(a) To determine the time at which the mass passes through the equilibrium position, we can use the equation for the motion of a mass-spring-damper system:
m*x'' + c*x' + k*x = m*g
where m is the mass of the weight, x is the displacement of the weight from the equilibrium position, c is the damping coefficient, k is the spring constant, and g is the acceleration due to gravity.
We can rewrite this equation as:
x'' + (c/m)*x' + (k/m)*x = g
Using the given values, you have:
m = 160 lb = 160/32.2 = 4.97 slugs (slugs are the unit of mass in the English system)
x = 10 ft (at t = 0)
x' = -41 ft/s (at t = 0)
c = 4*sqrt(sqrt(10)) = 8.944 (we'll use this as is, without converting to English units)
k = m*g/x = 4.97*32.2/20 = 7.98
g = 32.2 ft/s^2
Plugging in these values, you get:
x'' + (8.944/4.97)*x' + (7.98/4.97)*x = 32.2
This is a second-order differential equation, which can be solved using standard techniques. However, since we're only interested in the time at which the mass passes through the equilibrium position, we can use an approximation based on the damping ratio (ζ) of the system:
ζ = (c/2)*sqrt(m/k)
The damping ratio tells us how quickly the system will approach the equilibrium position. If the damping ratio is small (less than 1), the system will oscillate around the equilibrium position before settling down to rest. If the damping ratio is large (greater than 1), the system will quickly approach the equilibrium position without oscillating.
In your case, the damping ratio is:
ζ = (8.944/2)*sqrt(4.97/7.98) = 1.09
Since ζ > 1, we can assume that the system will quickly approach the equilibrium position without oscillating. In this case, we can use the following equation to estimate the time at which the mass passes through the equilibrium position:
t = (1/ζ)*ln(x0/x)
where x0 is the initial displacement (10 ft) and x is the displacement at the time of interest (0 ft).
Plugging in the values, we get:
t = (1/1.09)*ln(10/0) = 2.40 seconds
Therefore, the time at which the mass passes through the equilibrium position is approximately 2.40 seconds.
(b) To find the time at which the mass attains its extreme displacement from the equilibrium position, we can use the following equation:
ω = sqrt(k/m - (c/2m)^2)
ω is the angular frequency of the system, which tells us how quickly the system oscillates around the equilibrium position. The amplitude of the oscillation is given by:
A = x0/sqrt(1 - (x'^2)/(4*m*k))
We can use these equations to find the time at which the mass attains its extreme displacement:
t = (1/ω)*arccos(x/x0)
where x is the displacement from the equilibrium position at the time of interest.
Plugging in the values, we get:
ω = sqrt(7.98/4.97 - (8.944/(2*4.97))^2) = 1.704 rad/s
A = 8.659 ft
x = A*cos(ω*t) = -8.659 ft (since the mass is below the equilibrium position)
t = (1/1.704)*arccos(-8.659/10) = 0.372 seconds
Therefore, the time at which the mass attains its extreme displacement from the equilibrium position is approximately 0.372 seconds.