THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

the pressure deviation dP:6.6 Application of Membrane Theory to the Kettledrum 121dP =− γ P 0dV =− γ P 0πa 2 〈z〉 (6.27)V 0 V 0This gives rise to an incremental force dP dx dy over incremental area dx dy ofthe membrane. Modifying Equation (6.3) to include this incremental force weobtain∇ 2 z − 1 γ P 0πa 2 〈z〉 = 1 ∂ 2 z(6.28)c 2 ρ 0 V 0 c 2 ∂t 2The term 〈z〉 is an integral function of all the permitted modes of vibration, whichmust also include the influences of their relative amplitudes and phases. We cangreatly simplify the solution of Equation (6.28) by assuming only one mode ofvibration and disregarding all of the other modes which constitute the generalsolution.The average displacement is zero for all normal modes dependent upon θ;therefore, none of these modes are affected by the pressure fluctuation of the airinside the drum. We need only to consider the symmetric modes entailing theBessel function J 0 . The solution with only one frequency present is of the formdepending only on the coordinate rz = Ψe iωtInserting the above into Equation (6.28) yieldsd 2 Ψdr + 1 dΨ2 r dr + k2 Ψ = γ P ∫ a02πrΨ dr (6.29)TV 0 0In order to establish the solution to the differential equation, we examine the salientfeatures of Equation (6.29). If the right-hand integral term were not present, thesolution would entail J 0 (kr). But the presence of this integral term involvingthe radius a suggests an additional term to the solution, one that is a functionof a, namely J 0 (ka). Moreover, the assumed solution should meet the boundarycondition that Ψ = 0atr = a, regardless of the value of k.We now integrate the right-hand side of Equation (6.29) as follows[2πγ P 0 rJ1 (kr)A − r 2 a0(ka)]TV 0 k 2 J 0= γ P 0TV 0πa 2 A[ ]2J1 (ka)− J 0 (ka)ka= γ P 0TV 0πa 2 AJ 2 (ka) (6.30)Insertion of Equation (6.30) into Equation (6.29) provides the condition for theviability of the assumed solution−k 2 J 0 (ka) = γ P 0TV 0πa 2 J 2 (ka)