振動習(xí)題課件15-1質(zhì)量為m=5.88kg的物體，掛在彈簧上，讓它在豎直方向上作自由振動。在無阻尼情況下，其振動周期為T=0.4πs；在阻力與物體運(yùn)動速度成正比的某一介質(zhì)中，它的振動周期為T=0.5πs；求當(dāng)速度為0.01m/s時(shí)，物體在阻尼介質(zhì)中所受的阻力。返回結(jié)束mkmΔlmbACBA→C應(yīng)用功能原理12mglxmmΔ2()+=v12kx212k()lΔ2=2gxm+kx2()lΔ22gmlΔmkm=22vglxmΔ()+kx2()lΔ2mkm返回結(jié)束2v=2gxm+kx2()lΔ22gmlΔmkm2gm=A()lΔ22gmlΔkmC==2×2=4kmB==100=1001=100×(0.07)24×0.07=0.21令：dxdt22v==x+x2ABC返回結(jié)束dxdt<0由于dxdt=x+x2ABCdtdx=x+x2ABCdxx+x2ABC=òl(fā)Δbt=dtòt0dxdt22v==x+x2ABC返回結(jié)束lΔbBBCxBA21sin=4A+2arclΔBBCBA21sin4A+2arc=BBCBA21sin4A+2arcbdxx+x2ABC=òl(fā)Δbt返回結(jié)束lΔBBCBA21sin2A+2arct=BBCBA21sin4A+2arcb4=AB=100C=0.21Δl=0.07b=0.030.214sin442arc=××2×100×0.07100()101sinarc0.214442××2×100×0.03100()101arcsin(1)-arcsin(-1)=101=1012π2π()π=10=0.314s返回結(jié)束=3.53(kg.s-1)=2×3×5.88=3(s-1)ωβT20=2π2T20=2πT22π=2π0.4π22π0.5π2=2516=rβ2mF=rv=3.5×1.0×10-2

=0.353N2ωβT20=2π2ωβT20=2π2解：返回結(jié)束15-2一擺在空中振動，某時(shí)刻，振幅為A0=0.03m，經(jīng)過t1=10s后，振幅變?yōu)锳1=0.01m，問:由振幅為A0時(shí)起，經(jīng)多長時(shí)間，其振幅減為A2=0.003m

？返回結(jié)束=20.9(s)=0.110βAe0=At0=lnAlnAβt1=ln100.030.01110=lnβtAA1120=lnβtAA2=10.11ln0.030.003解：返回結(jié)束15-3試用最簡單的方法求出下列兩組簡諧振動合成后所得合振動的振幅：第一組：第二組：0.05cos(3t+π/3)mx1=0.05cos(3t+π/3)mx1=0.05cos(3t+7π/3)mx2=0.05cos(3t+4π/3)mx2=返回結(jié)束==3π73ππΦΔ2==3π43ππΦΔA

=A1+A2=0.05+0.05=0.10(m)A

=A1-A2=0解：0.05cos(3t+π/3)mx1=0.05cos(3t+7π/3)mx2=(1)0.05cos(3t+π/3)mx1=0.05cos(3t+4π/3)mx2=(2)返回結(jié)束15-4一質(zhì)點(diǎn)同時(shí)參與兩個(gè)在同一直線上的簡諧振動:

試求其合振動的振幅和初相位(式中x以m計(jì),t以s計(jì))。0.04cos(2t+π/6)mx1=0.03cos(2t-5π/6)mx2=返回結(jié)束=0.01mφ2A1A2cos()++=A22A1A22φ1=(0.04)2+(0.04)2+2×0.04×0.03cos(-π)arctg+=φ1A1sinφ2A2sinφ1