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Sneakatone
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Homework Statement
I do not know how to find f(t) with the given Ampliture 40 and a=pi
Homework Equations
The Attempt at a Solution
I have the solution above.
my set up was 1/2y''+y'+5=f(t)
1/2S^2* Y(s) + Y(s)+5=f(t)
Sneakatone said:Homework Statement
I do not know how to find f(t) with the given Ampliture 40 and a=pi
View attachment 76100
Homework Equations
The Attempt at a Solution
I have the solution above.
my set up was 1/2y''+y'+5=f(t)
1/2S^2* Y(s) + Y(s)+5=f(t)
SteamKing said:Have you studied periodic functions? Do you know how to obtain the Laplace transform of a periodic function?
Here is a brief article discussing how:
http://academic.udayton.edu/LynneYengulalp/Solutions219/LaplacePeriodicSolutions.pdf
Wrong. There is an ##f(t)## in the formula for the transform, so everything about ##f(t)## matters.Sneakatone said:From the looks of this does amplitude not matter , only the period?
A differential equation with Laplace transform is a mathematical tool used to model the behavior of a system, such as a spring, over time. The Laplace transform converts the differential equation into an algebraic equation, making it easier to solve and analyze the system's response.
In a differential equation with Laplace transform, springs are represented as a second-order linear differential equation. This takes into account the spring's displacement, velocity, and acceleration, and how they change over time.
Yes, the Laplace transform can be used to solve any linear differential equation related to springs. However, for more complex or nonlinear systems, other techniques may be necessary.
The Laplace transform allows us to analyze the behavior of springs in the frequency domain, rather than the time domain. This makes it easier to understand the system's response to different frequencies of input and to determine its stability and resonance.
One limitation is that the Laplace transform assumes that the system is linear, which may not always be the case for springs. It also requires initial conditions and assumes that the system is in equilibrium. Additionally, the Laplace transform may not provide a physical understanding of the system's behavior, as it is a purely mathematical tool.