Course Instructor: Dr. KHAZAR HAYATSubmitted By: WAQAR ALI

Assignment no. 11

Qualitative Analysis of Non-Linear Vibrating System

Quantitative Analysis of Non-Linear Vibrating SystemProblemA non-linear spring for a single degree of freedom s ystem is given by k (x) = 10x + 2000 x3. C for viscous damping is 1.5 kg sec/cm. A harmonic force 5 kg amplitude acts on the mass of 1 kg. Find the steady state response using the direct integration method.

SolutionThe differential equation of motion is of the form

M x+c x+kx=F0cosωt

Here m = 1; c = 1.5 kg sec/cm

k ( x )=10 x+2000 x3 ,Fο=5x+1.5 x+(10x+2000 x3) x=5cosωt

Let x1= A cos ωt be the first approximate steady state solution

Thenx2=−1.5 Aωsinωt−10 A2 cos2ωt−2000 A4cos4ωt+5cosωt

Now we know that cos2ωt=1+cos 2ωt2

And cos4ωt=(1+cos 2ωt2

)(1+cos2ωt2

)

¿ 14 {1+2cos2ωt+ 1+cos4ωt2 }

¿ 13(4cos 2ωt+cos 4ωt+3)

Substituting the results, we get

x2=−1.5 Aωsinωt−10 A2( 1+cos2ωt2

)−2000 A4(4 cos2ωt+cos4ωt+3)+5cosωt

¿5 cosωt−1.5Aωsinωt− (5 A2+8000 A4 )cos2ωt−2000 A4 cos4ωt−(5 A2+6000 A4 )

Integrating the above we get,

x2=( 5 sinωtω )+(1.5 Aωcosωtω )−( (5 A2+8000 A4 ) sin 2ωt2ω )−( 2000 A4 sin 4ωt4ω )−(5 A2+6000 A4 )t+c1

Integrating once again, we get

x2=( 5cosωtω2 )+( 1.5A sinωtω )−( (5 A2+8000 A4 )cos2ωt4ω2 )−(2000 A4 cos4ωt16ω2 )−(5 A2+6000 A4 ) t

2

2+c1 t+c2

If the constants of integration c1 and c2=0 so that the motion x1 and x2 are periodic.

x2=(−5cosωtω2 )+( 1.5 A sinωtω )−( (2.5 A2+4000 A4 )cos2ωt2ω2 )−(125 A4cos4ωtω2 )−(2.5 A2+3000 A4 ) t2

This is the second approximate steady state vibration