Question

A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 6 ft/s. Determine the time at which the mass passes through the equilibrium position. (Use g = 32 ft/s2 for the acceleration due to gravity.)

Answer 1

t = 025 s

Step-by-step explanation:

We know

weight, W = 4 pounds

spring constant, k = 2 lb/ft

Positive damping, β = 1

Therefore mass, m = W / g

m = 4 / 32

= 1 / 8 slug

From Newtons 2nd law

`(d^(2)x)/(dt^(2))=-kx-\beta .(dx)/(dt)`

where x(t) is the displacement from the mean or equilibrium position. The equation can be written as

`(d^(2)x)/(dt^(2))+(\beta )/(m).(dx)/(dt)+(k)/(m)x=0`

Substituting the values, the DE becomes

`(d^(2)x)/(dt^(2))+8(dx)/(dt)+16x=0`

Now the equation is

`m^(2)+8m+16=0`

and on solving the roots are

`m_(1)` =
`m_(2)` = -4

Therefore the general solution is
`x(t)=e^(-4t)\left ( c_(1)+c_(2)t \right )`

Now for initial condition x(0) = -1 ft

x'(0)= 8 ft/s

Now we can find the equation of motion becomes,

`x(t)=e^(-4t)\left ( -1+4t \right )`

Therefore, the mass passes through the equilibrium when

x(t) = 0

`e^(-4t)\left ( -1+4t \right )` = 0

-1+4t = 0

t =
`(1)/(4)`

= 0.25 s