Question

A 1-kilogram mass is attached to a spring whose constant is 24 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 11 times the instantaneous velocity. Determine the equations of motion if the following is true. (a) the mass is initially released from rest from a point 1 meter below the equilibrium position x(t) = m (b) the mass is initially released from a point 1 meter below the equilibrium position with an upward velocity of 13 m/s x(t) = m

Answer 1

Explanation:

for a damping system you have the regenerative force exerted by the spring and the term given by the resistance of the damper which is proportional to the speed of the block equal mass times acceleration

let m be the mass of the object, b the damper constant, and k the spring constant, so we have:

`m*(d^2)/(dx^2)x + b*(d)/(dx)x + kx = 0`

we replace in the previous equation to get:

`(d^2)/(dx^2)x + 11*(d)/(dx)x + 24*x = 0`

we are going to find the roots of the previous equation, let
`(d)/(dx)x = D`

`D^2 + 24*D + 11 = 0 =>  (D+8)(D+3)=0\\=>x(t)=C1e^(-8t)+C2e^(-3t)`

now for case a)
`x(0) = -1; v(0)=0`

so we have to differentiate the expression:

`v(t)=-8C1e^(-8t)-3C2e^(-3t)`

for case a just replace the first initial condition in x(t) and the second in v(t)

`-1=C1+C2\\ 0=-8C1-3C2`, then we have
`-8/3C1=C2`

and -1= C1 -8/3C1 => C1 = 3/5 and C2 = -8/5

now for case b)
`x(0) = -1; v(0)=13`

you have
`-1=C1+C2\\ 13=-8C1-3C2`

in this case C1 = -1-C2, then you replace in the second equation and you get

13 = -8*(-1-C2)-3C2 = 8+5C2 => C1 = -2 and C2 = 1