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emailanmol
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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
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