Taking displacement along the x-direction shown to be from the static equilibrium position, then applying along the shown direction, we obtain
which is the equation of motion. To obtain the natural frequency, we consider free vibration , which implies that , hence we see that the natural frequency is independent of
We see that gravity has no effect on the spring mass system, this is because we use to be from the static equilibrium position of the spring.
First we need to derive the equation of motion. Considering the following diagram
Using as generalized coordinates , we obtain
Notice that in the calculation of above, we assumed that the spring stretches by in the horizontal direction only, which we are allowed to do for small .
Now we can find Lagrangian
Hence the equation of motion is
The above is nonlinear equation. Linearize around (equilibrium point) using Taylor series, and for small we obtain and , hence the above becomes
Hence effective can be found from
Hence
Compare the above to the natural frequency of pendulum with no spring attached which is =, we can see the effect of adding a spring on the natural frequency: The more stiff the spring is, in other words, the larger is, the larger will become, and the smaller the period of oscillation will be. We conclude that a pendulum with a spring attached to it will always oscillate with a period which is smaller than the same pendulum without the spring attached. This makes sense as a mass with spring alone has
We need to solve for and
The characteristic equation is which has roots
Hence the solution is
is the general solution. Now we use I.C. to find . When
Hence , and and at , we obtain
Then
This is a plot of the solution for up to 50 seconds
We need to solve for and
The characteristic equation is which has roots
Hence the solution is
is the general solution. Now we use I.C. to find . When
Hence , and
and at , we obtain
Hence
This is a plot of the solution for up to 12 seconds
This is a single degree of freedom linear system. Assume from static equilibrium, then (using parallel springs) we obtain
Hence and the Lagrangian equation is
Hence equation of motion is
Solution
Part(a)
Assuming small angle oscillation, , hence
and for the mass, since it losses potential, we have
Hence Lagrangian is
Now find the Lagrangian equation
Hence
And the equation of motion is
Linearize by setting we obtain equation of motion
| (1) |
Hence
Part (b)
To discuss stability, we need to determine the location of the roots of the characteristic equation of the homogeneous EQM, hence from equation (1), we see that
And assuming solution leads to the characteristic equation
Since , then
Since roots of the characteristic equation on the imaginary axis, this is a marginally stable system regardless of the values of .
Since we are looking at the linearized system, there is only one equilibrium point, and the system is either stable or not. Here we found it is marginally stable. The effect of changing is to change the period of oscillation around the equilibrium point.