Question

A mass weighing 16 lb stretches a spring 3 in. The mass is attached to a viscous damper with a damping constant of 2 lb-s/ft. If the mass is set in motion from its equilibrium position with a downward velocity of 2 in/s, find its position u at any time t. Assume the acceleration of gravity g=32 ft/s2.

Answer 1

`u(t)=(1)/(12√(31))e^(-2t)sin(2√(31)t)`

Step-by-step explanation:

Our values given are,

`L=3in=0.25ft`

Ramping constant

`r=2lb.sec/ft`

By this way we can calculate k and m.

`m=(mg)/(g) = (16)/(32)=(1)/(2) lb/ft/sec^2`

`k=(mg)/(L) = (16)/(1/4) = 64lb/ft`

We can now create the differential equation, which is given by,

`mu''+ru'+ku=0`

`\Rightarrow (1)/(2)u''+2u'+54u=0`

The initial conditions are:

`u(0) = 0`

`u'(0) = (1)/(6)`

*Note that the velocity war given in inches and transformated to feet.

With that we can create our characteristic equation:

`(1)/(2)r^2+2r+64=0`

With two roots

`r_1 = -2+2√(31)i`

`r_2 = -2-2√(31)i`

Our general solution of the Differential equation is then,

`u(t) = e^(2t)[c_1 cos(2√(31)t)+c_2 sin(2√(31)t)]`

Applying the initial conditions,

`u(t)=0\Rightarrow 0=c_1`

`u'(t)=(1)/(6) \Rightarrow (1)/(6) = e^(0)(2√(31)c_2)`

The position of u at any time t is,

`u(t)=(1)/(12√(31))e^(-2t)sin(2√(31)t)`