Answer: (1) _ADouble pendulum with spring:System equation of motion are: ̈θ1+(gl+km1)θ1−(km1)θ2=0 ̈θ2+(gl+km2)θ2−(km2)θ1=0Characteristic equation θ1=X1cos(ωt−∅)θ2=X2cos(ωt−∅) ̈θ1=−X1ω2cos(ωt−∅) ̈θ2=−X2ω2cos(ωt−∅)Equation of motion M ̈x+Kx=0(−ω200−ω2)(X1X2)cos(ωt−∅)+[(gl+km1)−km−km(gl+km1)](X1X2)cos(ωt−∅)=0h2−2hω2+ω4−(km)2=0whereh=(gl+km1)Natural frequency and natural modes:ω2=h±km=(gl+km1)±kmω1=√gl, ω2=√gl+2km(−ω2+gl+km1)X1−kmX2=0For natural mode formula,1 | P a g e
X2X1=−mω2+2kkPut ω1=√glX2X1=−mω2+2kkX2X1=1ω2=√gl+2kmX2X1=−1Answer:1_ ( C) sinθ1=θ1=X1l........(1)sinθ2=θ2=X2−X1l..............(2)T2cosθ2=mgAnd T1cosθ1=mg+T2cosθ2¿thevalueofθ1∧θ2aboverelationsreduces¿T2=mg.................(3)And T1=mg+T2.............(4)Write differential equation of two masses for motion in horizontal direction, We have m ̈x1=T2sinθ2−T1sinθ1m ̈x2=−T2sinθ2From the equation of (1), (2), (3) and (4) following relation establish.m ̈x❑+[3ml]gx1=mlgx2...........(5)m ̈x❑+[ml]gx2=mlgx1....................(6)Consider harmonic motion. 2 | P a g e
x1=X1sinω❑t..........................(7)x2=X2sinω❑t..........................(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)3gl−ω2X1X2=(gl)−ω2gl(gl)3gl−ω2=(gl)−ω2glω2g2−4ω4lg+2l2=0ωn1=√(gl)(2−√2)ωn1=√(gl)(2+√2)Corresponding mode shape (X1X2)1=−1+√2=+0.41(X1X2)2=−1−√2=−2.41Answer: 1_ (E)Equation of motion using x (t) and θ(t)3 | P a g e
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