6
$\begingroup$

What actually is a natural frequency for an object and what makes it vibrate with increased amplitude when coupled with an external oscillator that matches the natural frequency?

$\endgroup$
  • $\begingroup$ My question is different.I want to know why objects have a natural frequency?Why is that particular frequency so special to the object? It's not about positive or negative work that I am asking, im asking why is the object so interested in that frequency. Link me any such answer. Thanks in advance $\endgroup$ – dushyanth Feb 14 '15 at 10:47
  • 1
    $\begingroup$ If you strike an object that is relatively solid, it will "ring" -- effectively, a shock wave from your striking the object propagates to the far end of the object and back, then bounces off the near end and does it all again. The rate at which it does this is it's resonant frequency. If you supply a steady input (eg, from a loudspeaker) at that frequency, each successive pulse from the loudspeaker will reenforce the shock wave traveling back and forth and it will get stronger and stronger. $\endgroup$ – Hot Licks Feb 15 '15 at 1:16
5
$\begingroup$

Here's a simpler answer.

Resonance is really all about the capture of energy into a system and its cyclic flow between potential and kinetic states. In mechanical systems we call these states potential energy and kinetic energy, but in electrical systems, as a another example, between magnetic and electrical fields. It's the rate of this cycling back and forth that results in the natural frequency.

For example the 'singing' aluminum rods that are often used to demonstrate standing wave resonance in the classroom capture energy from your fingers as they rub over the outside of the rod. The energy excites the atomic lattice causing the lattice to expand,relax, and compress at the rate of the natural frequency - the speed at which energy moves from a fully potential state: when fully stretched or compressed and at the lowest velocity to a fully kinetic state - halfway between stretching and compressing when the lattice is at maximum velocity. If the energy flow has only a small amount of losses - for example heat generated in the rod, then we say there is a low impedance to energy flow, and so the rod will have a tendency to suck up more energy than it loses and maintain the state of resonance.

The rate of energy flow is dependent on the material properties, but also the particular geometry of object. If the rate of energy loss from the object is greater than the rate of energy entering the object, the cycling will be 'damped' and therefore lack resonance.

That's resonance in a nutshell.

$\endgroup$
3
$\begingroup$

Natural frequency depends on the physical properties of a system. Some of the classical examples are masses attached to springs and pendula. For the former class, the basic model is based on Hooke's law, which translates into the differential equation (1D) $$m\ddot x + kx = 0,\qquad m,k > 0$$ where any sort of dissipative effect has been neglected. Dividing by $m$ one also gets $$\ddot x + \omega^2 x = 0,$$ where $\omega^2 = \frac km$ is the (square of the) so-called natural frequency. As it can be seen from its defining expression, it depends on the mechanical properties of the objects involved, namely the mass of the oscillating body $m$ and the elastic constant of the spring $k$. This can be generalised to systems of couples oscillators, which are basically systems of couples masses and spring, and such abstractions are used to model complicated physical system. The most general form is then something like $$M\ddot{\mathbf x} + K\mathbf x = 0,$$ where $\mathbf x\in\mathbb R^n$, and $M$ and $K$ are positive definite $n\times n$ matrices (assumptions on $\Omega^2$ may be weakened a bit but I will ignore this here). Since then $M$ is invertible, one can rewrite this system in the form $$\ddot{\mathbb x} + \Omega^2\mathbb x=0,$$ where $\Omega^2$ (which at this point is an abuse of notation) is the matrix given by $$\Omega^2 = M^{-1}K.$$ The problem becomes "interesting" when $M$ and $K$ commute, for in this case $\Omega^2$, as the notation suggests, is a positive definite matrix that can be diagonalised (physically this corresponds to the decoupling of the oscillators), and the diagonal entries are again the squares of the natural frequencies of the system under consideration. The eigenvectors (which form an orthonormal basis in this case) give the direction where the natural oscillations at the corresponding natural frequency occur.

If we relax the condition on commutativity between $M$ and $K$ one might still succeed in diagonalising $\Omega^2$, but this could fail to be positive definite. Though, when this happens, the eigenvectors won't be orthogonal to each other in general (and this happens in practice) and therefore the natural modes won't be mutually perpendicular.

Observe that we can also perform the change of coordinates $\mathbf y = M^{\frac12}\mathbf x$, with which we would obtain the new equation $$\ddot{\mathbf y} + M^{-\frac12}KM^{-\frac12}\mathbf y=0,$$ and now $\Omega^2 = M^{-\frac12}KM^{\frac12}$ is guaranteed to be positive definite, since it is of the form $B^TB$, with $B=K^{\frac12}M^{-\frac12}$.

$\endgroup$
1
$\begingroup$

when the force frequency equals the natural frequency, the displacement wave exceeds the force wave by 90 degree, you can represent that if the force wave is cosine and the displacement wave is sine. this means that when the vibrating part reaches the top value and tends to change its velocity downwards it is exposed to maximum value of the force, and vice versa. This means that the force exerts the maximum energy to the mass, and the displacement will be the maximum. If the damping is not large enough, the displacement will exceed the ultimate strength limit of the material and the failure occurs.

$\endgroup$

Your Answer

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

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