Query10Phys

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course Phy 232

010. `query 10 *********************************************

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:

Since the wavelength is the distance for the wave to do a cycle and the frequency is the amount of cycles per unit of time. Then we can maultiply the two to find the velocity of the wave in units distance divided by units of time.

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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 wave is the amount of time it takes for a wave to complete one wavelength. It is obvious that the length of the wavelength divided by how fast the wave is traveling is the period of that wave.

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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:

This question is a little confusing to me, but I will give a try at solving it.

y=Asin(`omega(t-x/v))

As stated above, the x value of 0 is A sin( `omega (t))

Since it says it is sinusoidal and we see it in the equation, we know that the motion stays between 1 and -1 in terms of y values.

I understand that x/v is the time it takes for the function to reach an x-value. However, I do not understand why this wouldnt create 0 within the function since you are creating a time and subtracting it by the time.

@&
t is variable and x/v is constant.

There is of course on specific value of t at which t - x / v is zero, that value of course being t = x / v. At that instant, the displacement is zero, just as it was at the left end when t was equal to 0.
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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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Query10Phys

@& t is variable and x/v is constant. There is of course on specific value of t at which t - x / v is zero, that value of course being t = x / v. At that instant, the displacement is zero, just as it was at the left end when t was equal to 0. *@

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

@&
t is variable and x/v is constant.

There is of course on specific value of t at which t - x / v is zero, that value of course being t = x / v. At that instant, the displacement is zero, just as it was at the left end when t was equal to 0.
*@