And how exactly do objects acquire "natural frequencies"? Is it due to the temperature and the lattice structure (the type of bonds they form with other atoms)? And thus, is resonance just a phenomenon that preserves energy within the lattice than leak it outside?

And is the below illustration accurate for explaining resonance?

Case 1: A hollow metal sphere. Say its natural frequency is 20 Hz. So, if we hit it with a hammer and the resultant energy (by accident/chance) makes it vibrate at 20 Hz, then it takes $n$ seconds to cool down and loses $x$ Joules to the atmosphere as heat energy (per second).

Case 2: Same sphere as above. We hit it with another hammer but more forcibly. It vibrates at 22 Hz in the beginning, and because that's not its natural frequency, it comes to a standstill at $n+4$ seconds, losing $x+4$ Joules as heat energy in the process (per second).

Case 3: Same sphere as above. We hit it with a smaller hammer, lightly. It initially vibrates at 18 Hz. Since that's also not the right frequency as the natural one, it loses energy fast but not as fast as Case 1, as the total energy, in this case, is also low (Law of Equilibrium says that one side has too much energy, it loses energy fast to reach the equilibrium). Say it takes $n+2$ seconds and loses $x+2$ energy per second.

What if we increase the impact in Case 1 by a thousand times (assuming the object doesn't break)? Will the resonant frequency still last longer than the frequency generated by the impact? Is there a threshold that tells the exact amount of energy that is required to cross this resonant threshold of losing energy?

  • $\begingroup$ IMO, if you want to understand "natural frequencies" (a.k.a., "resonance") you should start by learning what makes a pendulum work and then, by learning the equations that govern a simple harmonic oscillator. Got to understand the basics before you move on to more complex systems like your ringing metal sphere. $\endgroup$ Aug 4, 2021 at 12:32
  • $\begingroup$ Why use a sphere, when you can use a mass on an ideal spring? Also, the resonance cases are unrelated to the title...is this 2 different questions? $\endgroup$
    – JEB
    Aug 4, 2021 at 14:18
  • $\begingroup$ When you whack your sphere, it's motion will be complex--a superposition of multiple, three dimensional vibration modes. When you are ready to understand the math behind that, then you will be ready to dive in to quantum mechanics. No Joke! It's literally almost the same thing. That's why you should start with something simpler. JEB is right: a mass on a spring is better than a pendulum. But note: you'll be working with differential equations--It was 12th grade math when I was a kid, but I don't know what schools teach today. $\endgroup$ Aug 4, 2021 at 15:22

1 Answer 1


We will take the case of mechanical vibrations and resonance. This will let me answer the question posed in the title line of your post.

If you attach something with mass to something with springiness and then kick it, you set the mass into motion and it compresses the spring until the spring has stored up all the kinetic energy in the mass, at which point the mass has stopped moving. Then the spring pushes back on the mass, reversing its direction of motion, and it stops pushing when the mass is back at its starting position. But now the mass has kinetic energy and it overshoots its starting position and is now pulling on the spring, which pulls back on the mass until the spring is stretched out and the mass has come to a halt... and this process goes on, over and over, until the friction in the system has dissipated the energy you put in with your first kick.

In a mechanical system, the atoms all move together as a single mass, and their electrons (necessarily) move with them in unison.

So the necessary conditions for imparting a resonant frequency to a system is that it contain an inertance coupled with a compliance. A big mass connected to a soft spring will yield a low resonant frequency and a small mass coupled to a stiff spring will yield a high resonant frequency.


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