Ph12a_s2_2009

# Ph12a_s2_2009 - Ph12a Solution Set 2 2.12 In a given mode...

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Unformatted text preview: Ph12a Solution Set 2 October 16, 2009 2.12 In a given mode with wavenumber k , the displacement is given as ψ = cos( ωt + φ )( A sin kz + B cos kz ) , where ω = c s k and c s = p T /ρ . Since the ends of the string are free, it must have horizontal tangents at both ends —compare with figure 2.9 in the book, and the adjacent discussion. So we get that ∂ψ ∂z = 0 at both boundary points z = 0 and z = L . For z = 0, cos( ωt + φ )( Ak cos k- Bk sin k 0) = cos( ωt + φ ) Ak = 0 , which gives us A = 0. For z = L , using now A = 0: cos( ωt + φ )(- Bk sin kL ) = sin kL = 0 , which means that kL = mπ for some integer m . Let’s look now into the three lowest modes. In the lowest mode, for m = 0, we have that k = 0, and since ω = c s k , the frequency is also zero. Replacing that into the expressiof for ψ , we have that ψ is a constant: ψ ( z,t ) = B cos φ . This is not the most general solution for zero frequency, though. The relevant equation of motion here is (14) in the book: ∂ 2 ψ ∂t 2 =- T ρ ∂ 2 ψ ∂z 2 . The frequency ω = 0 means that for this mode ∂ 2 ψ ∂t 2 = 0 everywhere and at all times. This equation has the general solution ψ ( z,t ) = A t + B , with A and B constants. The uniform motion solutions with A 6 = 0 were lost by assuming that the solution’s dependence in time was like a 1 Ph12a Solutions Problem Set 2 October 16, 2009 sinusoidal wave ψ ( z,t ) = A ( z ) cos( ωt + φ ): this assumption is not general enough for the ω = 0 mode. For the next mode, m = 1, the wavenumber is k 1 = π/L and the wave- length is λ 1 = 2 π/k 1 = 2 L . The mode dependence in space and time is ψ 1 ( z,t ) = B cos( ω 1 t + φ ) cos k 1 z with ω 1 = c s k 1 , and B and φ constants. For m = 2, the wavenumber is k 2 = 2 π/L and the wavelength is λ 2 = 2 π/k 2 = L . The mode dependence in space and time is ψ 2 ( z,t ) = B cos( ω 2 t + φ ) cos k 2 z with ω 2 = c s k 2 , and constants B and φ . 2.13 The equation of motion for each bead n = 1, 2, 3 is M ¨ ψ n = ( T /a ) ( ψ n +1- 2 ψ n + ψ n- 1 ) . We need to define also ψ and ψ 4 , which will be used to carry the boundary condition information. For the free boundary condition, that is the con- dition that the boundaries exert no force on the beads. If the string going from bead 1 to the wall makes no force, it must be not stretched. So the boundary condition can be expressed by saying that ψ = ψ 1 . Similarly, ψ 4 = ψ 3 . Let’s assume that the system is in a normal mode, with frequency ω , such that ψ n = A n cos( ωt + φ ) for each of the three particles. Equation (65) in the book applies here again with no changes: A n +1 + A n- 1 = A n 2- Ma T ω 2 ....
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## This note was uploaded on 01/28/2011 for the course PH 2 taught by Professor Dudko during the Spring '09 term at UCSD.

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Ph12a_s2_2009 - Ph12a Solution Set 2 2.12 In a given mode...

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