Forced Mass-Spring System - Diff Eq

In summary, This conversation discusses a forced, mass-spring system without damping. The spring constant is 4 N/m, the weight of the mass is 9.8 N, and the mass is 1 Kg. The motion X(t) of the mass is found using ω = 1.5 (Hz), 1.9 (Hz), and 3 (Hz). The maximum elongation of the spring is deduced and the vibrations X(t) are sketched for each ω. The range of all safe frequencies ω is also found, which are those that do not cause the system to break (elongate more than 0.06 m). The equations for the external force and the external frequency
  • #1
DualCortex
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0

Homework Statement



You have a forced, mass-spring system, without damping.
Spring constant = 4 N/m
weight of mass = 9.8 N
mass = 1 Kg

Find the motion X(t) of the mass if ω = 1.5 (Hz) and deduce the
maximum elongation of the spring. Sketch the vibrations X(t). Do same for ω = 1.9, 3.

Find the range of all safe frequencies ω. That is, frequencies that do
not destroy the system (spring breaks if elongated more than 0.06 m.At t = 0, system is at equilibrium/rest.
Therefore X(0) = 2.45 and X'(0) = 0.

Homework Equations


External force: 0.04*cos(ω*t)
" ... the external frequency ω can be adjusted ... "

The Attempt at a Solution


Diff eq should be

[tex]X\text{''}+4*X=0.04*\cos (\omega *t)[/tex]

Therefore,
[tex]X(t) = \left(2.45-\frac{0.04}{4-\omega ^2}\right) \text{Cos}[2*t]+\frac{0.04 *\text{Cos}[t *\omega ]}{4-\omega ^2}[/tex]

So, plugging in ω isn't that bad for the first part. However, I'm not sure how I am supposed to find the maximum elongation at each of the frequencies.

Thanks for any hints/help.FORGET ABOUT THIS PART: Manually (the dirty method of plugging in values), using Matlab I found ω = (19.97653, 20.0560) to be the range where the amplitude went over 0.06 cm.

Turns out the professor made a mistake on one of his given variables, so the spring constant is actually 4 instead of 400, which makes a lot more sense. Just noticed the problem now is that either the spring should be able to be stretched longer or the mass weighs less ... will have to email my professor.
 
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  • #2
I'm assuming you got the solution correctly

If you just plug in the frequencies like you've done you get X(t)=some time varying function

The maximum elongation will occur when the velocity is 0, right?
 
  • #3
Well yeah, problem is that then you'll have something like
-Sin(1.5t) = Sin(20t)
if we let ω = 1.5 and not all waves have the same amplitude. Or, well, the amplitude will be pretty much constant unless ω gets close to 20.
 
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  • #4
Again, assuming you did it right(took dx/dt, set = 0)you'd solve for t to find the time that the velocity equals 0, and then you can use that time in your position equation to find the corresponding value of x
 
  • #5
Hey blochwave:

if you could please explain to me how to solve:

[tex]\frac{0.04*\omega}{4 - \omega^{2}}*sin(\omega * t) = 2*\left(\frac{0.04*\omega}{4 - \omega^{2}}-2.45\right)*sin(2*t)[/tex]

[tex]\omega = 1.5, 1.9, 3[/tex], as the problem asks me.
 
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  • #6
Ok to backtrack

I'm looking at the part where you're asked to find the maximum elongation. This occurs when the velocity of the spring is 0, as you can imagine(it's changing direction when the velocity is 0 so it's as far as it goes)

You had found an expression for its displacement from equilibrium in terms of time

by differentiation with respect to time, you find the velocity as a function of time. You set that velocity equal to 0, then you can solve for TIME(you're given the values of omega to use, just plug them in one by one), this will be the time(s) that the maximum elongation occurs, and then you can plug that into the X(t) equation to find what the actual elongation is
 
  • #7
blochwave, I fully understand what you are telling me, and it is the method I tried at first right after I solved the diff. eq.

The equations above is X'(t) set equal to 0.
So I plug in, say, 1.9. Then what?

I don't know how to solve it mathematically, but apparently the max is 2.45.
As w grows or decreases, terms go to zero and the only one that is left is 2.45*cos(2t)

On the other hand as w approaches 2, terms cancel out and again the only one left is 2.45*cos(2t)

ACTUALLY, forget about this problem for now. There's various issues with the given conditions. After he told us of the mistake, the spring constant is way too low (4, compared to 400 previously), first the weight on the system is too high (which I contacted my professor about) stretching the string too much, and the external force is way too low
 
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Related to Forced Mass-Spring System - Diff Eq

1. What is a forced mass-spring system?

A forced mass-spring system is a physical system consisting of a mass attached to a spring that is being acted upon by an external force or forces. This system can be used to model various real-world phenomena, such as the motion of a car's suspension or a swinging pendulum.

2. How is a forced mass-spring system described mathematically?

The motion of a forced mass-spring system can be described using a second-order differential equation, known as the forced mass-spring equation. This equation takes into account the mass of the object, the stiffness of the spring, and the magnitude and direction of the external force(s) acting on the system.

3. What is the difference between a free and a forced mass-spring system?

A free mass-spring system is one that is not being acted upon by any external forces, and its motion is solely determined by the properties of the mass and the spring. A forced mass-spring system, on the other hand, is subject to external forces that affect its motion in addition to the properties of the mass and the spring.

4. How does changing the parameters of a forced mass-spring system affect its behavior?

The behavior of a forced mass-spring system can be altered by changing the parameters that describe it, such as the mass, spring constant, and external force(s). For example, increasing the mass will result in a slower oscillation, while increasing the spring constant will result in a faster oscillation.

5. What real-life applications does a forced mass-spring system have?

Forced mass-spring systems have many practical applications, such as in engineering and physics. They are used to model the behavior of mechanical systems, such as car suspensions, bridges, and buildings. They are also used in fields such as seismology, where they can be used to study earthquake motion and vibrations.

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