LaPlace transform method to find the equation of motion

  • #1
1
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< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

So this is the problem

Here is the question:
A 32lb weight strecthes a spring 2ft.The weight is released from rest at the equilibrium position. beginning at t=0, a force equal to f(t)= sint acts on the system and is removed at t=2pi. Assuming no damping forces are present, use LaPlace transform method to find the equation of motion. Hint: Write f(t) in terms of the unit step function.

So far this is all i got done


Let W = the force of weight = -32 lbs
Let g = the acceleration due to gravity = -32 ft/s²
Let m = the mass of the weight = W/g = -32 lbs/-32 ft/s² = 1 slug
Let fs(y(t)) = the force of the spring = ky(t)

Given:

y(0) = -2 ft

Let k = the spring constant = 32 lb/2 ft = 16 lb/ft

Let fs(y(t)) = the force of the spring = -k(y(t))

Let f(t) = the driving force

Let u(t) = the unit step function

f(t) = u(t)sin(t) - u(t - 2π)sin(t - 2π)

sin(a - b) = sin(a)cos(b) - cos(a)sin(b)

sin(t - 2π) = sin(t)cos(2π) - cos(t)sin(2π)

sin(t - 2π) = sin(t)

f(t) = u(t)sin(t) - u(t - 2π)sin(t) lbs

Summing the forces in y direction

(m)a(t) = - W + f(t) + (-k)y(t)

The acceleration is y"(t)


(m)y"(t) = f(t) + k{2 ft - y(t)} - W


We are dividing by a mass that is equal to 1 so the equation does not really change except that the units for f(t) become acceleration units (ft/s²) and the units for k become lbs/{slug•ft}:

y"(t) = f(t) + (-k)y(t) - g

The Laplace transform for f(t) is

F(s) = 1/(s² + 1) - e^{-2πs}/(s² + 1)

The initial conditions are y'(0) = 0 y(0) = -2
_____________________________________________________________________

Can someone help me finish this problem? or at least explain what is next because I am totally confused
 

Answers and Replies

  • #2
21,261
4,720
< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

So this is the problem

Here is the question:
A 32lb weight strecthes a spring 2ft.The weight is released from rest at the equilibrium position. beginning at t=0, a force equal to f(t)= sint acts on the system and is removed at t=2pi. Assuming no damping forces are present, use LaPlace transform method to find the equation of motion. Hint: Write f(t) in terms of the unit step function.

So far this is all i got done


Let W = the force of weight = -32 lbs
Let g = the acceleration due to gravity = -32 ft/s²
Let m = the mass of the weight = W/g = -32 lbs/-32 ft/s² = 1 slug
Let fs(y(t)) = the force of the spring = ky(t)

Given:

y(0) = -2 ft

Let k = the spring constant = 32 lb/2 ft = 16 lb/ft

Let fs(y(t)) = the force of the spring = -k(y(t))

Let f(t) = the driving force

Let u(t) = the unit step function

f(t) = u(t)sin(t) - u(t - 2π)sin(t - 2π)

sin(a - b) = sin(a)cos(b) - cos(a)sin(b)

sin(t - 2π) = sin(t)cos(2π) - cos(t)sin(2π)

sin(t - 2π) = sin(t)

f(t) = u(t)sin(t) - u(t - 2π)sin(t) lbs

Summing the forces in y direction

(m)a(t) = - W + f(t) + (-k)y(t)

The acceleration is y"(t)


(m)y"(t) = f(t) + k{2 ft - y(t)} - W


We are dividing by a mass that is equal to 1 so the equation does not really change except that the units for f(t) become acceleration units (ft/s²) and the units for k become lbs/{slug•ft}:

y"(t) = f(t) + (-k)y(t) - g

There is a mistake here. There should be no -g in this equation, since W = 2k
The Laplace transform for f(t) is

F(s) = 1/(s² + 1) - e^{-2πs}/(s² + 1)

The initial conditions are y'(0) = 0 y(0) = -2
_____________________________________________________________________

Can someone help me finish this problem? or at least explain what is next because I am totally confused

What is the LT of y"(t)?
 

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