Timeline for Why are the resonant frequencies for displacement, velocity and acceleration different in a damped oscillator?
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| Jan 9, 2022 at 9:28 | history | bounty ended | user246795 | ||
| Jan 6, 2022 at 14:37 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 6, 2022 at 14:31 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 5, 2022 at 18:02 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 5, 2022 at 17:47 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 5, 2022 at 0:13 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 4, 2022 at 20:50 | comment | added | Cosmas Zachos | I'm not sure I did not reverse the order of the above crossings! t= π/2 we are discussing is the maximum of the displacement, at the very top of the cycle. v crosses 0 at t= π/2 but F bit later, at t= π/2ω. So the arrangement favors ω<1 , where 1 is the natural undamped frequency (your $\omega_0$). The swing wants to start falling, but the driver pulls it back, stretching the period and the potential energy, the surplus to be expended on matching frictional energy losses, in the steady state. | |
| Jan 4, 2022 at 20:09 | comment | added | user246795 | That makes more sense. The last thing is: why does the arrangement that absorbs maximal power not give a maximum in the amplitude (potential energy). That is, why should the maximum amplitude of the displacement occur when $F$ opposes $v$ a little bit and increases the period? Thanks for your help so far. | |
| Jan 4, 2022 at 18:18 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 4, 2022 at 16:31 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 4, 2022 at 16:08 | comment | added | Cosmas Zachos | Apologies for the peremptoriness. It would hep if you schematically plotted the cycles for x, v, and F (and a, if you wanted, but it is irrelevant in the energetics) for ω near 1, with x(0)=0, so F is in phase with v, enhances it rather than damp it, and the power input is maximized: the oscillator is absorbing maximal power to offset its frictional losses. In the "optimal energy absorption" case, note F crosses 0 at t= π/2, slightly before v, that is, it starts opposing it a bit and so prolongs the swing, thus increasing the period and ever so slightly decreasing ω. | |
| Jan 4, 2022 at 15:16 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 4, 2022 at 12:13 | comment | added | user246795 | The link you originally refer to is similarly close. I just don't understand what is meant by "For the former [maximum AMPLITUDE], you would like the frequency to be slightly lower (because you dissipate a certain amount of power per cycle)." | |
| Jan 4, 2022 at 10:02 | comment | added | user246795 | Could you clarify what you mean by "since you pump/replenish energy into the system by prolonging its cycle, i.e. stretching it longer at the point of minimum kinetic energy."? | |
| Jan 4, 2022 at 7:34 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 4, 2022 at 3:50 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 4, 2022 at 3:27 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 4, 2022 at 3:20 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 4, 2022 at 1:19 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 3, 2022 at 23:25 | history | edited | Cosmas Zachos | CC BY-SA 4.0 |
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| Jan 3, 2022 at 22:34 | history | answered | Cosmas Zachos | CC BY-SA 4.0 |