0
$\begingroup$

In a forced oscillation, what exactly is happening? My textbook says that: The oscillator, initially, oscillates with the natural frequency. When we apply external periodic force, the oscillation with natural frequency die out; body oscillates with the frequency of the periodic force. Also, why is the amplitude maximum when the frequency of the external force is equal to natural frequency? I can understand it mathematically.

$$ A = \frac{F_0}{\sqrt{m^2(ω_d - ω)^2 + (ω_db)^2 }}$$ where, $ω_d$ is the frequency fo the driving force

So when, $$ω_d = ω $$ $$A = \frac{F_0}{(ω_db)}$$

But can you explain what exactly is happening when the frequency of, say the swing, is the same as the frequency of the force with which my friend is pushing me? This is confusing.

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ Why is it confusing? Just follow the math. Isn't it intuitive that amplitude maxes when both frequencies are the same? It is to me. $\endgroup$
    – Gert
    Commented Feb 11, 2020 at 18:43
  • $\begingroup$ When you push the swing at the frequency it wants to swing at, doesn’t your friend go higher? $\endgroup$
    – G. Smith
    Commented Feb 11, 2020 at 18:56
  • $\begingroup$ When the oscillator is undamped, you will get beats when the driver frequency is near resonance. $\endgroup$
    – user137289
    Commented Feb 11, 2020 at 18:57
  • $\begingroup$ @Gert I don't understand the following: The oscillator, initially, oscillates with the natural frequency. When we apply external periodic force, the oscillation with natural frequency die out; body oscillates with the frequency of the periodic force. Why does the oscillation with natural frequency die out? My mind says that it should be something like 'net' frequency. After we apply external periodic force, it should oscillate with a frequency that constitutes the frequency of the force as well as its natural frequency. $\endgroup$
    – Kaushik
    Commented Feb 11, 2020 at 19:41
  • $\begingroup$ Look at your equation with the damping term, but without the driving term--the homogeneous one. No driving frequency around. The solution is a decaying oscillation with the natural frequency. You may add this homogeneous equation solution to the one discussed here, which solves the inhomogeneous equation. The natural frequency term is doomed out of the problem, and so uninteresting. $\endgroup$
    – Cosmas Zachos
    Commented Feb 11, 2020 at 23:16

0

Reset to default

Browse other questions tagged or ask your own question.