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MathematicalPhysicist
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There is a chain of springs, such that qk is the displacement of the kth mass in this chain.
We have the next equation of Newton second law, where T is the tension force:
[tex]\frac{d^2q_k}{dt^2}=T(q_{k+1}-q_k)-T(q_k-q_{k-1})[/tex]
Let Qk=qk+1-qk
I showed that if Q(t)k=Q(k-ct)=Q(s)
then we get the next ODE:
(1)[tex] c^2 Q''(s)=T(s+1)-2T(s)+T(s-1)[/tex]
I need to show that the solution is:
[tex]Q(s)=\int_{s-1}^{s+1} (1-|s-z|)T(z)dz[/tex]
by direct integration i.e integrating (1) or by Fourier transform.
I tried both but I get stuck, by direct integration I get:
[tex]c^2Q(s)= \int^{s} \int^{x} [T(z+1)-2T(z)+T(z-1)] dz dx[/tex]
don't know how to proceed from here, can anyone give a helping hand? (-:
We have the next equation of Newton second law, where T is the tension force:
[tex]\frac{d^2q_k}{dt^2}=T(q_{k+1}-q_k)-T(q_k-q_{k-1})[/tex]
Let Qk=qk+1-qk
I showed that if Q(t)k=Q(k-ct)=Q(s)
then we get the next ODE:
(1)[tex] c^2 Q''(s)=T(s+1)-2T(s)+T(s-1)[/tex]
I need to show that the solution is:
[tex]Q(s)=\int_{s-1}^{s+1} (1-|s-z|)T(z)dz[/tex]
by direct integration i.e integrating (1) or by Fourier transform.
I tried both but I get stuck, by direct integration I get:
[tex]c^2Q(s)= \int^{s} \int^{x} [T(z+1)-2T(z)+T(z-1)] dz dx[/tex]
don't know how to proceed from here, can anyone give a helping hand? (-: