What is the derivation of the wave equation for traveling waves?

[emailanmol]

Recently I was going through the derivation of wave equation
I want to discuss it to get my concepts fully clear by deriving and comparing the two major type of eqtns i came across.
I found two equations

1) When initial positon is x' and t=0

a) y=f(x-vt) for +ve direction
b) y=f(x+vt) for -ve direction.

2) When initial position is 0 and t=t'

c) y=f(t-x/v) for +ve direction
d) y=f(t+x/v) for -ve direction


So i would request the people who know this to go through this and see the derivation
-----------------------------------------------
Deriving equation (1)

If i take the intial postion to be x' and time=0 then y=f(x' + phase constnt) initially

Now let me consider a later time where horizontal position be x and velocity of wave be v.

Since there is no acceleration

x-x'=vt

So x'=x-vt (this is for +ve direcion of travel)

Now if the vertical displacement is same as initial case then

y=f(x-vt + phase constant)

However if the wave had been traveling backward in the -ve direction

then still x' = x-vt

but since x'>x (as its traveling in -ve direction i conclude v is -ve.)

So the equation becomes x=x+vt

so y=f(x+vt + phse constant)

but here when using this equation I will put only the absolute value of velocity as its -ve sign has already been considered in derivation.
X can be both –ve and +ve depending on the position.
Right? (This is my first question)

So Conclusions


If wave was traveling in +ve direction

y=f(x-vt + phse cnstnt)--------------(A)

if -ve direction

y=f(x+vt +phse cnstnt)---------------(B)


However the value of v in both cases is absolute

X can be both -ve and positive depending on direction.














Deriving equation (2)

Intially postion is x=0 and time =t'

so y=f(t' + phse constnt)

Later position is x,time is t.

So now x-0=v*(t-t')

so t'=t-x/v (for =+ve direction)

so y=f(t-x/v = phse cnstnt)


However if the wave had been traveling in -ve direction we still need to ensure that t >t'(as time always increases)
The eqtn still will be t’=t-x/v

Now since t>t’ so x/v is +ve .We know x has to be –ve as I am starting at x=0 and going backward.
So v is -ve

t’=t –x/v where both x and v is –ve.
So as t’=t+x/v with v as =ve that is I put the absolute value of v(for –ve direction)

Y=f(t+x/v + phse cnstnt)

So CONCLUSIONS

Y=f(t-x/v) for +ve x direction travel---------------------------------------------(C)

X is always +ve
V is +ve

Y=f(t+x/v) for -ve x direction travel------------------------------------------------(D)

X is always –ve
V is +ve (meaning we put the positive value of v)


Right? ------------------------(My 2nd Question)



Now let's consider the function to be sinusoidal.
W=Lambda(Wavelength)
T=Time period
P1-phase cnstnt 1
P2-phse cnstnt 2


So A becomes y=Asin{(x-vt)*(2pi/W) + P1}
& B becomes y=Asin{(x+vt)*(2pi/W) +P1}
We multiply by 2pi/W to make it dimensionless.Also then the dunction has the period W,T

So finally A is y=Asin(kx-wt+p1)
B is y=Asin(kx+wt +p1)

Here in both eqtns V is +ve and x can be both –ve and +ve.

Similarly C become y=Asin{(t-x/v)*(2pi/T)+P2}
And D becomes y=Asin{(t+x/v)*(2pi/T) +P2}

Here 2pi/T us multiplied for making it dimensionless and to make the function periodic at W,T

Finally C is y=Asin(wt-kx +p2)
And D is y=Asin(wt+kx +p2)

In C x is always +ve,v is always +ve
In D x is always –ve and v is always +ve

However here my main doubt comes when P1=P2 = 0 which just signifies a fixed origin

we see eqtn A and C are –ve of each other(for x as +ve and eqtn B=D for x as –ve)

So how do we know which eqtn to use
Different equations will give different value of y

 

[emailanmol]

Also what's the difference b/w the eqtns?

