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Resonance watch

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    I understand that resonance occurs when the frequency of the periodic driving force is equal to the natural frequency (or resonant frequency) of the oscillator. I also understand that the periodic driving force is pi/2 out of phase with respect to the displacement of the oscillator. So my question is, why is it that the oscillator, when exposed to a driving force at resonant frequency, "attunes" itself to become pi/2 out of phase? Is it because it is the state of lowest energy? As it is doing the minimal amount of work, if any, against the driving force.
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    (Original post by Protoxylic)
    I understand that resonance occurs when the frequency of the periodic driving force is equal to the natural frequency (or resonant frequency) of the oscillator. I also understand that the periodic driving force is pi/2 out of phase with respect to the displacement of the oscillator. So my question is, why is it that the oscillator, when exposed to a driving force at resonant frequency, "attunes" itself to become pi/2 out of phase? Is it because it is the state of lowest energy? As it is doing the minimal amount of work, if any, against the driving force.
    Where did you get this from? I don't think there is anything that suggests the driving force will always necessarily be \frac{\pi}{2} out of phase with the oscillator?
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    (Original post by pleasedtobeatyou)
    Where did you get this from? I don't think there is anything that suggests the driving force will always necessarily be \frac{\pi}{2} out of phase with the oscillator?
    That is common knowledge that the driving force is pi/2 out if phase with the displacement of the oscillator when at resonance.

    Edit: My original post makes it seem like I'm talking about forced vibrations also. I am strictly talking about resonance.
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    (Original post by Protoxylic)
    I understand that resonance occurs when the frequency of the periodic driving force is equal to the natural frequency (or resonant frequency) of the oscillator. I also understand that the periodic driving force is pi/2 out of phase with respect to the displacement of the oscillator. So my question is, why is it that the oscillator, when exposed to a driving force at resonant frequency, "attunes" itself to become pi/2 out of phase? Is it because it is the state of lowest energy? As it is doing the minimal amount of work, if any, against the driving force.
    With a phase displacement other than pi/2, some of the driving force (or all dependent on the phase angle), will be working against the natural restoring force which therefore acts to attenuate displacement amplitude.

    Thus the driving force will cause the original natural oscillation energy to dissipate (because the restoring force is working in opposition to the driving force) and the oscilllation is increasingly dominated by the driving force until pi/2 is achieved.

    At resonance, all of the driving force is then working with the natural restoring force to maximise displacement amplitude which settles at a new amplitude where system energy loss equals the input driving energy.
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    (Original post by uberteknik)
    With a phase displacement other than pi/2, some of the driving force (or all dependent on the phase angle), will be working against the natural restoring force which therefore acts to attenuate displacement amplitude.

    Thus the driving force will cause the original natural oscillation energy to dissipate (because the restoring force is working in opposition to the driving force) and the oscilllation is increasingly dominated by the driving force until pi/2 is achieved.

    At resonance, all of the driving force is then working with the natural restoring force to maximise displacement amplitude which settles at a new amplitude where system energy loss equals the input driving energy.
    So the driving force is pi/2 out of phase with the displacement of the oscillator so as to limit the amount of work the oscillator will do against the driving force, if any, such that the amplitude becomes a maximum with respect to frequency of the applied force.
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    (Original post by Protoxylic)
    So the driving force is pi/2 out of phase with the displacement of the oscillator so as to limit the amount of work the oscillator will do against the driving force, if any, such that the amplitude becomes a maximum with respect to frequency of the applied force.
    One way of visualising it is to realise first that, with no damping (the exact pi/2 difference only applies when damping is zero) the mass on the spring, say, will oscillate anyway as a result of an elastic driving force which is maximum at maximum displacement.
    The most efficient way of applying another force would be to make it such that this applied force is greatest when the natural restoring force is zero and zero when the restoring force is maximum. This inevitably means the applied force is 90 degs out of phase with the oscillation.

    Oscillation: 0 + 0 - 0 + 0 -
    Applied F:: + 0 - 0 + 0 - 0
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    (Original post by Stonebridge)
    One way of visualising it is to realise first that, with no damping (the exact pi/2 difference only applies when damping is zero) the mass on the spring, say, will oscillate anyway as a result of an elastic driving force which is maximum at maximum displacement.
    The most efficient way of applying another force would be to make it such that this applied force is greatest when the natural restoring force is zero and zero when the restoring force is maximum. This inevitably means the applied force is 90 degs out of phase with the oscillation.

    Oscillation: 0 + 0 - 0 + 0 -
    Applied F:: + 0 - 0 + 0 - 0
    I understand that principal, the situation that I am struggling to understand is this scenario: Say you have an oscillator that is immediately exposed to a driving force of frequency that is equal to the oscillator's natural frequency. How so does the oscillator's displacement adjust such that it is pi/2 out of phase with the frequency of the driving force. And my view was that it did so such that the minimal amount of work was done in opposing the driving force, such that the total energy of the system is maximised.
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    (Original post by Protoxylic)
    I understand that principal, the situation that I am struggling to understand is this scenario: Say you have an oscillator that is immediately exposed to a driving force of frequency that is equal to the oscillator's natural frequency. How so does the oscillator's displacement adjust such that it is pi/2 out of phase with the frequency of the driving force. And my view was that it did so such that the minimal amount of work was done in opposing the driving force, such that the total energy of the system is maximised.
    At resonance the driving force doesn't oppose the natural restoring force. This only happens when there is the phase difference of pi/2. It always acts with it. If for some reason the system finds itself such that the phase difference isn't pi/2, then the driving force at some point is acting against the restoring force and acting to slow down the mass. If this happens then the mass oscillates at a lower frequency and its amplitude will reduce. If the frequencies are different then you no longer have resonance. If the frequencies are different then the phase angle will change. I presume it changes until it's once again at pi/2. Once at pi/2 it has no reason to change any further.

    Edit: to add

    It's a sort of "self adjusting" system.
 
 
 

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