Assingment 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 at

http://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:

If we know the wavelength then simply multiplying with the given frequency will give us the velocity

since we already know how many wavelengths will pass in a second.

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:

The given solution has the same idea as my solution

Your Self-Critique Rating:Ok

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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:

If we are given the velocity and wavelenght then we know Velocity=lambda/Period then if we rearrange

the

equation and solve for Period=lambda/Velocity

confidence rating #$&*:

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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:

The given solution has a different approach to the problem but a similar idea.

Your Self-Critique Rating: Ok

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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:

Since the quantity t is the time required for the disturbance to propagate from x=0 point to a certain

given position, therefore it takes until t=time for the motion of the particle to reach the same phase

as the motion of particle located at x=0, thats why A sin omega(t-x/v) or at x=0 Asin(omega t).

confidence rating #$&*:

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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:

The given solution has a different approach but similar idea.

Your Self-Critique Rating:Ok

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Self-critique (if necessary):

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Self-critique (if necessary):

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