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Benny

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Could someone help me out on the following questions?

Q. Consider the free vibrations of a string of length L clamped at x = 0 and constrained at x = L such that [tex]u_x \left( {L,t} \right) = - ku\left( {L,t} \right),k > 0[/tex].

(a) Show that the eigenvalues are given by the positive roots of: [tex]k\tan \left( {\omega L} \right) + \omega = 0[/tex]. How many positive solutions for omega does this equation have?

(b) The string is released from rest so that [tex]u_t \left( {x,0} \right) = 0[/tex], and the string is given an initial deformation given by u(x,0) = (hx)/L, where h is constant. Use the principle of linear superposition to find the solution u(x,t).

Wave equation: [tex]\frac{{\partial ^2 u}}{{\partial t^2 }} = c^2 \frac{{\partial ^2 u}}{{\partial x^2 }}[/tex]

For part (a) I just tried a separable solution [tex]u\left( {x,t} \right) = T\left( t \right)\phi \left( x \right)[/tex]. Where lambda is a constant I obtain the two ODEs:

[tex]

T''\left( t \right) + \lambda c^2 T\left( t \right) = 0

[/tex]

[tex]

\phi ''\left( x \right) + \lambda \phi \left( x \right) = 0

[/tex]

The boundary conditions (the string being clamped at x = 0 and constrained at x = L in the specified way) results in the following ODE problem.

[tex]

\frac{{d^2 \phi }}{{dt^2 }} + \lambda \phi = 0,\phi \left( 0 \right) = 0,\phi '\left( L \right) + k\phi \left( L \right) = 0

[/tex]

I found that if lambda is not positive then the ODE has trivial solutions so I take lambda to be positive.

[tex]

\phi \left( x \right) = A\cos \left( {\omega x} \right) + b\sin \left( {\omega x} \right),\lambda = \omega ^2 > 0

[/tex]

Plugging in the boundary conditions of the ODE above I get a A = 0 and [tex]F\left( {\sin \left( {\omega L} \right) + k\omega \cos \left( {\omega L} \right)} \right) = 0[/tex]. I take F not equal to zero to avoid a trivial solution to the ODE. So I get

[tex]

{\sin \left( {\omega L} \right) + k\omega \cos \left( {\omega L} \right)}

[/tex] = 0

which does not correspond to the given result. I also can't figure out how many positive solutions for omega the equation that I needed to derive in part (a) has.

I don't think I can do part (b) without having completed part (a) because I need to sum over a sequence of functions whose arguments contain omega to find the solution to the PDE.

Can someone give me some help? Thanks.

Q. Consider the free vibrations of a string of length L clamped at x = 0 and constrained at x = L such that [tex]u_x \left( {L,t} \right) = - ku\left( {L,t} \right),k > 0[/tex].

(a) Show that the eigenvalues are given by the positive roots of: [tex]k\tan \left( {\omega L} \right) + \omega = 0[/tex]. How many positive solutions for omega does this equation have?

(b) The string is released from rest so that [tex]u_t \left( {x,0} \right) = 0[/tex], and the string is given an initial deformation given by u(x,0) = (hx)/L, where h is constant. Use the principle of linear superposition to find the solution u(x,t).

Wave equation: [tex]\frac{{\partial ^2 u}}{{\partial t^2 }} = c^2 \frac{{\partial ^2 u}}{{\partial x^2 }}[/tex]

For part (a) I just tried a separable solution [tex]u\left( {x,t} \right) = T\left( t \right)\phi \left( x \right)[/tex]. Where lambda is a constant I obtain the two ODEs:

[tex]

T''\left( t \right) + \lambda c^2 T\left( t \right) = 0

[/tex]

[tex]

\phi ''\left( x \right) + \lambda \phi \left( x \right) = 0

[/tex]

The boundary conditions (the string being clamped at x = 0 and constrained at x = L in the specified way) results in the following ODE problem.

[tex]

\frac{{d^2 \phi }}{{dt^2 }} + \lambda \phi = 0,\phi \left( 0 \right) = 0,\phi '\left( L \right) + k\phi \left( L \right) = 0

[/tex]

I found that if lambda is not positive then the ODE has trivial solutions so I take lambda to be positive.

[tex]

\phi \left( x \right) = A\cos \left( {\omega x} \right) + b\sin \left( {\omega x} \right),\lambda = \omega ^2 > 0

[/tex]

Plugging in the boundary conditions of the ODE above I get a A = 0 and [tex]F\left( {\sin \left( {\omega L} \right) + k\omega \cos \left( {\omega L} \right)} \right) = 0[/tex]. I take F not equal to zero to avoid a trivial solution to the ODE. So I get

[tex]

{\sin \left( {\omega L} \right) + k\omega \cos \left( {\omega L} \right)}

[/tex] = 0

which does not correspond to the given result. I also can't figure out how many positive solutions for omega the equation that I needed to derive in part (a) has.

I don't think I can do part (b) without having completed part (a) because I need to sum over a sequence of functions whose arguments contain omega to find the solution to the PDE.

Can someone give me some help? Thanks.

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