Chapter 5 Atmospheric Oscillations 5.1. Show that the Fourier component F .x/ Re [C exp .imx/] can be written as F .x/ C cos m .x x0/, where x0 1 1 D D j j C D m sin .Ci= C / and Ci stands for the imaginary part of C: − − j j Solution: +F .x/ Re [C exp .imx/] Cr cos mx Ci sin mx, where C Cr iCi. Define Cr C cos mx0; Ci D 1 D − D C D j j D C sin mx0. Thus, mx0 sin− .Ci= C /, and F .x/ C .cos mx cos mx0 sin mx sin mx0/ C cos [m .x x0/]. j j D j j D j j − D j j − 5.2. In the study of atmospheric wave motions, it is often necessary to consider the possibility of amplifying or decaying waves. In such a case we might assume that a solution has the form A cos .kx νt kx0/ exp .αt/, where A is the initial D − − amplitude, α the amplification factor, and x0 the initial phase. Show that this expression can be written more concisely as Re Beik.x ct/, where both B and c are complex constants. Determine the real and imaginary parts of B and c in terms D − of a, α, k, ν, and x0. kc t Solution: Re B exp [ik .x ct/] Re B exp [ik .x crt/] e i , but from Problem 7.1 we can write this as D f kc t − g D f αt− g 1 D B cos [k .x crt/ kx0] e i A cos .[kx νt kx0]/ e , where A B , α kci, ν kcr, and kx0 sin− .Bi= B /. j j − − D − − D j j D D D − j j Thus, Br A cos kx0; Bi A sin kx0. D D − 5.3. Several of the wave types discussed in this chapter are governed by equations that are generalizations of the wave equation @2 @2 c2 @t2 D @x2 This equation can be shown to have solutions corresponding to waves of arbitrary profile moving at the speed c in both the positive and negative x directions. We consider an arbitrary initial profile of the field ; f .x/ at t 0. If the profile is translated in the positive x direction at speed c without change of shape, then f .x /, whereD x is a coordinateD moving at D 0 0 speed c so that x x0 ct. Thus, in terms of the fixed coordinate x, we can write f .x ct/, corresponding to a profile that moves in theD positiveC x direction at speed c without change of shape. Verify thatD − f .x ct) is a solution for any arbitrary continuous profile f .x ct). D − Hint: Let x ct x and differentiate− f using the chain rule. − D 0 @ @f @x0 @f Solution: f x0 ; x0 x ct;@x0 @x 1;@x0 @t c, Thus, by the chain rule, . D D − D D − @x D @x 0 @x D @x0 2 2 @ @ f @ @f @x0 @f Differentiating by x gives 2 2 . Similarly, @t @x @t c @x , which, when differentiated by t, gives @x D @x0 D 0 D − 0 @2 2 @2f 2 c 2 . Substituting into the wave equation then shows that f x0 is indeed a solution. @t D @x0 5.4. Assuming that the pressure perturbation for a one-dimensional acoustic wave is given by (7.15), find the corresponding solutions of the zonal wind and density perturbations. Express the amplitude and phase for u0 and ρ0 in terms of the amplitude and phase of p0: Solution: From the momentum equation: @ u @ u 1 @p0 ik Aeik.x ct/. Substituting in the assumed @t C @x 0 D − ρ @x D − ρ − ik.x ct/ 1 2 solution u Be gives ik .c u/ B .ik=ρ/ A, or B A [ρ .c u/]− . But, .c u/ γ .p=ρ/. Thus, u 0 D − − D D − − D 0 D p .c u/.γ p/ 1. Now linearizing (5.15) yields @ u @ ρ ρ @u0 ikρBeik.x ct/. Letting ρ Ceik.x ct/ gives 0 − − @t C @x 0 D − @x D − − 0 D − ik .c u/ C .ikρ/ B. Thus, C ρB .c u/ 1 A .c u/ 2. Hence, ρ p .c u/ 2. − D D − − D − − 0 D 0 − − An Introduction to Dynamic Meteorology, Solutions Manual Copyright © 2013 Elsevier Inc. All rights reserved. 