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Answer :
Mass weighing 32 lb stretches a spring 6 inches. A mass is in a medium that exerts a viscous resistance of 93 lb when the mass has a velocity of 6 ft/s. The object is displaced an additional 5 inches and released. Let us determine the spring constant, k.
Using Hooke's law,
F = -kx …(1)
Where,
F = force = mass * acceleration
k = spring constant
x = extension of the spring = 6 inches = (6/12) ft = 0.5 ft
We know that Force = mass * acceleration
a = F/m = (-kx)/m
m = 32 lb = 32/32.2 slugs
Acceleration = a = -kx/m
k = -ax/m = -(32/32.2)(0.5)/(32/32.2)
k = -0.0784 lb/ft
Displacement, x = 5 inches = 5/12 ft
Time constant, τ = m/k
τ = 32/0.0784 = 408.1633
The differential equation of motion of a body with viscous resistance is given by
mdv/dt + cv = -kx, where v is the velocity of the body, and c is the viscous resistance.
Since v = dx/dt, we have,
md²x/dt² + cdx/dt + kx = 0
md²u/dt² + cd/dt(u) + ku = 0 …(2) [Let u = displacement]
As per the given data,
c = 93 lb/kv
u = Ae^(rt) …(3) [Characteristic equation of the differential equation]
Where A and r are constants, which can be found from the initial conditions.
At t = 0, u = 0
u(0) = 0, u'(0) = 0
u(0) = Ae^0 = 0, i.e., A = 0
u'(0) = Ar = 0, i.e., r = 0
Hence, u(t) = 0
[tex]u(t) = c1e^0t + c2te^0t = c1 + c2 …(4)[/tex]
Since, u(0) = 0
u(0) = c1 + c2 = 0 …(5)
At t = τ, u = 0
[tex]u(τ) = c1e^(0) + c2τe^(0) = 0 …(6)[/tex]
On solving (5) and (6), we get,
c1 = 0, c2 = 0
Therefore, u(t) = 0. Hence, the displacement of the object is zero.
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