1. (20) When two parallel springs, with constants k1 and k2, support a single we

Question

1. (20) When two parallel springs, with constants k1 and k2, support a single weight W, the effective spring constant on the system is given by 4k1k2 k1 + k2 ? A 24-pound weight stretches one spring 18 inches and another spring 6 inches. The springs are attached to a common rigid support and then to a metal plate (as shown). A 48-pound weight is then attached to the center of the plate in double-spring arrangement. If the resistance is three times the instantaneous velocity, and if there is a k, driving force of f (t) sint, set up and solve a differential equation to find the equation of motion of the weight given that the system begins from rest at the equilibrium position. (Use g 32ft/s2)

Explanation / Answer

weight=24 pound=106.757 N

elongation =18 inches=45.72 cm=0.4572 m

then spring constant for the first spring=force/elongation

=233.5 N/m

for the second spring, elongation=6 inches=15.24 cm=0.1524 m

spring constant for the second spring=force/elongation=700.51 N/m

then effective spring constant =4*k1*k2/(k1+k2)

=700.5 N/m

weight attached=48 pound=213.515 N

mass of the weight=m=213.515/9.8=21.787 kg

damping coefficient=c=3

driving force=(7/6)*sin(t) pound=5.1896*sin(t) N

equation of motion:

m*x’’+c*x’+k*x=driving force

==>21.787*x’’+3*x’+700.5*x=5.1896*sin(t)

solving for homogenous portion:

21.787*x’’+3*x’+700.5*x=0

let x=e^(p*t)

then 21.787*p^2+3*p+700.5=0

==>p=-0.0688+5.67 i. or

p=-0.0688 -5.67 i

then homogenous solution;

xh=e^(-0.0688*t)*(A*cos(5.67*t)+B*sin(5.67*t))

let particular solution be :

xp=C*sin(t)+D*cos(t)

xp’=C*cos(t)-D*sin(t)

xp’’=-xp

substituting,

21.787*(-C*sin(t)-D*cos(t))+3*(C*cos(t)-D*sin(t))+700.5*(C*sin(t)+D*cos(t))=5.1896*sin(t)

equating coefficients of sin(t):

-21..787*C-3*D+700.5*C=5.1896

==>678.71*C-3*D=5.1896…(1)

equating coefficients cos(t):

-21.787*D+3*C+700.5*D=0

==>3*C+678.71*D=0….(2)

solving equation 1 and 2:

C=7.6461*10^(-3)

D=-3.3797*10^(-5)

so xp=7.6461*10^(-3)*sin(t)-3.3797*10^(-5)*cos(t)

total solution:

x=xh+xp

=e^(-0.0688*t)*(A*cos(5.67*t)+B*sin(5.67*t)) +7.6461*10^(-3)*sin(t)-3.3797*10^(-5)*cos(t)

at t=0, x=0 and at t=0 ,x’=0

==>A-3.3797*10^(-5)=0

==>A=3.3797*10^(-5)

and -0.0688*A+5.67*B+7.6461*10^(-3)=0

==>B=-1.3481*10^(-3)

hence complete solution of motion:

x=e^(-0.0688*t)*(3.3797*10^(-5)*cos(5.67*t)-1.3481*10^(-3)*sin(5.67*t)) +7.6461*10^(-3)*sin(t)-3.3797*10^(-5)*cos(t)

meters

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