Solutions A2-1 Official(English)EQ2:Acoustic black box solution1 A.1(0.2 pt)x(t)=ust cos(β)+R cos(⑴t+φ)+XC(1)y(t)=ust sin(β)+R sin(⑴t+φ)+YC(2)A.2(1.2 pt)Figure below shows the graph obtained for the data point interval 0.02.Sr not(s)fmin16.26545.36211.52544.4316.82544.03422.14543.85527.46543.75632.8543.69738.14543.65843.5543.62948.84543.591054.18543.58 1 Siddharth Tiwary(IIT Powai,Mumbai),Siddhant Mukherjee(The University of Cambridge,UK),Chandan Relekar(IISc,Banga-lore),Charudutt Kadolkar(IIT Guwahati),Praveen Pathak(HBCSE-TIFR,Mumbai),were the principal authors of this problem.The contributions of the Academic Committee and the International Board are gratefully acknowledged.fmin 545.5 545 544.5 544 543.5 10 20 30 40 50 t A.2(cont.)A.3(1.0 pt)We take a general case in which both detector and the source are moving with velocities Ud and Us respectively.Also,the line joining source and detector makes angleαwith the x-axis as defined infig.1 of the question.Note thatαis a function of time.Let"(人)be the vector joining the source and the detector.For the case when the source is approaching the detector,frequency detected by the detector is(6)Similarly,for the source moving away from the detector The expression of minimum frequency in the asymptotic limit(t→∞)is A.4(1.4 pt)Initial location of the source:Keep the detectorfirst onthe x−axis(say x1,0∘)and then on they−axis(say y1,90∘)and from the graph,note down the time taken to reach thefirst signal to the detector.Lets denote these timings asΔtx1 andΔty1 respectively.Then,(x−x1)2+y2=(cΔtx1)2(9)x2+(y−y1)2=(cΔty1)2(10)Solving above two equations will give the coordinates of the source.From the simulation,for x1=y1=500m,Δtx1=1.5344 s andΔty1=1.2727 s.Above equations have two solutions.We can keep the detector at third location to choose the correct pair.The answer is xA=419.99,yA=499.99 A.5(2.1 pt)Let the detector beat such a position where the source approaches the detector from a large distance(say from left side),crosses it and then moves away at a large distance(to the right side).In the asymptotic limits(far left and far right,β≈α),two pairs of the frequencies will be detected by the detector.We take Ud=0.On the far left side On the far right side Eqs.(11)and(12)yields Eqs.(13)and(14)yields It is also given that at t=0,there is afinite distance between the source and the detector.This will cause a signal delay.LetΔt be the time interval between two peaks(fmax).In this case Eqs.(15-17)can be solved together to obtain the values of Us,⑴,and R.It is necessary to keep the stationary detector at such coordinates(say XD,YD),so that the source approaches the detector from a large distance,crosses it and then moves away to a large distance.Note that the asymptotic behaviour can be identified in the region where the extrema in the graph remains almost constant.Also,we expect a sharp change in the graph if the detector’s distance from the origin is such that the angleα≈β.Keeping the distancefixed at 8000 m,we try with various values ofθ.A.5(cont.)20。30。10。50。60。40。80。90。70。
We can see that atθ=30。,far left and right parts of the graph show asymptotic behaviour.In these regions,peak frequencies do not show appreciable change.Notice that the values of the peak frequencies in the left side of the graph is higher than the values of the peak frequencies in the right side of the graph in this region.This indicates that the source is moving away from the detector in the right side of the graph.Detector is placed somewhere in the transient region.Expand the graph for a far left region this gives with a decreased data point interval(say 0.001)for a more accurate fmax and fmin numbers.fmin=788.24 Hz and fmax=5569.59 Hz.Inserting this in Eq.(11)Far right region gives fmin=543.96 Hz and fmax=1353.45 Hz.Inserting this in Eq.(12)equations(18-19)yields Us=91.1 m/s and⑴R=179.66 m/s.Also,for any two peaks in asymptotic case We use the value of Us=91.1 m/s to get⑴=1.49 rad s__1.From⑴R=179.66 m/s,R=120.57 m.To obtain f0,insert fmin=5327.82Hz on the far right side in Eq.(8)and solve for f0.This gives f0 to be 990.26 Hz.A.5(cont.)f0(Hz)⑴(s__1)R(m)US(m/s)990.26Hz1.49S__1120.57m91.1m/s A.6(2.0 pt)CalculatingβFigure below represents a schematic picture,where S is a source at a very large distance.P and Q represent two different positions of detectors placed at different instants.
S y xβP QAt large distances.let the time taken for the sound signal to reach at P detector:t0=1009.61 let the time taken for the sound signal to reach at Q detector:t1=1007.85 The distance between P detector and Q detector is 660m and corresponding time taken by sound to reach their respective detectors are 1009.61s and 1007.85s respectively.The expression for time difference is given by which givesβ=28.36。A.6(cont.)Alternate solution forβ:Fβy 2 C H E C’L 1 D B A B’θ2θ1 x ORed line AF depicts the direction of the velocity us of the circle.We aim to determineβwhich u s(→)makes with the x-axis.Value of the frequency detected by the detector depends on two aspects,first from which lo-cation on the cycloid,the source emitted the signal and second,on the location of the detector.Points H and L during one cycle of the source’s trajectory depict the location where the source’s speed is maximum and minimum respectively.This is due to u s(→)being parallel or anti-parallel to the tangential velocity component of the rotation on these points.As the source takes ntℎturn on the cycloid,detector on different angular positions on circular arc 1 will detect different values of fmax corresponding to those positions.Starting from the angular position near the x-axis(0o),fmax will keep increasing till the detector is kept on point D at(θ1).In fact,for any position on line BC which is parallel to AF,the detector will detect maximum of all fmax.Similarly,if the detector is placed anywhere on line B/C/which is also parallel to AF,it will detect minimum of fmin.In the simulation,you can change the angle by changing x,y coordinates and keeping the velocities zero.We repeat this exercise by changing the detector distance to arc 2.Scanning across the arc,angleθ2 can be obtained for which the detector detects maximum of fmax.Once we have the angular positionsθ1 andθ2 determined,we can use the coordinates of point D and E to calculate the angle of segment DE which it makes with the x-axis.This is the angleβ.If the coordinates of point D and E are(x1,y1)and(x2,y2)respectively.Thenβ=arctan y2__y1(23)x2__x1 A.6(cont.)This process is illustrated in table below and the corresponding graph.First we place the detector at 8000 m away from the origin and change the coordinates for the corresponding angular position 0o-90o.We record fmax for anyfixed cycle,(10th in this case).It can be seen from the plot of fmax vsθthat theθ1 is between 25o-35o.θ。
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