Pressure Wave Analysis of a Cylindrical Drum

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1 Pressure Wve Anlysis of Cylindricl Drum Chris Clrk, Brin Anderson, Brin Thoms, nd Josh Symonds Deprtment of Mthemtics The University of Rochester, Rochester, NY 4627 (Dted: December, 24 In this pper, hypotheticl drum consisting of vibrting circulr membrne with hollow cvity ws nlyzed. A coupling reltion ws used to link the vibrtion of the drum with the pressure wves inside the cvity. The boundry vlue problem with differentil equtions ws nlyzed nd produced generl solution s well s specific solution under certin resonble ssumptions. After nlysis, our model produced single tone. INTRODUCTION We re curious bout the time-dependent behvior of pressure wves inside drum. To investigte this problem, we crete n idelized drum model nd use techniques of fourier nlysis to obtin simplified representtion of ir motion in the cvity. ANALYSIS AND RESULTS The drum model is constructed with tught circulr membrne ffixed to one end of hollow cylindricl tube s shown in Figure. The rdius of the membrne is nd the length of the cvity is L. The speed of sound in the membrne is c M nd the speed of sound in ir is. The men ir density in the drum cvity is ρ. The being struck, begins vibrting nd cretes pressure wves in the djcent ir. The vibrtions in the membrne re governed by the wve eqution 2 u = c2 M 2 u, where u is the displcement of the membrne. Under the ssumption tht the initil strike is circulrly symmetric, the solution u will be function of only r nd t. The generl solution to this problem is u(r, t = n J ( λ n rcos(c M λn t+ b n J ( λ n rsin(c M λn t, where the coefficients n nd b n re determined by pplying the initil conditions.[] Initilly, the membrne is flt, so u(r, = n =. The initil velocity is determined by the nture of the strike tht hits the membrne. A resonble function is cosine curve tht peks t r = nd reches zero t the boundries. For mthemticl convenience, it is ssumed tht initil velocities re given by Bessel function, u ( t (r, = β(r = J z r, (where z is the first zero of the zero th Bessel function which is very similr to the cosine curve s evidenced by Figure 2. The membrne hs the boundry condition u(, t =, since it is fixed to the drum tube. Therefore, J ( λ n =. FIG. : Drum Model drum opertes vi the principle tht the membrne, fter The lest solution of this eqution corresponds to the first eigenfunction, so λ = z

2 where v is the velocity of the membrne.[2] To mke the problem trctble, we will mke the dditionl ssumption tht ech verticl column of ir is independent of ll others, which mens tht only the z component of the membrne s motion is relevnt, so the coupling eqution becomes p 2 u z = ρ z=. ( The pressure creted ccording to this eqution then propgtes through the ir in the cvity following the wve eqution 2 p = 2 p c2 A z 2. We seek solutions for p of the form p(r, z, t = f(rg(zh(t. As shown in Appendix B, the generl solution is [ ( ( ] p(r, z, t = f(r C sin z + C 2 cos z [C 3 sin( t + C 4 cos( t] We tke the pressure to be zero initilly: p(r, z, =, 2 FIG. 2: Comprison of Cosine nd Bessel Function In order to pply the initil velocity condition, we use the result from Appendix A tht the b n re given by b n = β(rj ( λ n rrdr J 2 ( λ n r(c M λn rdr. Now substituting the initil condition in with the replcement λ = z / gives b n = J ( λ rj ( λ n rrdr J 2 ( λ n r(c M λn rdr. Since the Bessel functions re orthogonl with weight r, b n = for n nd b = c M λ = c M z. The complete solution is then u(r, t = ( z ( J c M z r z sin c M t. The next step is to clculte the effect of this membrne displcement on the ir below it. By symmetry, the pressure in the ir will be function of only r, z, nd t, so we will write the pressure s p(r, z, t. The coupling t the interfce is determined by the differentil eqution p(r,, t = ρ v t, since there is no displcement of the membrne t t =, so C 4 =. The coupling eqution hndles one boundry condition, nd for the second, we use the ssumption tht the pressure is zero t the end of the drum where it meets open ir: p(r, L, t =, since the lrge volume of ir quickly dmps ny pressure. For this to hold without trivil solution, the g function will hve to be zero t z = L: C sin ( L + C 2 cos ( L C 2 = C tn Inserting these results, we obtin the form = ( L p(r, z, t = Cf(r [ ( ( ( ] sin z tn L cos z sin( t. Now pplying the coupling eqution (, [ ] Cf(r sin( t = ρ ( z ( J c M z r z ( sin c M t z c M 2

