Question

A 10 10 kilogram object suspended from the end of a vertically hanging spring stretches the spring 9.8 9.8 centimeters. At time ?=0 t = 0 , the resulting mass-spring system is disturbed from its rest state by the force ?(?)=150cos(10?) F ( t ) = 150 cos ( 10 t ) . The force ?(?) F ( t ) is expressed in Newtons and is positive in the downward direction, and time is measured in seconds.

Solve the initial value problem for ?(?)y(t)

Answer 1

mass is m=10 kg

spring stretches x=9.8 cm = 0.098 m

the spring constant is given by

F = kr

mg = kr

10.9.8 = k (0.098)

k = 1000

there is no damping so damping constant is c=0

.

external force is

f(t) = 150 cos(100)

DE is given by

my+cy + ky = f(t)

104 + Oy + 1000y = 150 cos (100)

104 + 1000y = 150 cos (100)

y" + 100y = 15 cos (100)

find the roots

? + 100 = 0

= -100

I= £10

for complex roots general solution is

yc = e(Cicos (100) + C2 sin (10t))

Yc = Ci cos (100) + C sin (10+)

.

here g(t)=15cos(10t) so assume that particular solution is in the form of

.....................(1)

y = aot sin 10t) + azt cos (100)

y' = ao (sin (10t) + 10t cos (100)) +a (cos (100) – 10t sin (10+))

y = ap 20 cos (107) – 100t sin (101)) + a (-20 sin (107) – 100t cos (100)

put all values in DE

ap (20 cos (100) – 100t sin (10+)) + a(-20 sin (100) – 100t cos (10t)) + 100(aot sin (10t) + ait cos (100)) = 15 cos(10t)

20ao cos (101) – 20asin (10+) = 15 cos (101)

compare coefficient both sides

0= -20a1...................= 0

15 = 2000.............. ...do =

put both constants in equation 1

-t sin (10t) +0. t cos (100)

(201) u = di

general solution is

y = yc + yp

...................(1)

00 y = ci cos (10t) + C2 sin (10t) + -t sin (10)

here mass is initially released from the equilibrium position.

so y(0)=0

3. O.sin (100) 0 = C1 cos (10 - 0) + c2 sin (100) +

0 = Cicos (0) + C2 sin (0) + 0

0=c1+0+ 0

..................put it back in equation 2

0 = 15

.

y=c1 cos (10t) + ca sin (10+) + -t sin (10+)

y = 0 cos (10t) + C2 sin (10t) + -t sin (10)

.....................(3)

y = ca sin (10t) + 3t sin (10+)

take derivative

AC (10t) + 10t cos (10t)) + C2 cos (10t) · 10

initial velocity is zero so y'(0)=0

C (sin (10-0) + 10.0.cos (100)) + C2 cos (10-0). 10

0= (sin (0) +0) + C2 cos (0). 10

0=0+ 10c

.................put it back in equation 3

C2=0

.

y = ca sin (10t) + 3t sin (10+)

y=0 sin (10+) + t sin (107)

سایت | شهر -t sin (10t