Assignment 10

course Phy 232

June 25 at 3:00

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:

V = lambda*f

Units of lambda (wavelength) – m : this is the space between the peaks of the wave

Units of f (frequency) – 1/s = Hz = this is the amount of wavelengths that occur in a second.

The product of the two is: m*Hz = m/s = which is the units of velocity.

If you know that the space between wavelengths, and the period of the wavelength, you can divide the space by the period to get the speed. This intuitively makes more sense to me.

Confidence Rating: 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: Ok

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:

The period of a wavelength is the amount of time between the peaks of the wavelengths. So if we know the distance between two peaks, and we know the speed at which the wave is moving, we can reason that by dividing the distance by the distance per time, we can determine the time necessary to travel from one peak to the next.

Confidence Rating: 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:OK

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:

The equation A sin( `omega t - x / v) takes into account the time it takes to travel from x = 0 to a particular point x. By dividing the initial point by the velocity, we can calculate the time it requires to get there. However, if the position is 0, this is not necessary, so then the equation becomes A sin(`omega t).

Confidence Rating: 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.

Your Self-Critique:

The point made that what happens at clock time t happened at x=0 when clock time was t = x/v.

This intuitively makes sense. The time lag simply carries the function from x = 0 to the specified point where we want to start, making that x = 0.

good statement

Self-Critique Rating: 2

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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:

The frequencies of the first few harmonics are found by dividing the velocity (distance/time) by the wavelength (distance). This gives us the amount of wavelengths which will occur over a specified period of time.

Confidence Rating: 3

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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: OK

Your Self-Critique Rating: Ok

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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:

The velocity of the string = (Tension*Length/Mass)^.5

Tension is measured in N, and Length is in meters. So this can be described as a unit of work.

Work = Force*Distance = Mass* Acceleration*Distance

Dividing the work by the mass gives us: Acceleration * Distance, which has the units of (distance/time)^2

The sqrt of this gives us the velocity in distance/time.

Confidence Rating: 3

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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: Ok

Your Self-Critique Rating: Ok

&#This looks good. Let me know if you have any questions. &#