# Springs and Sinusoidal Forcing

1. Jun 14, 2005

### Cyrus

I have a question that needs solving please. I have a stand that has a forced oscillator on it. The oscillator can force the stand to move at different harmonics. On this stand is a rig that a spring is clamped rigidly onto. ( I will provide a simplified drawing.). The movement is detected by an interferometer. The calculated resonance frequency should be somewhere around 300htz; however, all sorts of weard data has been obtained. My question is that, in essence, we have a spring within a spring. I.e. the stand is one object of mass oscillating, of which there is a second massed spring object that is free to move as well. Given enough time, should BOTH springed objects move back and forth with the same frequency as the driving force, or is it possible for the frequencies to be different?

Last edited: Dec 26, 2005
2. Jun 15, 2005

### SGT

Since your model has no damping, the movement of both masses will be a combination of the forcing sinusoid and the natural frequencies of both mass-spring sets.
Of course, the big mass will have a component of the higher natural frequency (small mass and spring) with very small amplitude, while the small mass will oscilate with greater amplitudes for the three frequencies.

3. Jun 15, 2005

### Cyrus

So the frequency will NOT be the same for both the whole stand, and the spring being held on the stand? Could you help me find a way to calculate what the frequency would be?

You mentioned 3 frequencies, but thats wrong. there are only TWO things moving. Sorry, my pic/or explanation is bad. Also, this picture is a simplification, but it is esentially whats going on.

See the big square, this is supported by two springs on either end. This whole big square is subjected to oscillations that are forced. Now contained within it is the red square. Its attacted within the big square. Think of the big square as being hollowed out, and putting a small mass with springs inside of it. So its really one big square being forced, but inside this big square is a second smaller (Red) square along for the ride. Now over time, will both of these squares have the same frequency? Its obvious that the big square will have the same frequency of the forced oscillator, but what about the smaller square?

Last edited: Dec 26, 2005
4. Jun 16, 2005

### SGT

Let's call M1 the mass of the big square, k1 the elastic constant of the springs attached to this square, x1 its displacement, M2 the mass of the small square, k2 the elastic constant of the springs attached to this square and x2 its displacement.
We have:
$$F = M_1\frac{d^2 x_1}{dt^2} + k_1 x_1 + M_2\frac{d^2 x_2}{dt^2} + k_2 x_2$$.
Since you have two sets mass-spring, you must have two natural frequencies. If you add the frequency of the driving force F it makes 3 frequencies.

5. Jun 16, 2005

### Astronuc

Staff Emeritus
Using SGT's convention, the smaller mass M2 on the inside affects only that smaller spring, and the larger box simply provides the forcing function for that spring. The larger mass M1 affects the outer spring set. Question - does M1 contain M2, or are they additive, as in M1 contain M2? Different masses mostly likely means different natural frequency, unless the spring stiffnesses are proportional. Think of how the natural frequency is expressed as a function of spring constant k and mass M.

Also, damping coefficients (due to internal and external friction), what are they? The damping coefficients will be different for each spring set.

6. Jun 16, 2005

### Cyrus

well, its really meaningless to speak of the natural frequency. Its not being struck my by an impluse force. The big mass system is being forced into oscillations, so it will take on the oscillations of the forcing device, which can be set to any number of my choosing. So think of this big mass as moving to some setting I choose. Now this second mass is held INSIDE FREE TO MOVE. Think of there being a cavity in the big mass. On each end of the wall of the cavity, I attach a spring, and a mass. So the big mass moves back and forth, so the cavity walls move back and forth, and the mass inside should be forced into moving back and forth. Ok? If not let me know and ill explain it again until we are on the same page.

7. Jun 16, 2005

### SGT

Why do you think there is need of an impulse force to start natural frequency oscillations?
Those oscillations arise in a system whenever the initial conditions are different from the initial value of the forcing function.

8. Jun 16, 2005

### Cyrus

It was my impression that given an impulse, any system has a "natural frequency" with which it will oscillate. But when you supply a forced oscillation, it takes on the frequency of the forcing device.

9. Jun 17, 2005

### SGT

No, natural oscillations arise when the initial conditions differ from the initial value of the excitation. The more common examples are when the excitation is an impulse or a step, but it can happen with any excitation.
In your case, both masses are initially with zero position and acceleration. So, if your driving force starts at zero and oscillates sinusoidally, you will have no natural oscillations. Of course, the smaller mass will lag the bigger one.
If during the operation of the system you change the amplitude or the frequency of the excitation, the new initial conditions and initial excitation will no longer match and you will have natural oscillations.

10. Jun 17, 2005

### panthera

i think this isnt a theoretical Q to be answered in this way...draw seperate block diagrams of the external stand and the inner string.Now draw the individual forces acting on each block...not necessarily the frequency of both will be same...it depends on the mass as well as spring const
w=sqrt K/M but u cant use blindly this formula..block diagram with forces is MUST or best way is solving by conservation of energy...if the Q is the two freqs will be same or not then ans is NOT always...

11. Jun 17, 2005

### Cyrus

Ok, but I am using a wave form generator to vary the oscillations, so I think its a smooth and continuous change when I increase or decrease the frequency. If that is the case, can I asume that it is always in phase with the system, so it will always match the reading on the scope? Will they always have the same initial conditions as the excitiation as I vary it?

12. Jun 18, 2005

### SGT

Let's consider only the inner mass-spring set, that is what interests you. The differential equation is:
$$M_2 \ddot x_2 + k_2 x_2 = A cos(\omega t)$$
At start you have $$x_2(0) = 0$$ and $$\dot x_2(0) = 0$$
The solution of the homogeneous equation is
$$x_{2H}(t) = B cos(\omega_1 t + \phi_1)$$
While the forced solution is
$$x_{2P}(t) = C cos(\omega t + \phi)$$

The general solution is $$x_2(t) = x_{2H}(t) + x_{2P}(t) = B cos(\omega_1 t + \phi_1) + C cos(\omega t + \phi)$$
So,
$$\dot x_2(t) = -\omega_1 B sin(\omega_1 t + \phi_1) -\omega C sin(\omega t + \phi)$$
Replacing the values at t = 0 we get:
$$x_2(0) = B cos(\phi_1) + C cos(\phi) = 0$$
$$\dot x_2(0) = -\omega_1 B sin(\phi_1) -\omega C sin(\phi) = 0$$
Solving the system, you will find that $$B$$ and $$\phi_1$$ are not zero, so you will have the natural frequency $$\omega_1$$ in your response.
Of course, if there is damping, the term containing $$\omega_1$$ will tend to zero and after some time you get only the forcing frequency $$\omega$$ in the steady state.
The same reasoning is valid when you switch the frequency after being in steady state. The initial conditions no longer match the excitation.
Of course your system has friction and, by consequence, damping. So you must elaborate your system better in order to get the right answer.
In any case, even if at steady state the frequencies of the inner and outer boxes are the same, the phases are not.

13. Jun 18, 2005

### Cyrus

Ok, I dont really care about the phase relationships. The only thing of interest is that the two masses have the same frequency as the forcing device.

14. Jun 18, 2005

### aura

they will not be the same as per me...

for forced oscillations

mx'' +kx = F coswt where F is the applied force...

so from here itself we can see that w depends on mass and spring const...even if u consider the displacement and acceleration to be same as the larger spring, still w will be different...

15. Jun 18, 2005

### SGT

As I said, if you have friction in your system (and you certainly have), the natural oscillations will be damped and disappear after some time (~5 time constants).
So, you have to wait some time in order to measure the oscillation frequency, not only at start but every time you change amplitude or frequency of oscillation.