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 Post subject: Spring SystemPosted: Mon, 5 Dec 2011 19:15:35 UTC
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Joined: Mon, 5 Dec 2011 18:50:51 UTC
Posts: 2
Is the approach correct for this sort of problem?

A 32 lb object stretches a spring 8 ft. If the resistive force due to damping is 5x' , determine the displacement function if the object is released from 1 ft below the equilibrium position with an upward velocity of 1 ft/sec.

Here is what I have:
w = m(32ft/s)
f = ma
f = k(s + x)
mx'' + cx' + kx = 0

x(0) = 8ft
x'(0) = 1ft/s
c = 5

w = m(32ft/s): 32lb = m(32ft/s) m = 1 slug
f = k(x + x): f = kx 32 = k*8 k = 4

1x'' + 5x' + 4x = 0

So I took the Laplace transform of that and got this:

s^(2)X - sx(0) - x'(0) + 5[sX - x(0)] + 4X = 0
s^(2)X - 8s - 1 + 5[sX - 1] + 4X = 0
s^(2)X + 5sX +4X - 8x - 2 = 0
X(s^2 + 5s + 4) = 8s + 2

X = (8s + 2)/(s+4)(s+1)

The inverse Laplace of X yeilded 10e^(-4t) - 2e^(-t)

Thanks!

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 Post subject: Re: Spring SystemPosted: Tue, 6 Dec 2011 11:01:30 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6003
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
Bryon0101 wrote:
Is the approach correct for this sort of problem?

A 32 lb object stretches a spring 8 ft. If the resistive force due to damping is 5x' , determine the displacement function if the object is released from 1 ft below the equilibrium position with an upward velocity of 1 ft/sec.

Here is what I have:
w = m(32ft/s)
f = ma
f = k(s + x)
mx'' + cx' + kx = 0

x(0) = 8ft
x'(0) = 1ft/s
c = 5

w = m(32ft/s): 32lb = m(32ft/s) m = 1 slug
f = k(x + x): f = kx 32 = k*8 k = 4

1x'' + 5x' + 4x = 0

So I took the Laplace transform of that and got this:

s^(2)X - sx(0) - x'(0) + 5[sX - x(0)] + 4X = 0
s^(2)X - 8s - 1 + 5[sX - 1] + 4X = 0
s^(2)X + 5sX +4X - 8x - 2 = 0
X(s^2 + 5s + 4) = 8s + 2

X = (8s + 2)/(s+4)(s+1)

The inverse Laplace of X yeilded 10e^(-4t) - 2e^(-t)

Thanks!

No. Note the initial condition highlighted in red, which your answer fails to satisfy.

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