How Do You Calculate Phase Difference and Equilibrium Timing in Harmonic Waves?

[late347]

Homework Statement

the velocity of a harmonic wave in a strung-wire is 1550m/s
and the wave's frequency is 740Hz

there are two oscillators (?) in the wave called A and B such that B is 2m further ahead of A in the direction of the wave's travela) calculate the phase difference between A and B

b) If A at the time of t=0, is at its equilibrium position (?), then at what time (any one moment in time), will B be at its own equilibrium position?

Homework Equations

v=fλ wave equation
ΔΘ=k*Δx phase difference
$k= (2\pi)/\lambda$

possibly useful
$y(x, t) = \hat{y}sin[\omega *t -kx]$

The Attempt at a Solution

a) the problem is easily calculated by using wave equation to get wavelength
then find wavenumber k

wavelength = v/f
=2,0946m

k= 2,9997 radian/m

ΔΘ= 2,9997 radian/m * 2m = 5,9994 radians

b) honestly I was a little bit stuck at even understanding the question here. It says that A will be somewhere at an equilibrium position. I have a vague memory that in this particular case it means that the y value = 0 at that particular time. Because if we use the function y(x,t), it measures the y.

If you had a spring and a weight system (without friction) and the weight would go back and forth with the spring, then magnitude of springn Force increases directly proportional to the displacement from equilibrium position (?)

1.) I'm not really sure how or if that function formula is supposed to be used though.
the function y(x,t) = something

2.) and furthermore can the function be used to solve the problem, because we don't t´know the value of $\hat{y}$ ?

any tips to the B part would be welcome

 

[TSny]

Draw a picture of the wave at some instant of time. Mark points A and B.

 

[late347]

Draw a picture of the wave at some instant of time. Mark points A and B.

it looks like I tried to draw a picture of the instant of t=0
the time is at the beginning for both oscillators obviously

drawing these is a little bit tricky because we rushed through this subject a little bit in our class. How would you possibly draw a simple sine function,
when it has two inputs like y(x,t) = (some function's clause). What do you put fo r the axes of the drawing? I think of course y-axis must be y, and there exists some amplitude $\hat{y}$ . It will be a transverse wave judging by the assignment. Or the wave can be drawn as such a wave at least.

 

[late347]

$y (x, t) = \hat{y} * sin [w * (t- \frac {x }{v} ) ]$
$y (2, t) = 0$
$y (2, t) = \hat{y} * sin [w * (t- \frac {x }{v} ) ]$
$0= \hat{y} * sin [w * (t- \frac {2} {1550} ) ]$
divide both by $\hat{y}$ assume it is non-zero
$0 = sin[w*(t- \frac {1} {775})]$take arcsin
$0 = wt - \frac {w} {775}$
$\frac {w} {775} = wt$divide by w assume it is non-zero
$t= \frac {1} {775}$

something like this ?

 

[TSny]

Your picture looks good. You should be able to get the answer just using the picture.

Can you see how far the wave must travel to the right before point B is at its equilibrium point (y_B=0)?

 

[late347]

Your picture looks good. You should be able to get the answer just using the picture.

Can you see how far the wave must travel to the right before point B is at its equilibrium point (y_B=0)?

so youre saying that the equilibrium position will shift over to the B's static location (somewhere)... I think it means that the wave should start at A's location, and up at B's place. The distance ought to be 2m.

I suppose using that logic then... the time it takes for the wave to travel from A->B is what is really being asked for . s= vt <=> s/v =t

t= 2m/(1550m/s)

 

[TSny]

t = 2m/(1550 m/s) would be a time at which B would be at its equilibrium position. But it is not the earliest time for which B will be at its equilibrium position. If you look at your picture, you should see a point between A and B that is at the equilibrium position at t = 0.

 

[TSny]

Your mathematical solution in post #4 gives the same answer as your graphical solution in post #6. But this solution does not give you the earliest time for B to reach the equilibrium position.

 

[late347]

Your mathematical solution in post #4 gives the same answer as your graphical solution in post #6. But this solution does not give you the earliest time for B to reach the equilibrium position.

you got me there, I'm drawing a blank right now

 

[late347]

Your mathematical solution in post #4 gives the same answer as your graphical solution in post #6. But this solution does not give you the earliest time for B to reach the equilibrium position.

you mean that... in the first loop of the wave (in the distance between A and B) there is a node point.

When that node point hits the B then B will be at equilibrium

 

[TSny]

you mean that... in the first loop of the wave (in the distance between A and B) there is a node point.

When that node point hits the B then B will be at equilibrium

Yes. Hopefully that makes sense.

 

[late347]

Yes. Hopefully that makes sense.

Im not so sure anymore about that. I may be wrong, though.

The thing which seems to happen in the problem is that in the beginning if there isn't any wave anywhere. Then the "pulse " travels up to the location at 2m. Wouldn't that first pulse be the origin point of A itself?... so you're supposed to essentially draw a sine graph such that the equilibrium point is at the location B. Then draw the sine graph from B towards the left. Point A would no longer be at equilibrium but instead it would have some phase difference .

After the pulse has reached B... then basically A's original equilibrium has shifted to B. However A will simply no longer be at equilibrium itself but it doesn't even have to be anymore at that point in time (?).

Or is this the wrong idea about wave propagation velocity?

 

[TSny]

You are not dealing with "pulses" in this problem. As I understand the wording of the problem, you have a single harmonic wave traveling toward the right (positive x direction). Your picture shows the wave at some instant of time that you can take to be t = 0.

The statement of the problem says that "there are two oscillators (?) in the wave called A and B". This is peculiar wording. Is it the exact wording of the problem as it was given to you?

But I interpret this to mean that A and B are two points of the wire with B located 2 m to the right of A. At t = 0, point B is not at its equilibrium position. Rather, it is as shown in your picture. I believe the question is asking for the time when point B of the wire will be at y = 0. Your comment in post #10 is correct. So, use that idea to solve the problem.

 

[late347]

You are not dealing with "pulses" in this problem. As I understand the wording of the problem, you have a single harmonic wave traveling toward the right (positive x direction). Your picture shows the wave at some instant of time that you can take to be t = 0.

The statement of the problem says that "there are two oscillators (?) in the wave called A and B". This is peculiar wording. Is it the exact wording of the problem as it was given to you?

But I interpret this to mean that A and B are two points of the wire with B located 2 m to the right of A. At t = 0, point B is not at its equilibrium position. Rather, it is as shown in your picture. I believe the question is asking for the time when point B of the wire will be at y = 0. Your comment in post #10 is correct. So, use that idea to solve the problem.

I had to use a physics dictionary to translate some of the words, but that does sound just about right. Our teacher gave us an answer that 1.3 milliseconds is one correct answer

Indeed the question did originally state that anyone correct answer will suffice.

But the idea with the trivial solution is still correct I assume? In order to find out one solution for B's equilibrium position, one simply finds out when does the A's "node point" travel forwads up to the location of B, (as in essentially how much time it takes for the wave to travel that far.)

But looking at my picture it does look like the first instance of the zero position for B to come, will come about when the wave travels a little bit less forwards.

A's and B's location in the x-axis remains the same (it's an oscillator of the wave) so B is located 2m from A in the x-axis. And the it would seem that the "node point" in question is currently at 1/2 * $\lambda$

So the wave has yet to travel some more distance 2m - (1/2 * $\lambda$m)

$t= \frac {2m - (1/2 * \lambda ~~m)} {1550~~m/s}$

 

[TSny]

Yes. That all sounds very good.