2

I am curious about that since damping will not affect frequency of SHM, then why it does affect on the natural frequency of the SHM. In the resonance damping graph the peak amplitude become lower but is shifted left. The more greater tge damping effect,the more the peak is shifted.

I have searched many reference but there are no clear explanation about this phenomenon. Can someone explain this phenomenon? enter image description here

Qmechanic
  • 201,751
Pck Tsp
  • 83
  • FYI the mathematical formula of the above is $$\omega = \omega_n \sqrt{1-\zeta^2} $$ where $\zeta$ is the damping ratio, $\omega_n$ is the undamped natural frequency and $\omega$ is the damped natural frequency. – John Alexiou Oct 11 '22 at 13:00
  • Intuitively, one can imagine that since there is damping, it takes longer for the object to complete one cycle, hence decreasing the period. This picture is clearly incomplete, however, because the object also travels a smaller distance (since its motion is damped), and of course, in an undamped oscillator, these two effects exactly "cancel" each other out (that is, increasing the amplitude of an SHO doesn't change the frequency due to the delicate balance between the object having a larger distance to travel but also feeling a larger force). – march Oct 11 '22 at 16:33
  • Possible duplicates: https://physics.stackexchange.com/q/228279/2451 , https://physics.stackexchange.com/q/186239/2451 and links therein. – Qmechanic Oct 11 '22 at 17:05
  • See also answers and links here: https://physics.stackexchange.com/q/749963/226902 – Quillo Feb 22 '23 at 10:53

1 Answers1

0

This is easy to understand if we take the example of a vibrating string with fixed constraints on both ends. We set the string in motion, and observe it vibrating.

We then add damping to the system by allowing one of the end constraints to slide up and down a bit against friction with the wall the string pulls up and down on it during its oscillation. This allows some of the vibrational energy contained in the string to "leak out" past the constraint and leave the system.

Now since the constraint point can move up and down, the "apparent" constraint point (where the vibrational displacement goes to zero) appears to be somewhere beyond the movable constraint, off the end of the string. You can see this by sketching a line tangent to the string through the movable constraint point that extends beyond it, and then sketching a second line that extends the centerline of the string axis beyond the movable constraint. These two lines in intersect at the "apparent" constraint point, which lies beyond the movable constraint.

The string therefore is vibrating with an apparent length which is slightly longer than the actual length of the string, and that extra length gives the string a lower natural frequency.

If we force the displacement of the movable constraint to go to zero, the natural frequency of the string rises back to its "normal" level as the leakage of energy out of the system is clamped off.

If we allow the displacement of the movable constraint to grow until it matches the maximum displacement of the vibrating string, then almost all the vibrational energy leaks out and there is then no reflection at the (movable) constraint with which to establish harmonic motion, and the system exhibits exponential decay instead.

niels nielsen
  • 92,630