0
$\begingroup$

In a mass-spring system with an oscillation generator, a force is supplied by the generator that causes the mass to oscillate. I know that the force is periodic/harmonic and I have defined it as $F=f\cos(wt)$. However, is there a way in finding the amplitude $f$ of the force?

I have experimental data, namely displacement-time graph $x(t)$ and velocities at certain times, also the frequency which the oscillation generator was set at.

$\endgroup$
0

2 Answers 2

0
$\begingroup$

You would not be able to do so at or near resonance. Once the oscillator reaches a stable resonance oscillation, the generator is exactly balancing the damping force. If there were no damping force for the generator to oppose, oscillation would eventually reach a high enough amplitude to break apart the apparatus.

At very low frequency, the entire structure (spring and mass) would just move with the oscillation generator. There would be essentially no damping. You could treat it as a solid object hanging from the generator and just use Newton's Second Law.

At very high frequency, the mass would not move. Wave patterns would form in the spring just as if the spring were anchored solidly at the bottom. With the very high frequencies, the wavelengths would be very small. Most frequencies would be at or near standing waves, again making the generator force counter any damping effects.

If the oscillator amplitude is very small, damping effects will be just as strong as the driving generator. Only at extremely low frequencies will you be able to measure the driving amplitude.

$\endgroup$
0
$\begingroup$

From the moment the external force was turned on, wait enough time until the transient vanishes. Then you have the steady state movement only. The formula of the steady state can be written as $x_{ss} (t) = A \cdot \cos(\omega t) + B \cdot \sin(\omega t)$. Count the time of the period $T$, then find $\omega = \frac{2\pi}{T}$. Now you have the ${(t_i, x_i)}$ series raw data to plug in $x_{ss}$ formula and find $A$ and $B$ (I'd recommend the Least Squares Method). Recall from the ODE of the mass-spring system: $$x'' + 2 \gamma x' + \omega_0^2 x = F_0 \cos(\omega t)$$ (where $\gamma = \zeta \omega_0$ is the damping rate and $\omega_0$ the resonant frequency).

And we have the steady state solution. Now you plug $x_{ss}$ in the ODE, and since $\sin$ and $\cos$ are L.I. you have the following system:

$$A \cdot(-\omega² -2\gamma + \omega_0^2) = F_0 $$ $$B \cdot(-\omega² + 2\gamma+ \omega_0²) = 0 $$

You have to find $\gamma, \omega_0^{2}, F_0$ with two equations only. You need to know some variable more. I can only think in two options: 1. If you can find the mass $m$ and the constant of the spring $k$, for example turning the mass-spring vertical and using the fact that $mg=k\Delta y$ to solve for $k$, then $\omega_0 = \sqrt{k/m}$. 2. You can suddenly turn off the external force and measure the settling time to estimate $\gamma$, then you solve the system.

$\endgroup$
2
  • $\begingroup$ "If the system starts from the rest the transient movement won't exist (check from the homogeneous solution of the ODE). Then you have the steady state movement only." That's not true; the transient is generally still needed to match the initial conditions. $\endgroup$
    – Buzz
    Feb 16, 2023 at 22:38
  • $\begingroup$ @Buzz Right, I don't know how I overlooked that. Thanks $\endgroup$
    – tac
    Feb 16, 2023 at 23:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.