A 20-lb weight is hung on an 18-in. spring stretching it 6 in. The weight is pulled down 5 in. and 5 lb are added to the weight. If the weight is now released with a downward velocity of y0 in./sec, find the position of mass relative to the equilibrium in terms of y0 and valid for any time t >= 0.
hookes law
To find the position of the mass relative to the equilibrium in terms of y0, we need to consider the forces acting on the system.
First, let's analyze the situation before any changes are made. The weight of 20 lb stretches the spring by 6 inches. Using Hooke's Law, we can calculate the spring constant (k) as follows:
k = F / x
where F is the force applied (20 lb = 20 lb * 32.2 ft/s^2) and x is the displacement (6 inches = 6/12 ft).
k = (20 lb * 32.2 ft/s^2) / (6/12 ft)
k = 107.33 lb/ft
Now, the weight is pulled down an additional 5 inches, and 5 lb are added to the weight. This new configuration creates a new equilibrium position. The displacement from the original equilibrium position is given by:
delta_x = 6 inches + 5 inches
delta_x = 11 inches = 11/12 ft
The new force applied to the spring is:
F_new = 20 lb + 5 lb
F_new = 25 lb
Now, let's consider the position of the mass relative to the equilibrium. We can use the equation of motion for a mass-spring system:
m * y'' + k * y = 0
where m is the mass, y'' is the second derivative of position (acceleration), and y is the position relative to the equilibrium.
Since the mass is released with a downward velocity of y0 in./sec, the initial conditions are:
y(0) = 0 (initial position at equilibrium)
y'(0) = -y0 (initial downward velocity)
Let's solve the differential equation to find the expression for the position of the mass (relative to the equilibrium) as a function of time:
m * y'' + k * y = 0
The characteristic equation is:
m * r^2 + k = 0
Solving this quadratic equation, we obtain:
r = ± sqrt(-k/m) = ± sqrt(-(107.33 lb/ft) / m)
The general solution of the differential equation is:
y(t) = c1 * cos(sqrt(-(107.33 lb/ft) / m) * t) + c2 * sin(sqrt(-(107.33 lb/ft) / m) * t)
Using the initial conditions, we can find the values of c1 and c2:
y(0) = 0
0 = c1 * cos(0) + c2 * sin(0)
0 = c1
y'(0) = -y0
-y0 = -sqrt(-(107.33 lb/ft) / m) * c1 * sin(0) + sqrt(-(107.33 lb/ft) / m) * c2 * cos(0)
-y0 = sqrt(-(107.33 lb/ft) / m) * c2
Simplifying, we get:
c1 = 0
c2 = -y0 * sqrt(-m / (107.33 lb/ft))
Therefore, the position of the mass relative to the equilibrium in terms of y0 and valid for any time t >= 0 is:
y(t) = -y0 * sqrt(-m / (107.33 lb/ft)) * sin(sqrt(-(107.33 lb/ft) / m) * t)
Note that the mass value (m) is required to obtain a numerical solution.