Simplify F(s) = X1(m1 s^2 + k) - KX2

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In summary, the conversation is about simplifying a function and finding the value of X2(s)/F(s). The steps involved include taking X1 to have a value, covering m1 and k by 1, expanding the squared term, and using partial fraction.
  • #1
hamadee
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hii guys
i have a problem on simplifying this

F(s) = X1(m1 s^2 + k) - KX2
0 = X2(m1 s^2 + k) - kx1

take x1 to have value
x1= X2(m1 s^2 + k)/k

F(s) = X2(m1 s^2 + k)/k . (m1 s^2 + k) - kX2

X2 [(m1 s^2 + k)^2 -k / k]


now we cover m1 and k by 1 because in the question they have detrmind them by 1
soo

X2 [(1 s^2 + 1)^2 - 1 / 1]

soo what can i do after


they need in the question X2(s)/F(s)



can you help me please on doing this
 
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  • #2
Turn the 1/1 into 1 (why on Earth would you leave it like that), expand the squared term and go from there.
 
  • #3
how to expand the squared

using this form (a^2 + 2ab + b^2) ?

1/ (s^4 + 2s^2 + 1)

then using partial fraction
 
  • #4
Shouldn't it be 1/(s^4 + 2s^2 + 1 - 1)?
 

FAQ: Simplify F(s) = X1(m1 s^2 + k) - KX2

What is the purpose of the function F(s)?

The function F(s) is used to simplify the equation X1(m1 s^2 + k) - KX2, which is commonly used in physics and engineering to describe the motion of a system with two masses connected by a spring. By simplifying the equation, it becomes easier to solve and understand the behavior of the system.

What do the variables represent in the function F(s)?

The variable s represents the Laplace transform variable, which is used to convert differential equations into algebraic equations. The variables X1 and X2 represent the displacements of the two masses, while m1 and k represent the mass and spring constant of the system, respectively.

How does simplifying the equation F(s) = X1(m1 s^2 + k) - KX2 help in solving problems?

Simplifying the equation makes it easier to manipulate and solve, as it reduces the number of terms and variables. This allows for a clearer understanding of the system's behavior and makes it easier to find important values such as the natural frequency, damping ratio, and equilibrium position.

Can the function F(s) be used to solve real-life problems?

Yes, the function F(s) can be used to solve real-life problems involving systems with masses and springs, such as analyzing the motion of a car's suspension or modeling the behavior of a diving board.

Are there any limitations to using the function F(s)?

While the function F(s) simplifies the equation, it does not take into account external factors such as friction and air resistance. In real-world situations, these factors may significantly affect the behavior of the system and must be considered separately.

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