2
$\begingroup$

I recently carried out an experiment varying different factors affecting simple harmonic motion, namely friction and air resistance. Whilst carrying out research for this, I found the relationship angular frequency $$ω = \sqrt{ \left(\frac{k}{m}\right)^2 - \left(\frac{b}{2m}\right)^2}$$ where $b$ is the viscous damping coefficient. However, my experiment found that throughout, the angular frequency stayed the same, seemingly independent of the drag/friction force.

What could the reason for this be?

$\endgroup$
6
  • 1
    $\begingroup$ Question, how large is the first term in your expression for $\omega$ as compared to the second part? As in what are some sample values of k, m, and b? Another thing it would help if you were more descriptive of your experiment in general. $\endgroup$
    – Triatticus
    Mar 28 at 15:07
  • 1
    $\begingroup$ Also, how precise are your measurements of the frequency? $\endgroup$ Mar 28 at 15:08
  • $\begingroup$ the value for k I calculated was 11.2 N/m and m was 0.235 kg. wouldn't b change throughout the experiment? $\endgroup$ Mar 28 at 15:40
  • $\begingroup$ moreover, my experiment was using differently sized sails on a tethered trolley connected to 3 springs on either side, to vary the drag force on the trolley. I also used a sandpaper on the track under the trolley to vary the friction for a different variation of the same experiment. $\endgroup$ Mar 28 at 15:43
  • $\begingroup$ You should be finding that $\omega=\sqrt{\omega_{0}^2 - (\frac{b}{2m})^2}=\sqrt{\frac{k}{m} - (\frac{b}{2m})^2}$ though right? Where $\omega_0 = \sqrt{\frac{k}{m}}$ is the undamped frequency of the system. It is my understanding that you can use the viscous damping coefficient $b$ to determine a damping ratio, or directly discover if the system is over/under/critically damped. So what frequency would you calculate for your system without damping? Does this correspond roughly to your experimental observation? $\endgroup$
    – bleuofblue
    Mar 28 at 17:38

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.