21 22 Solutions Manual 5.5. Show that for isothermal motion (DT=Dt 0) the acoustic wave speed is given by (gH)1=2, where H RT/g is the scale height. D D Solution: Taking the log of the ideal gas law (p ρRT/ gives ln p ln ρ ln R ln T. For isothermal motion, D ln T=Dt 0, so that D ln p=Dt D ln ρ=Dt 0. This differs fromD the isentropic caseD (5.17)C byC the absence of γ . Thus, the solution is givenD by (7.16), with γ−replaced byD unity: c u .RT/1=2 .gH/1=2. − D ± D ± 5.6. In Section 5.3.1, the linearized equations for acoustic waves were developed for the special situation of one-dimensional propagation in a horizontal tube. Although this situation does not appear to be directly applicable to the atmosphere, there is a special atmospheric mode, the Lamb wave, that is a horizontally propagating acoustic mode with no vertical velocity perturbation (w0 0). Such oscillations have been observed following violent explosions such as volcanic eruptions and atmospheric nuclearD tests. Using (5.20), (5.21), plus the linearized forms of the hydrostatic equation and the continuity equation (5.15) derives the height dependence of the perturbation fields for the Lamb mode in an isothermal basic state atmosphere assuming that the pressure perturbation at the lower boundary (z 0) has the form (5.23). Determine the vertically integrated kinetic energy density per unit horizontal area for this mode.D Solution: When vertical velocity vanishes, the hydrostatic approximation (2.29) applies for the disturbance fields: @p =@z 0 D ρ g. Linearizing (5.15), and combining with (5.21) to eliminate u yields @ u @ p γ p @ u @ ρ 0, which − 0 0 @t C @x 0 − ρ @t C @x 0 D 2 2 implies that p0 cs ρ0. Using this result to eliminate ρ0 in the hydrostatic equation gives @p0=@z g=cs p0. Thus, p0 2D D − 1 D A exp gz=cs exp [ik .x ct/]. Substitution into the perturbation momentum equation (7.12) gives ρu0 .c u/− p0 −1 2− D − D .c u/− A exp gz=cs exp [ik .x ct/], where c u cs, and ρ ρ0 exp . z=H/. Averaged over a wave period, the − − − 2 2 −h D ± i D − ρu0 A 1 2g 1 2g kinetic energy density is given by 2 2 exp z H 2 , where we have assumed that A is real. But H 2 D 2ρ0cs − cs − cs D 1 2 3 2 R ρu 2 1 , where we have used that facts that c γ RT, γ 7=5, and H RT=g. Thus, 1 0 dz H − γ D − 7H s D D D 0 2 D A2 7H 5A2 2 3 6ρ g . 2ρ0cs D 0 5.7. If the surface height perturbation in a shallow water gravity wave is given by h Re Aeik.x ct/, find the corresponding 0 D − velocity perturbation u0(x,t/. Sketch the phase relationship between h0 and u0 for an eastward propagating wave. Solution: In rotated coordinates, from (5.37), νkg u0 h0 D ν2 f 2 − From (5.32) in rotated coordinates, ν2 ghk2 D and ν ck D Therefore, c u0 h0 D h Figure 5.8 shows the relationship between u0 and h0 for an eastward propagating wave. Solutions Manual 23 5.8. Assuming that the vertical velocity perturbation for a two-dimensional internal gravity wave is given by (5.65), obtain the 2 corresponding solution for the u0, p0, and θ 0 fields. Use these results to verify the approximation ρ0θ 0=θ p0=cs , which was used in (5.58). 2 u 3 6w 7 Solution: Since the system is linear all the disturbance fields must have solutions of similar form: 6 0 7 4 p0 5 D θ 0 82 u 3 9 > O > <6w7 = Re 6 O 7 exp [i .kx mz νt/] . Substitution into (5.61) then yields u .m=k/ w, substitution into (5.62) gives 4 p 5 C − O D − O > O > : θ ; O i .ν uk/ θ wdθ=dz, and substitution into (5.59) gives i .ν uk/ u .ik=ρ0/ p. The first and third of these − − O D − O 2 − − O D − O can be combined to eliminate u: p=ρ0 m .ν uk/ k w. These may be checked by substitution into (5.60) and use O O D − − O of the dispersion relation (5.66). To verify the inequality, note that from (5.62) we can deduce (using the definition of N2/ 2 that ρ0θ0 iρ .ν uk/ 1 N w , while from the relation between p and w given above, together with the definition θ 0 − g 0 D − − O O 2 p0 mρ0 cs γ RT γ gH, we obtain, 2 .ν uk/ 2 w0. Forming the required ratio and using the dispersion rela- D D cs D − − k γ gH 2 2 2 ρ0θ . p N g .h m.v uk/ i k m tion (5.66): 0 0 − γ H C .