A body weighing 9.25 grams force hangs from a spring stretching it 1.4 centimete

Question

A body weighing 9.25 grams force hangs from a spring stretching it 1.4 centimeters. Initially the body part
of rest 2.6 centimeters below its equilibrium position. The medium where the body moves offers a force
of resistance to movement that is numerically equal to 1/2
of its instantaneous speed. Knowing that there is a force
external, changing in time, which is defined by the formula:
f (t) = cos (t)
grams strength. Find the position in centimeters of the body after 5 seconds. Take as positive above the
balance position. Consider positive negative and upward magnitudes.

Explanation / Answer

1 gram-force=0.0098 N

weight of the body=9.25 grams force

=9.25*0.0098=0.09065 N

mass of the body=m=weight/g

=0.00925 kg

with this amount force applied, elongation=1.4 cm=0.014 m

let spring constant be k N/m.

then force=k*elongation

==>0.0098=k*0.014

==>k=0.7 N/m

damping constant c=1/2=0.5

external force=f(t)=cos(t) gram force=cos(t)*0.0098 N

writing force balance equation:

m*x’’=f(t)k*x-c*x’

==>m*x’’+c*x’+k*x=cos(t)*0.0098

==>0.00925*x’’+0.5*x’+0.7*x=cos(t)*0.0098

==>x’’+54.054*x’+75.676*x=1.0595*cos(t)

solving for homogenous equation:

let x=e^(m*t)

m^2+54.054*m+75.676=0

solving for m ,

we get m=-52.6157 and m=-1.4383

then homogenous solution:

xh=A*e^(-52.6157*t)+B*e^(-1.4383*t)

particular solution is of the form:

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

==>xp’=-C*sin(t)+D*cos(t)

==>xp’’=-C*cos(t)-D*sin(t)

xp’’+54.054*xp’+75.676*xp=1.0595*cos(t)

==>-C*cos(t)-D*sin(t)-54.054*C*sin(t)+54.054*D*cos(t)+75.676*C*cos(t)+75.676*D*sin(t)=1.0595*cos(t)

writing equation for coefficients of cos(t) and sin(t):

-C+54.054*D+75.676*C=1.0595

==>74.676*C+54.054*D=1.0595….(1)

and -D+54.054*C+75.676*D=0

==>54.054*C+74.676*D=0….(2)

solving 1 and 2 together,we get

C=0.03

D=-0.0216

so complete solution :

x=xh+xp

==>x=A*e^(-52.6157*t)+B*e^(-1.4383*t)+0.03*cos(t)-0.0216*sin(t)

at t=0, x=-0.026

==>-0.026=A+B+0.03

==>A+B=-0.056 …(3)

at t=0, speed=0==>dx/dt=0

==>-52.6157*A-1.4383*B+0.0216=0

==>52.6157*A+1.4383*B=0.0216….(4)

solving 3 and 4 together,

A=0.002

B=-0.058

then complete solution:

x=0.002*e^(-52.6157*t)-0.058*e^(-1.4383*t)+0.03*cos(t)-0.0216*sin(t)

at t=5 seconds,

x=0.03 m

so position at 5 seconds is 3 cm.

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