Dynamics and Vibrations - Assignment

Added on - 21 Apr 2020

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Answer: (1) _ADouble pendulum with spring:System equation of motion are:̈θ1+(gl+km1)θ1(km1)θ2=0̈θ2+(gl+km2)θ2(km2)θ1=0Characteristic equationθ1=X1cos(ωt)θ2=X2cos(ωt)̈θ1=X1ω2cos(ωt)̈θ2=X2ω2cos(ωt)Equation of motionM̈x+Kx=0(ω200ω2)(X1X2)cos(ωt)+[(gl+km1)kmkm(gl+km1)](X1X2)cos(ωt)=0h22hω2+ω4(km)2=0whereh=(gl+km1)Natural frequency and natural modes:ω2=h±km=(gl+km1)±kmω1=gl,ω2=gl+2km(ω2+gl+km1)X1kmX2=0For natural mode formula,1|P a g e
X2X1=mω2+2kkPutω1=glX2X1=mω2+2kkX2X1=1ω2=gl+2kmX2X1=1Answer:1_ ( C)sinθ1=θ1=X1l........(1)sinθ2=θ2=X2X1l..............(2)T2cosθ2=mgAndT1cosθ1=mg+T2cosθ2¿thevalueofθ1θ2aboverelationsreduces¿T2=mg.................(3)AndT1=mg+T2.............(4)Write differential equation of two masses for motion in horizontal direction,We havem̈x1=T2sinθ2T1sinθ1m̈x2=T2sinθ2From the equation of (1), (2), (3) and (4) following relation establish.m̈x+[3ml]gx1=mlgx2...........(5)m̈x+[ml]gx2=mlgx1....................(6)Consider harmonic motion.2|P a g e
x1=X1sinωt..........................(7)x2=X2sinωt..........................(8)Value of Equation (7) and (8) put into equation (5) and (6)Cancel common sinωt[ -mω2+[3ml]g¿X1=mlgX2[-mω2+mlg¿X2=mlgX1Re arrange above both equationX1X2=(gl)3glω2X1X2=(gl)ω2gl(gl)3glω2=(gl)ω2glω2g24ω4lg+2l2=0ωn1=(gl)(22)ωn1=(gl)(2+2)Corresponding mode shape(X1X2)1=1+2=+0.41(X1X2)2=12=2.41Answer: 1_ (E)Equation of motion using x (t) andθ(t)3|P a g e
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