A10 - OpenQuery

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course PHY 232

10/19/10 23:08

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:

Velocity units are meters per second (m/sec)

Wavelength ['lambda] units are meters (m)

Frequency ['omega] units are seconds (sec)

It is logical that dividing Wavelength/Frequency gives a result in m/sec

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:

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self-critique rating #$&*:

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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 velocity = wavelength/period, using algebra the period can be isolated to show that

wavelength/velocity = period

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:

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self-critique rating #$&*: 2

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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:

This is what I had trouble with in this problem set.

THe following is a guess:

Assume that the equation is true:

position(x) = A sin( `omega (t - x / v) )

for position(0) the equation becomes:

= A sin( `omega (t - 0 / v) )

= A sin( `omega (t - 0) )

= A sin( `omega (t) )

= A sin( `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:

This explanation helped a lot. The relationship between wavelength, period, and frequency is very

easy to understand but this was more of a challenge.

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self-critique rating #$&*: 2

Good work. That time-delay idea, and the equation of the wave, does bring together a number of concepts, so it's far from trivial.

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