A mass that weighs 8 lb stretches a spring 6 in. The system isacted on by an external force of 8 sin 8t lb. If the mass is pulleddown 3 in and then released, determine the position of the mass atany time. Determine the first four times at which the velocity ofthe mass is zero.

t= 1/8, pi/8, 2pi/8,3pi/8

Step-by-step explanation:

Given

m=(8/32) lb s^2/ft

K=8/(6/12)=16 lb/ft

Use the following equation and plug in values

mu''+ku=f(t)

1/4u''+16u=8sin8t

u''+64u=32sin8t

This equation corresponds to the following homogeneous equation

u''+64u=0

r=+/-8i

uc(t)=c1cos8t+c2sin8t

Now find the particular solution

u(t)=Atcos8t+Btsin8t

u'(t)=-8Atsin8t+Acos8t+B8tcos8t+Bsin8t

u''(t)=-8tAsin8t-64Atcos8t-8Asin8t+B8cos8t-64Btsin8t+8Bcos8t

Substitute these values into the original equation and solve for Aand B

A=-2 B=0

the particular solution is u(t)=-2tcos8t

the general solution is u=u1(t)+u(t)

u=c1cos8t+c2sin8t-2tcos8t

Use the initial conditions to solve for c1 andc2

c1+0=(1/4)   8c2-2=0

c1=(-1/4) c2=(1/4)

u=(1/4)[cos8t+sin8t-8tcos8t]

To solve the next step differentiate u

u'=-2sin8t+2cos8t-2cos8t+16tsin8t

= -2sin8t+16sin8t

= 2sin8t(8t-1)

Velocity=2sin8t(8t-1)

Set this equation equal to zero to solve for zero velocity

8t-1=0 t=1/8

t= 1/8, pi/8, 2pi/8,3pi/8

$12. that is my answer.

take the logarithm of both sides (the base of the log doesn't matter):

apply the exponent property - this says that :

distribute the log terms on both sides:

group like terms together:

divide through both sides by the coefficient on :

you can do some additional rewriting here using the properties of the logarithm to simplify things if you like: