Need some help with spring-mass-damper problem

  • Thread starter blitzzz
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In summary, the problem is trying to determine the motion of a mass on an oscillating spring system if the mass changes abruptly.
  • #1
blitzzz
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Hi.

I'm working with a spring-mass-damper-system problem but the thing is.. i can't figure out whether it is a single degree of freedom system(SDOF) or multi-degree of freedom system(MDOF). The question begins with a seat that is supported by a spring and also a damper. A man of 60 kg then sits on it. If the man slips off the seat withour imparting any vertical impulse to it, determine the expression for the ensuing motion of the seat against time.

Is the 60kg weight counted as an excitation force? Any form of help will be greatly appreciated.

Regards,
Joel
 
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  • #2
This is a one dimensional problem. Basically, they want to you determine what happens if the mass on an oscillating spring system changes abruptly.
 
  • #3
Tide said:
This is a one dimensional problem. Basically, they want to you determine what happens if the mass on an oscillating spring system changes abruptly.

Okay..i'm still lost. Do i need to find out the natural frequency of the system?
 
  • #4
blitzzz said:
Okay..i'm still lost. Do i need to find out the natural frequency of the system?

Of course! The presence of the man tells you how much the spring is compressed initially and now you have an oscillator starting with that amount of compression with only the seat itself as the mass. The system will undergo damped harmonic oscillation and ultimately approach a new equilibrium.
 
  • #5
Tide said:
Of course! The presence of the man tells you how much the spring is compressed initially and now you have an oscillator starting with that amount of compression with only the seat itself as the mass. The system will undergo damped harmonic oscillation and ultimately approach a new equilibrium.

Hmm but do i calculate the natural frequency using the spring stiffness or the damping coefficient, C?
In the case of spring stiffness, natural freq, w = squareroot(K/m)
In the case of damping coefficient, natural freq, w = C/2*m*damping factor,
I assume its not the latter since I'm not given the damping factor.
Or do we calulate the natural frequencies of the combined system of the man and the seat together?
 
  • #6
It all becomes clear if you write down the differntial equaiton of the system and solve it.

You can write the force equation (with x = position, x'=velocity, x''=acceleration) as

mx'' = -dx' - kx

or mx'' + dx' + kx = 0

the solution of which is [tex] x = C e^{\frac{-d +/-\sqrt{d^2-4 m k}}{2m} t}[/tex]

This oscillates only if d^2-4 m k < 0, in which case the magnitude of the imaginary part is sqrt(d^2 - 4*m*k)/2m). You can see that in the limit as d=0, the oscillation frequency is sqrt(4*m*k)/2m = sqrt(m/k).
 
  • #7
pervect said:
It all becomes clear if you write down the differntial equaiton of the system and solve it.

You can write the force equation (with x = position, x'=velocity, x''=acceleration) as

mx'' = -dx' - kx

or mx'' + dx' + kx = 0

the solution of which is [tex] x = C e^{\frac{-d +/-\sqrt{d^2-4 m k}}{2m} t}[/tex]

This oscillates only if d^2-4 m k < 0, in which case the magnitude of the imaginary part is sqrt(d^2 - 4*m*k)/2m). You can see that in the limit as d=0, the oscillation frequency is sqrt(4*m*k)/2m = sqrt(m/k).

I was given Damping coeeficient = 355 ,mass of seat = 40kg, mass of man = 60kg and spring constant = 19620.
Would my equation of motion be 40x'' = -355x' - 19620x + 60*9.81 (inclusive of the man's weight)? This problem would be relatively easy but the addition of the man sitting down and slipping off the seat part makes it all difficult to understand for me. But i do realize this is a SDOF problem.
 
  • #8
re-write your equation as

mx'' + dx' + k(x-a) = 0

then substitute u=x-a

You'll get rid of the constant term. This change of variables represents the measurement of the distance x as a departure from it's equilbrium position.
 
  • #9
pervect said:
re-write your equation as

mx'' + dx' + k(x-a) = 0

then substitute u=x-a

You'll get rid of the constant term. This change of variables represents the measurement of the distance x as a departure from it's equilbrium position.
Umm, i understand x is the vertical displacement of the seat from its equilibrium position, but wot is a?
 

1. What is a spring-mass-damper system?

A spring-mass-damper system is a type of mechanical system that consists of a mass attached to a spring and a damper. It is commonly used to model the behavior of vibrating systems in engineering and physics.

2. How do I solve a spring-mass-damper problem?

To solve a spring-mass-damper problem, you will need to use the equations of motion for the system. These equations can be derived using Newton's laws of motion and the properties of the spring and damper. Once you have the equations, you can use mathematical techniques such as differential equations to find the solution.

3. What are the key parameters in a spring-mass-damper system?

The key parameters in a spring-mass-damper system are the mass of the object, the stiffness of the spring, and the damping coefficient of the damper. These parameters determine the behavior of the system and can be adjusted to change its response.

4. How does the stiffness of the spring affect the system's behavior?

The stiffness of the spring affects the system's behavior by determining how much force is required to stretch or compress the spring. A stiffer spring will result in a higher natural frequency and a faster response, while a less stiff spring will result in a lower natural frequency and a slower response.

5. Can a spring-mass-damper system be used to model real-world systems?

Yes, a spring-mass-damper system can be used to model many real-world systems, such as car suspensions, earthquake-resistant buildings, and musical instruments. However, it is important to note that these models are simplified versions of the real systems and may not capture all of their complexities.

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