Assignment 10

course Phy 202

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 frequency is the amount of peaks that pass in a second, and the wavelength is the length of each portion, then we need to multiply the two factors to find out how fast they are going by.

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:

Your Self-Critique Rating:3

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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:

By definition, the period of a wave is the time it takes for one complete cycle, or the time between peaks. Since the wavelength is the length of each cycle and the velocity is the speed at which that cycle goes by, we need to divide the wavelength by the velocity to get time.

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:

Your Self-Critique Rating:3

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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:

We are told that the position is x. The period of the wave is x (position)/v (velocity). Although I don’t really understand what is meant by a delay in the time, I know that we subtract x from omega t because of this time delay.

Confidence Rating:1

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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.

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Question: Query introductory set six, problems 11-14

given the length of a string how do we determine the wavelengths of the first few harmonics?

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Your Solution:

The length of a string is equal to ½ wavelength, so 2* the string length is equal to a whole wavelength. So we need to multiply the wavelength by 2 in order to get the string length, or divide the string length in half.

Confidence Rating:1

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Given Solution:

** As wavelength decreases you can fit more half-waves onto the string. You can fit one half-wave, or 2 half-waves, or 3, etc..

So you get

1 half-wavelength = string length, or wavelength = 2 * string length; using `lambda to stand for wavelength and L for string length this would be

1 * 1/2 `lambda = L so `lambda = 2 L.

For 2 wavelengths fit into the string you get

2 * 1/2 `lambda = L so `lambda = L.

For 3 wavelengths you get

3 * 1/2 `lambda = L so `lambda = 2/3 L; etc.

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Your wavelengths are therefore 2L, L, 2/3 L, 1/2 L, etc.. **

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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 frequency of a wave by definition is the wave velocity divided by the wave length. Thus is we take a look at the number of peaks found in the length of one segment and divide this by the length of the entire wavelength, we will be able to find the frequency.

Confidence Rating:2

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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. **

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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:

We can find the velocity of the wave in the string by using the equation:

V= (sqrt) [tension/(mass/length)]

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) ). **

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Question: gen phy explain in your own words the meaning of the principal of superposition

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Your Solution:

When two waves meet, the wave created is the sum of the size of each wave individually.

Confidence Rating:

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Given Solution:

** the principle of superposition tells us that when two different waveforms meet, or are present in a medium, the displacements of the two waveforms are added at each point to create the waveform that will be seen. **

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Question: gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?

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Your Solution:

This is the law of reflection. If a ray’s angle of incidence is equal to its angle of reflection, this means that the ray is striking a surface and being reflected back at the exact same angle.

Confidence Rating:3

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Given Solution:

** angle of incidence with a surface is the angle with the perpendicular to that surface; when a ray comes in at a given angle of incidence it reflects at an equal angle on the other side of that perpendicular **

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Assignment 10

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