In: Physics
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.
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.