Simple sinosudial wave in string. What will be velocity of the particle?

[vkash]

Homework Statement

consider a simple sinosudial wave traveling in a string with equation y=A*sin(wt-kx). Draw it's graph for t=0. It's simple sin curve. Now i am required to find out the direction of velocity of particle in between x=λ/4 to x=λ/2.

IF T is time period of oscllations then
w is 2*pi/T. k=2*pi/λ, λ is wavelength.

2. The attempt at a solution
velocity of particle is dy/dt.doing partial differentiation of the above equation.
v=dy/dt=Aw*cos(wt-kx). If we draw it's curve it will like a cos curve for all the points in betwen x=λ/4 to x=λ/2 it is in negative direction so velocity of these particles should negative but it is not so. Velocity of particle should +ve. where am i doing it wrong.
thanks...

 

[PeterO]

Homework Statement

consider a simple sinosudial wave traveling in a string with equation y=A*sin(wt-kx). Draw it's graph for t=0. It's simple sin curve. Now i am required to find out the direction of velocity of particle in between x=λ/4 to x=λ/2.

IF T is time period of oscllations then
w is 2*pi/T. k=2*pi/λ, λ is wavelength.

2. The attempt at a solution
velocity of particle is dy/dt.doing partial differentiation of the above equation.
v=dy/dt=Aw*cos(wt-kx). If we draw it's curve it will like a cos curve for all the points in betwen x=λ/4 to x=λ/2 it is in negative direction so velocity of these particles should negative but it is not so. Velocity of particle should +ve. where am i doing it wrong.
thanks...

Note that you were only required to find the direction of motion of the particles.
One way to find that is to draw the wave when t > 0 , but only a little more.
Many parts of that new wave will be below the first - showing the points are moving down, and many will be above - showing they are moving up.

When doing that you should find that all the points between x=λ/4 to x=λ/2 are moving in the same direction.

 

[vkash]

Note that you were only required to find the direction of motion of the particles.
One way to find that is to draw the wave when t > 0 , but only a little more.
Many parts of that new wave will be below the first - showing the points are moving down, and many will be above - showing they are moving up.

When doing that you should find that all the points between x=λ/4 to x=λ/2 are moving in the same direction.

I know that's correct..
what's wrong with my method.
that seems to be correct but gving incorrect answer.

 

[PeterO]

what's wrong with my method.
that seems to be correct but giving incorrect answer.

I could be glib and say your method gave the wrong answer - but you realize that.

I always find the maths over complicates this simple sort of question - and even when you get a positive or negative answer, you have to interpret whether that means moving up or moving down.

I always prefer a simple diagram.

Note: I didn't say whether that section of the wave was moving up or down, just that the entire section is moving in the one direction.

I am not exactly sure what the wave should look like when drawn.

If the wave moving left or right?
Do the particles on the string first go up, then down then back to where they were, or is it down first, as the wave passes?
Is the λ/4 position one quarter of a wavelength from the leading edge of the wave or the trailing edge of the wave? - left hand end of a full cycle or right hand side of a full cycle?

I don't know the answer to any of those questions.

 

[vkash]

If the wave moving left or right?
Do the particles on the string first go up, then down then back to where they were, or is it down first, as the wave passes?
Is the λ/4 position one quarter of a wavelength from the leading edge of the wave or the trailing edge of the wave? - left hand end of a full cycle or right hand side of a full cycle?

I don't know the answer to any of those questions.

yes the wave is moving from left to right.

rest of the answer are in attachment mage.
take intersection of axises (0,0) then λ/4 and λ/2 distances are measured from origin(along x axis).
I think + means velocity in upward direcion and - means velocity in downward direction(for cosine curve of velocity).

 

[PeterO]

yes the wave is moving from left to right.

rest of the answer are in attachment mage.
take intersection of axises (0,0) then λ/4 and λ/2 distances are measured from origin(along x axis).
I think + means velocity in upward direcion and - means velocity in downward direction(for cosine curve of velocity).

I like your interpretation of + - up - down.

Never mind the maths - just draw the wave ever so slightly to the right and you will see all the parts are moving up.

Perhaps the partial derivative should in fact have been wrt x; ie -kw*cos(wt-kx) Then the velocity would be positive. [Just trying to get a mathematical answer that matches what the answer has to be.]

 

[vkash]

I like your interpretation of + - up - down.

Never mind the maths - just draw the wave ever so slightly to the right and you will see all the parts are moving up.

Perhaps the partial derivative should in fact have been wrt x; ie -kw*cos(wt-kx) Then the velocity would be positive. [Just trying to get a mathematical answer that matches what the answer has to be.][/color]

Oh wow u are breaking rules to get answer; Is it good? Being retired physics teacher(an experienced ex teacher) u should not do so to get your answer.
How ever this time i am here with a correct answer for this question.

wave equation of a particle(as in figure) is for a specified time so one can not draw a graph for y=sin(wt-ks+p). If you want to draw graph for y and x then u have to fix t; If not then there will a lot of curves. OK;
So let me say that displacement curve(as in diagram) is at t=0. then phase constant became p=π(to get the displacement graph as in figure).
Now put p=π and t=0, in second curve(velocity function) u will get Aωcos(kx-π)=-Aω*cos(kx).
This graph will tell velocity displacement of the particle.(got the answer without breaking any rule)
that's called answer!

 

[PeterO]

Oh wow u are breaking rules to get answer; Is it good? Being retired physics teacher(an experienced ex teacher) u should not do so to get your answer.
How ever this time i am here with a correct answer for this question.

wave equation of a particle(as in figure) is for a specified time so one can not draw a graph for y=sin(wt-ks+p). If you want to draw graph for y and x then u have to fix t; If not then there will a lot of curves. OK;
So let me say that displacement curve(as in diagram) is at t=0. then phase constant became p=π(to get the displacement graph as in figure).
Now put p=π and t=0, in second curve(velocity function) u will get Aωcos(kx-π)=-Aω*cos(kx).
This graph will tell velocity displacement of the particle.(got the answer without breaking any rule)
that's called answer!

As a retired physics teacher, I was using physics rather than mathematics to solve the problem.

 

[PeterO]

yes the wave is moving from left to right.

rest of the answer are in attachment mage.
take intersection of axises (0,0) then λ/4 and λ/2 distances are measured from origin(along x axis).
I think + means velocity in upward direcion and - means velocity in downward direction(for cosine curve of velocity).

Your included graph is a displacement vs time graph (or at least it is labelled that way). λ/4 and λ/2 are both distances, as you said.

How did you measure along a time axis to get a distance?

 

[vkash]

Your included graph is a displacement vs time graph (or at least it is labelled that way). λ/4 and λ/2 are both distances, as you said.

How did you measure along a time axis to get a distance?

Just take y-axis as y(particle displacement) and x as displacement of wave.
I did not draw that graph i place that from google images. So it is not so as i want.Oh sorry sir; i did not see that.
Newer and better image in attachment. We are required to find velocity of particle between t₁ and t₂.
thanks for taking interest in my post;