query 10

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course Phy 232

7/132:30

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 10

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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:

Frequency (s^-1) tells us how many waves pass by in that time period. If you know the

length (m) of the wave, you can multiply length and frequency to find velocity (m/s).

The units verfiy this when they are multiplied.

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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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:

The period=wavelength/velocity. If you know the distance between peaks (wavelength) and how fast it's going (velocity)

then you can solve by using the equation above because you know how fast it takes for one wavelength to go a complete

cycle, which is 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. **

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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:

It has to do with the time taken to move from one point to another and the time the disturbance takes to get from

x=0 to x. This is x/v. When x is the distance down the wave, x/v is how fast it takes the wave to get to that

distance. When it's time t at position x, it's the same as t-x/v when x=0. This is like a sinusoidal wave because

it's a cycle like one.

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:

Your Self-Critique Rating:

8 9 10

1 8 9 10

2 16 18 20

3 24 27 30

4 32 36 40

We see directly from this sample space that 10 of the 12 possible outcomes are even. **

Self-critique

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Self-critique Rating:3

8 9 10

1 8 9 10

2 16 18 20

3 24 27 30

4 32 36 40

We see directly from this sample space that 10 of the 12 possible outcomes are even. **

Self-critique

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Self-critique Rating:3

#*&!

query 10

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