 

[Studiot]

You seem to be basically talking about D'Alembert's Solution.

http://www-solar.mcs.st-and.ac.uk/~alan/MT2003/PDE/node12.html

The point is that y = f(x-vt) etc are not complete solutions to the wave equation.

A complete solution is a (linear) combination of solutions as in the link above.

so y = f(x-vt) + g(x+vt) is a complete solution.

Don't worry if you are not familiar with the partial differentials in the link, just proceed to the worked examples with the sines.

 

[emailanmol]

Okay..

I have another query.

If we take take the equation as y=asin(kx-wt) for +ve direction and y=asin(kx+wt) for -ve direction then it means that if the wave is traveling ahead from x'(initial place) then the particle velocity (dy/dt) will be -awcos(kx-wt) and if its traveling behind then it will be awcos(kx-wt).

So if I am standing and observing the wave then I would see that If the wave would have gone ahead the particle would have gone up(or maybe down,i am taking it to go up as an example) but if the wave would have gone back then the particle would have gone down(or up) meaning opposite to the direction.

So like such callibration of writing x-vt as x+vt is also changing the particle velocity.

Thankyou for your help.

 

[Studiot]

Remember if waves of both direction are present then the result is a standing wave.

 

[emailanmol]

Okay.
And one more doubt.

Why is it that the PE of a particle in standing waves has its maximum value at peak while in a traveling wave its value is 0 or minimum at the peak.

Why does this difference come?

Also in a traveling wave why is the string more stretched at zero displacement.?
How do we know its most stretched?

Also when we derive the potential energy like here

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/AnalyzingWaves.htm

Section Energy and Power in a Traveling Harmonic Wavewhy are they taking only the force exerted by left particle on right particle.
Why aren't we taking work done by the net force on a particle which is Fd(tanx)

 

[Studiot]

Why is it that the PE of a particle in standing waves has its maximum value at peak while in a traveling wave its value is 0 or minimum at the peak.

Depends where you measure displacement from. This also determines whether you have a sin or cos wave.

Mechanical examples are easiest. So think about a pendulum.

As it passes through its lowest point its velocity is max so its kinetic energy is a x. Since the oscillation is a continual exchange of KE and PE its PE is a min.

Similarly as it passes through its highest point in the swing its velocity is momentarily zero. So its KE is a min and its PE is a max.

Now do you measure displacement from the highest or lowest point in the wave?

 

[emailanmol]

I take a sine wave and a traveling wave.I am measuring the displacement from the equilibrium position ie the string if wud hv been undisturbed.

For it both the KE and PE at peak is zero and max at zero displacement.
Here DU/dt=dK/dt...ie its nt that PE and KE are interchanging.
they are both changing equally.both are increasing simultaneously and decreasing simultaneously...so comparing it to pendulum is wrong i guess.

in a standing wave however energy is interchanging into potential and kinetic.

Also

pls answer these Sir if u can

"Also in a traveling wave why is the string more stretched at zero displacement.?
How do we know its most stretched?""why are they taking only the force exerted by left particle on right particle.
Why aren't we taking work done by the net force on a particle which is Fd(tanx)"

 

[Studiot]

For it both the KE and PE at peak is zero and max at zero displacement.

This is not correct. Make sure you get the basics right before trying the more difficult stuff.



http://en.wikipedia.org/wiki/Simple_harmonic_motion

 

[emailanmol]

No Sir,i guess there's a confusion here.

I now posted a new thread on this entire thing to give the mathematical view.
In simple harmonic motion we have the system as isolated.

In a traveling wave there's is an external source creating the wave
So the dK/dT=-dU/dT

You can view this at
http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/AnalyzingWaves.htm

or even in the famous Resnick Halliday Krane

I am sorry if I am wrong.

 

[emailanmol]

I mean the dK/dt= -dU/dt in simple harmonic motion not traveling waves

for traveling waves dK/dt=dU/dT
(Source Resnick Halliday Krane)