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Self-critique Rating: (If you believe your solution matches the given solution then just type in 'OK'.

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

Self-critique (if necessary):

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Self-critique rating:

Otherwise evaluate the quality of your self-critique by typing in a number between 0 and 3.

• 3 indicates that you believe you have addressed all discrepancies between the given solution and your solution, in such a way as to demonstrate your complete understanding of the situation.

Self-critique (if necessary):

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Self-critique rating:

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Assignment 10

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