Solved Problems I (Wave Phenomenon)

1. A tension F exercises in a rope with mass per unit length ρ_{o$\left[ {}^{kg}/{}_{m} \right]$}; we neglect the gravitational force in this case. A transverse harmonic wave (in term of $\cos (\omega t+\phi (x))$ ) travels with frequency $f={}^{2\pi }/{}_{\omega }$ along the rope. The wave travels in the positive x-direction, while the disturbance takes place in the y-direction.

a.       Which transverse force is working on the small part (length ds) of the rope during the disturbance?

b.      We provide now an extra mass with mass per unit length ρ₁ on the rope. The transition lies at x=0 (ρ=ρ_o for x<0 and ρ=ρ₁ for x>0). Which boundary conditions can be applied with regard to the disturbance at x=0?

c.       At x=0, we introduce now a point mass m$\left[ kg \right]$ on the first uniform rope (with mass per unit length ρ_o). We neglect the spatial dimensions of the mass in this case. Define the boundary conditions at x=0 for transverse wave movement.

Solution:

a.

             $\sum{F=ma}$, dissolving the disturbance in y-direction: $F\sin (\alpha +d\alpha )-F\sin \alpha ={{\rho }_{o}}ds\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$

$F(\sin \alpha \cos d\alpha +\cos \alpha \sin d\alpha )-F\sin \alpha ={{\rho }_{o}}ds\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$ since dα is small: $\cos d\alpha \approx 1$ and $\sin d\alpha \approx d\alpha $. Therefore: $F\cos \alpha d\alpha ={{\rho }_{o}}ds\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$. We simplify further this equation with: $\tan \alpha =\frac{\partial y}{\partial x}$ differentiating to x, we obtain: $\frac{1}{{{\cos }^{2}}\alpha }\frac{d\alpha }{dx}=\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}$  -> $d\alpha ={{\cos }^{2}}\alpha dx\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}$

Thus: $F\cos \alpha {{\cos }^{2}}\alpha dx\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}={{\rho }_{o}}ds\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$ applying Taylor expansion ($\cos x=1-\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}-...$ ) of cos α for small angle α: $\cos \alpha \approx 1$ and $ds\approx dx$ ,finally we obtain: $\frac{{{\partial }^{2}}y}{\partial {{x}^{2}}}=\frac{{{\rho }_{o}}}{F}\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$.

b.  

Assuming that there is no nod (smooth rope), the boundary conditions are ${{y}_{o}}={{y}_{1}}$ and $\frac{d{{y}_{o}}}{dx}=\frac{d{{y}_{1}}}{dx}$ at x=0.

            

 c.

The boundary conditions are: ${{y}_{o}}={{y}_{1}}$ ; $\frac{d{{y}_{o}}}{dx}=\frac{d{{y}_{1}}}{dx}$ and $Fd\alpha =m\frac{{{\partial }^{2}}y}{\partial {{t}^{2}}}$.  

2.   A transverse force F_tr(t) applies on starting point O of  a long rope, a harmonic transverse wave is formed with angular frequency ω. The amplitude of the periodics force is y_o. ρ is the mass per meter of the rope and F_s is the tension force applied on the rope.

 

a.       Give the general expression of lateral deflection y(x,t) of the rope as a function of space and time as a result of the transverse wave and give also the expression of transverse velocity v_tr(x,t).

b.      Apply now the specific expression for the oscillatory rope. The resultant of the horizontal tractional force F_h and the driving force F_tr on the starting point of the rope is in balance with the tension F_s of the rope at all time. Give a sketch of the equilibrium of forces apply on the rope, notice the transverse components of the forces and define the relationship between F_tr and F_s.

                 Solution:

a.       $y(x,t)={{y}_{o}}\cos (kx-\omega t+{{\phi }_{o}})={{y}_{o}}\cos (kx+{{\phi }_{o}})\cos \omega t+{{y}_{o}}\sin (kx+{{\phi }_{o}})\sin \omega t$

Complex representation:

$Y(x,\omega )={{y}_{o}}\left[ \cos (kx+{{\phi }_{o}})+i\sin (kx+{{\phi }_{o}}) \right]={{y}_{o}}{{e}^{i(kx+{{\phi }_{o}})}}={{y}_{o}}{{e}^{i{{\phi }_{o}}}}{{e}^{ikx}}$ 

$y(x,t)=$ Re$\left[ Y(x,\omega ){{e}^{-i\omega t}} \right]=$ Re$\left[ {{y}_{o}}{{e}^{i{{\phi }_{o}}}}{{e}^{i(kx-\omega t)}} \right]$ 

${{v}_{tr}}(x,t)=\frac{\partial y}{\partial t}=$Re$\left[ Y(x,\omega )\frac{\partial }{\partial t}({{e}^{-i\omega t}}) \right]=$ Re$\left[ Y(x,\omega )\cdot -i\omega {{e}^{-i\omega t}} \right]=$ Re$\left[ -i\omega Y(x,\omega ){{e}^{-i\omega t}} \right]$   

b.     

Balance of the forces: ${{F}_{s}}\sin \alpha +{{F}_{tr}}=0\to {{F}_{tr}}=-{{F}_{s}}\sin \alpha $

                                   ${{F}_{s}}\cos \alpha -{{F}_{h}}=0\to {{F}_{h}}={{F}_{s}}\cos \alpha $

For small α : ${{F}_{tr}}\approx -{{F}_{s}}\alpha $ and $\tan \alpha \approx \alpha =\frac{\partial y}{\partial x}$ therefore:

${{F}_{tr}}\approx -{{F}_{s}}\alpha =-{{F}_{s}}\frac{\partial y}{\partial x}=-{{F}_{s}}\frac{\partial }{\partial x}\left( {{y}_{o}}\cos (kx-\omega t+{{\phi }_{o}}) \right)={{F}_{s}}k{{y}_{o}}\sin (kx-\omega t+{{\phi }_{o}})$