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Equation for transverse wave 
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#1
Dec1712, 07:53 AM

P: 16

What is [an] equation for a transverse wave with no boundary conditions, as a function of x and t? I want to model a fluctuation string where neither of the ends are bound.



#2
Dec1712, 08:12 AM

P: 5,462

The wave equation and the boundary conditions are separate matters.
The simplest transverse wave equation is [tex]\frac{{{\partial ^2}y}}{{\partial {x^2}}} = {c^2}\frac{{{\partial ^2}y}}{{\partial {t^2}}}[/tex] You need to apply boundary conditions to establish the two arbitrary functions that appear in the solution. 


#3
Dec1712, 09:03 AM

P: 16

That's what I'm confused about. I know that I can express the motion of a vibrating string using sin and cos terms, but I don't know what BC's to apply if the ends of the string are free to move.
I want to express the shape of a string using sin and cos, but I am not sure what combination of terms is appropriate here. Would y=Acos(kxwt) + Bcos(kx+wt) + Csin(kxwt) + Dsin(kx+wt) fully describe the motion? 


#4
Dec1712, 10:21 AM

P: 5,462

Equation for transverse wave
Let the string be ABCD with A, D the ends of the string.
If both A and D are 'free to move' then how do you apply tension to the string? If you clamp at intermediated points, say B and C the AB and CD play no part in the wave motion. 


#5
Dec1712, 10:53 AM

P: 5,462

It does not really matter but I am going to put the c^{2} in its conventional place in what follows. Sorry that is what happens when you trust to an aging memory.
[tex]\frac{{{\partial ^2}y}}{{\partial {x^2}}} = \frac{1}{{{c^2}}}\frac{{{\partial ^2}y}}{{\partial {t^2}}}[/tex] Now for fixed ends the boundary conditions are [tex]y(0,t) = y(l,t) = 0[/tex] If the ends are not 'free' but still participating in the wave then they can be attributed initial displacement and velocity conditions [tex]\begin{array}{l} y(x,0) = f(x) \\ {\left( {\frac{{\partial y}}{{\partial t}}} \right)_{t = 0}} = g(x) \\ \end{array}[/tex] The wave equation itself may be solved by the method of separating the variables [tex]y(x,t) = F(x)G(t)[/tex] F is a function of x only and G a function of t only. Substituting and dividing through by y=FG [tex]\frac{1}{F}\frac{{{d^2}F}}{{d{x^2}}} = \frac{1}{{G{c^2}}}\frac{{{d^2}G}}{{d{t^2}}}[/tex] Both sides of this equation can only be equal if they are constant. Convention has this constant as λ^{2}. Some algebra on the resultant pair of ordinary diffrential equations will lead to your required trigonometric solution ( not the one you offered ) where F has the form [tex]{F_n}(x) = \sin \frac{{n\pi x}}{l}[/tex] and G has the form [tex]{G_n}(t) = {A_n}\cos {\omega _n}t + {B_n}\sin {\omega _n}t[/tex] Thus [tex]{y_n}(x,t) = {F_n}(x){G_n}(t) = \sin \frac{{n\pi x}}{l}\left[ {{A_n}\cos {\omega _n}t + {B_n}\sin {\omega _n}t} \right][/tex] Where A and B are determined by the intial conditiions. In general the solution above will not be complete since it depends upon n. To obtain a complete solution you need to sum solutions over n from n=1 to ∞ Thus [tex]\begin{array}{l} y(x,0) = f(x) = \sum\limits_{n = 1}^\infty {{A_n}} \sin \frac{{n\pi x}}{l} \\ {\left( {\frac{{\partial y}}{{\partial t}}} \right)_{t = 0}} = g(x) = \sum\limits_{n = 1}^\infty {{B_n}} {\omega _n}\sin \frac{{n\pi x}}{l} \\ \end{array}[/tex] 


#6
Dec1712, 12:21 PM

P: 16

Thank you, that makes sense. But why does Fn(x) only consist of sin terms, instead of both sin and cos terms?
What I originally had was y = Ʃ sin(nπx/L)*[An cos([itex]\omega[/itex]t) + Bn sin([itex]\omega[/itex]t] + cos(nπx/L)*[Cn cos([itex]\omega[/itex]t) + Dn sin([itex]\omega[/itex]t]. Is there a reason not to include the second term? ps, the reason why the ends are unbound is because there is not supposed to be any tension this is a string in Brownian motion (so there are no real initial conditions that I can apply either, since it does not have to be in any specific configuration at t=0). I guess it would have been better to talk about it as an abstract wave than an actual string. 