3 3 By equting time prts, sin( ( z t = sin c M t z = c M By equting rdil prts[3], f(r = J ( z r. And finlly, equting coefficients, c M z c M z C = ρ C = ρ = ρ. Therefore, the desired solution for p is: ( z p(r, z, t = ρ J r [ ( ( cm z sin z cm z tn L sin ( c M z t ( ] cm z cos z This expression is plotted t z = L/2 in Figure 3. wves. In generl, our solution provides physiclly plusible result. The model suggests tht the drum cretes single tone of frequency f = c M z 2π. It is importnt to note tht the zero of pressure ws defined to be the initil pressure, so negtive pressures re observed. During the course of the derivtion, mny simplifying ssumptions were mde tht hve corresponding effects on the ccurcy of the result. The ssumption tht the initil velocity of the membrne is Bessel function ws very fortuitous, but not terribly unrelistic becuse the velocities creted by striking drum t its center decrese with incresing rdius, which is consistent. The boundry conditions tht we used introduced more ssumptions including the ssumption tht the pressure is zero t the open end of the drum. This ssumption implies tht no wves propgte through the end of the drum becuse longitudinl wves cnnot penetrte region of zero mplitude like trnsverse wves cn. Therefore, you would hve to ctully be inside the drum in order to her it ccording to our model. The most unrelistic ssumption ws tht the ir in the drum ws divided into mny independent columns of ir, ech with pressure function creted by the piece of vibrting membrne bove it. In n ctul drum these different pressures long the rdius would influence ech other nd produce much more complicted solution. This ssumption is responsible for why the solution suggests tht the drum emits single frequency. CONCLUSION FIG. 3: Pressure Plot From this figure we cn see tht the pressure distribution plot is clerly dependent on r nd oscilltory with respect to time. The pressure wve hs gretest mplitude t r =, nd goes to zero t the edge of the drum tube. This is consistent with the initil conditions on the membrne. DISCUSSION The results of our nlysis re consistent with wht would be expected physiclly. The rdil prts of our solution re Bessel functions, which is wht one would expect for circulr membrne. The time solutions re sinusoidl, which is consistent with the oscilltion of sound We were ble to find functions tht describe pressure inside the cvity. Through coupling, we were ble to connect the vibrtion of the drum to the pressure wves inside the cvity. This gve us resonble understnding of the behvior of this system. By crefully selecting our initil conditions, we were ble to hrness the problem. Proceeding through the differentil equtions we rrived t the result tht our drum produces single frequency. As such, our model of the cylindriclly shped drum produces physiclly sensible result given simple initil conditions. [] R. Hbermn, Applied Prtil Differentil Equtions [2] Professor Gns, Lecture [3] Including multiplictive constnt is possible, but is not necessry

