Question #138205
A vibrational system consisting of mass 1/10 kg is attached to a spring (spring constant= 4kg/m). The mass is released from rest 1m below the equilibrium position. the motion is damped (damping constant=1.2) and is being driven by an external force 5sin4t, beginning at t=0. Write the governing equations of the system and intrepet the type of motion. Hence find the position of mass at time t.
1
Expert's answer
2020-10-14T17:04:27-0400

Given,

Mass m=0.1kg

spring constant k=4kg/m

displacement of mass s=1m

Damping constant c=1.2

External force F=5sin4t


Damping force , Fd=csF_d=-cs' (where s' is velocity)


Spring force f=k(x+s)=-k(x+s)


Force due to gravity on the block= mg


When the Mass is in equillibrium

According to hook's law

mg=ks


Now the equation that governs the system is given by,


mx=mgk(s+x)+fd+Fmx''=mg-k(s+x)+f_d+F


mx=kxcx+5sin4tmx''=-kx-cx'+5sin4t ( since mg=ks)


mx+cx+kx=5sin4tx''+cx'+kx=5sin4t ....(1)


Now Damping ratio is given by,

ζ=c2mk\zeta=\frac{c}{2\sqrt{mk}}

=1.22×0.1×4\frac{1.2}{2\times \sqrt{0.1\times 4}}


=0.60.4\frac{0.6}{\sqrt{0.4}}


=0.94<1

ζ<1\zeta<1

So The motion is underdamped.

for position solving equation equation 1,

0.1x''+1.2x'+4x=5sin4t


on solving the left side part,

0.1m2+1.2m+4=00.1m^2+1.2m+4=0

Roots of above equation is given by,

r1=c2miω1r_1=-\frac{c}{2m}-i\omega_1 and r2=c2m+iω1r_2=-\frac{c}{2m}+i\omega_1


where ω1=4mkc22m\omega_1=\frac{\sqrt{4mk-c^2}}{2m} =4×0.1×4(1.2)22×0.1\frac{\sqrt{4\times 0.1\times 4-(1.2)^2}}{2\times0.1}


=1.61.440.2\frac{1.6-1.44}{0.2} =0.160.2\frac{0.16}{0.2} =0.8

The complimentary function is given by,

c.f=ect2m(c1cos0.8t+c2sin0.8t)e^{-\frac{ct}{2m}}(c_1cos0.8t+c_2sin0.8t)



= e1.2t0.2(c1cos0.8t+c2sin0.8t)e^{-\frac{1.2t}{0.2}}(c_1cos0.8t+c_2sin0.8t)


Particular integral is given by,

P.I.=5sin4t0.1m2+1.2m+45\frac{sin4t}{0.1m^2+1.2m+4}


=5sin4t0.1×16+1.2m+4\frac{5sin4t}{-0.1\times16+1.2m+4}

=5sin4t1.2m+2.4\frac{5sin4t}{1.2m+2.4}

=5sin4t1.2m+2.4×1.2m2.41.2m2.4\frac{5sin4t}{1.2m+2.4}\times \frac{1.2m-2.4}{1.2m-2.4}


=(1.2m2.4)5sin4t1.44m25.76\frac{(1.2m-2.4)5sin4t}{1.44m^2-5.76}


=(1.2m2.4)5sin4t1.44×165.76\frac{(1.2m-2.4)5sin4t}{1.44\times -16-5.76}


=(24cos4t12sin4t)23.045.76\frac{(24cos4t-12sin4t)}{-23.04-5.76}


=(24cos4t12sin4t)28.8\frac{(24cos4t-12sin4t)}{-28.8}


Complete solution is given by,

x= e6t(c1cos0.8t+c2sin0.8t)e^{-6t}(c_1cos0.8t+c_2sin0.8t) -(24cos4t12sin4t)28.8\frac{(24cos4t-12sin4t)}{28.8}

This is the position at time t.



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