#7
Dec1712, 02:18 PM

P: 5,462

Without tension there is no wave in the string.
To derive the wave equation from first mechanical principles you require a restoring force. There is none in your situation. You string takes up the classic random walk line under the influence of brownian motion. This crosses and recrosses itself many times so takes up a but of an undulatory shape. But it is not a wave. Are you familiar with the mathematics of a 'random walk'? 


#8
Dec1712, 02:36 PM

P: 16

You are correct. It is definitely not a real wave in the classical sense. I am familiar with the random walk, and I agree that that is what this is. But since the string will have an internal resistance to bending, that rigidity brings it back to equilibrium (so the resistance to bending is the restoring force). It will fluctuate away from this equilibrium (and I am neglecting the fact that the entire string is diffusing, rotating or doing anything else more complicated) and the time scale is determined by drag forces, and amplitude of these fluctuations is determined the rigidity.
But the dynamics of the fluctuation should be able to be modeled as a simple wave equation I just wanted to check what that wave equation is, since most of the time (for obvious reasons) there actually are boundary conditions. I've modeled the shape at t = 0 as Ʃ Asin(nπx/L) +Bcos(nπx/L). Can I just multiply this by (cos(wt) + sin(wt)) to make it a function of time? 


#9
Dec1712, 05:32 PM

P: 5,462

I obtained my expression by carrying out the algebra (separating the partial equation into two ordinary differential equations, one in x, one in t) as stated in post#5 Using [tex]\frac{1}{F}\frac{{{d^2}F}}{{d{x^2}}} =  {\lambda ^2}[/tex] and the boundary conditions F(0)=F(l)=0 leads to the particular solution [tex]{F_n}(x) = \sin \frac{{n\pi x}}{l}[/tex] Where [tex]{\lambda _n} = \frac{{n\pi }}{l}[/tex] If we took any value but zero for the cosine term constant in the general solution we could not satisfy the boundary conditions. So we are left with only a sine function in x. Similarly the initial ordinary differential equation in t leads to [tex]\frac{1}{{G{v^2}}}\frac{{{d^2}G}}{{d{t^2}}} =  {\lambda ^2}[/tex] with intitial conditions as stated. Since this is an initial value problem that does not require G to vanish we put in our values to determine A and B in the general solution. You cannot avoid putting up some values for these to determine A and B 


#10
Dec1712, 06:09 PM

P: 16

I thought post # 3 and post 6 were equivalent through some trig identity.
I see why you applied the boundary conditions and set the coefficient on the cos(n[itex]\pi[/itex]x/L) term to zero. But if I want my wave to look and act like the progressive wave shown here: http://www.mta.ca/faculty/science/ph...e/Twave01.html I thought that I can't apply that boundary condition (because the ends aren't actually pinned). I remember deriving the equation that you showed for a particle in the infinate square well and always throwing out the spacial cos term. But I thought that in this case we need to keep it. 


#11
Dec1712, 06:41 PM

P: 5,462

What I am saying to you is that you need to find some initial conditions to put into f(x) and g(t) in post#5.
Alternatively you need to find some other points on your string. You cannot solve for the arbitrary functions/constants without this. The whole point (trick) of the exercise is to choose easy to solve positions on the string. For sure the ends of the string must start and be somewhere. 


#12
Dec1712, 09:06 PM

P: 741

You can assume that the displacement and velocity are periodic in distance with a period of length, κL. Here l This can be expressed by the following two equations. f(x,t)=f(x+κL,t), and, ∂f(x,t)/∂t=∂f(x+κL,t)/∂t, for all x and all t. Here, κ is an arbitrary constant that can be any real number. L is a scale length. 


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