4 ppendix_.nb Appendix A Derivtion of membrne displcement function u Hr, tl We strt with the wve eqution, 2 u Hr, tl = c 2 2 u Hr, tl where c is the wve speed in the membrne. Due to the symmetry in the membrne, we will use cylindricl coordintes. Applying the Lplcin for cylindricl coordintes, we get, 2 u Hr, tl = c 2 t 2 r i jr k u Hr, tl y z { We ssume u Hr, tl is seprble into two functions, f HrL nd g HtL u Hr, tl = f HrL g HtL Seprtion gives us, f HrL 2 g HtL = c 2 g HtL r i k jr f HrL After diving both sides of the eqution by f HrL g HtL, the left side is purely function of time nd the right side purely function of position. We set both sides equl to λ. y z { g HtL c 2 2 g HtL = f HrL The ordinry differentil eqution for time is r i k jr f HrL y z = λ { which gives us solutions of the form 2 g HtL +λc 2 g HtL = g HtL = C sin Ic è!!! λ tm + C 2 cos Ic è!!! λ tm The ordinry differentil eqution for position is f HrL r i f HrL j + r 2 f HrL y k 2 z = λ { Mnipultion gives us n eqution in the form of Bessel differentil eqution

5 ppendix_.nb 2 r 2 f HrL 2 r 2 2 f HrL 2 We hve solutions to the Bessel function s + r2 2 f HrL = λr 2 f HrL 2 + r f HrL +λr 2 f HrL = f HrL = C 3 J m I è!!! λ rm + C 4 Y m I è!!! λ rm But since the vlue of our function must be finite for ll r, the second term of the solution to the Bessel eqution must go wy. Thus we re left with f HrL = C 3 J I è!!! λ rm Next we wnt to solve for the coefficients. We now know tht f HrL = C J I è!!! λ rm g HtL = C 4 sin Ic è!!! λ tm + C 5 cos Ic è!!! λ tm Which mens u Hr, tl = A n sin I è!!!!!!!! λ n ctm J I è!!!!!!!! λ n rm + B n cos I è!!!!!!!! λ n ctm J I è!!!!!!!! λ n rm We hve our two initil conditions u Hr, L = u Hr, L t =β HrL Applying the first initil condition, we re left with u Hr, L = = B n J I è!!!!!!!! λ n rm which implies tht B n = Applying the second initil condition, we hve u Hr, L t = A n I è!!!!!!!! λ n cm J I è!!!!!!!! λ n rm =β HrL Multiplying both sides by J I è!!!!!!!! λ m rm r nd integrting from zero to with respect to r, we get

6 ppendix_.nb 3 A n I è!!!!!!!! λ n cm J I è!!!!!!!! λ n rm J I è!!!!!!!! λ m rm r r = β HrL J I è!!!!!!!! λ m rm r r Using orthognlity nd solving for A n, we get Null A n = Ÿ β HrL J I è!!!!!!!! λ n rm r r IŸ J 2 I è!!!!!!!! λ n rm r rm I è!!!!!!!! λ n cm We now pply the boundry condition to solve for λ. Knowing tht u H, tl = gives us J I è!!!!!!!! λ n M = So if z n is the n th zero of the order Bessel J function then, λ n = J z n N2 nd our finl solution is : u Hr, tl = A n sin I è!!!!!!!! λ n ctm J I è!!!!!!!! λ n rm

7 ppendix_b.nb Appendix B Seprtion of vribles for the pressure function p Hr, z, tl Given the wve eqution : 2 p Hr, z, tl = c 2 2 p Hr, z, tl We ssume p Hr, z, tl is seprble into p Hr, z, tl = f HrL g HzL h HtL We plug in nd get f HrL g HzL 2 h HtL = c 2 f HrL h HzL 2 g HzL z 2 2 h HtL = c2 2 g HzL h HtL g z 2 And set these equtions equl to n eigenvlue 2 h HtL = h HtL c 2 2 g HzL = g z 2 These ordinry differentil equtions yield solutions h HtL = IC 3 sin I è!!! tm + C 4 cos I è!!! tmm g HzL = i j C sin i j k k And the complete generl solution is : p Hr, z, tl = f HrL i j C sin i j k k è!!! c z y { è!!! c z y { z + C 2 cos i è!!! j z y y c k zz {{ z + C 2 cos i è!!! j z y y c k zz IC 3 sin I è!!! tm + C 4 cos I è!!! tmm {{

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