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

Answer:

t = 025 s

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

$\frac{d^{2}x}{dt^{2}}=-kx-\beta .\frac{dx}{dt}$

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

$\frac{d^{2}x}{dt^{2}}+\frac{\beta }{m}.\frac{dx}{dt}+\frac{k}{m}x=0$

Substituting the values, the DE becomes

$\frac{d^{2}x}{dt^{2}}+8\frac{dx}{dt}+16x=0$

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 = $\frac{1}{4}$

  = 0.25 s