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course Phy 232

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 distance that something travels, the wavelength, and we know how many of these distances pass through in one second, the frequency, we can multiply them together to get the velocity of wavelength segments.

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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 **

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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 periodic wave is in seconds. Therefore, the wavelength (m) divided by the velocity (m/s), gives us final units in seconds. This proves that the units are correct for this equation. The wavelength is a distance, and if we know that distance and the velocity during that distance, we can calculate the time in seconds at any given point during that distance. That is why period = wavelength / velocity.

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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:

A certain point located further away down a wave is at position d = x. This means that this point is x meters away from the origin, which is located at d = 0. In order to find the time it took that point to travel from the origin to the point d = x, is calculated by the distance x / the velocity of that point. Dividing these gives you units in seconds, which is in units for time. To find the time at which the point was at the origin, we must say that t - (x / v). This represents the time at the point minus the time it took from the origin to get to that point. This then can give us the time at the origin when the point was located there. This means at the clock time t - (x / v), the position of the point is described by the equation, A sin (`omega (t - x / v)).

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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)).

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