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I How do we excite specific standing waves on a string?

  1. Jan 13, 2017 #1
    Hello,

    I am aware that there are specific oscillatory patterns than can form on on a string. These patterns are called normal modes and represent standing waves. Each standing wave has an associated frequency f which indicates the speed at which the string's points are moving up and down. Higher harmonics (modes) have higher frequency.

    How do we solely excite a specific mode without exciting the other modes? For instance, can we excite the 3rd mode without exciting the fundamental or the other higher modes? Do we need to pull the string where the mode has one of its antinodes or make sure we force the string point where a node would not to displace? Usually, a higher harmonic has multiple nodes and multiple antinodes...

    Clearly, we could excite a higher mode, the 4th one which has 4 antinodes, by pulling the string at rest where those points are on the string and keep the nodes from moving. We practically stretch the string so that is assume the shape of the mode. After that, the mode will continue its existence on its own and oscillate...

    Thanks!
     
  2. jcsd
  3. Jan 13, 2017 #2

    Henryk

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    Gold Member

    Yes, you can excite only one mode. To do that, you have to apply a sinusoidal force at the frequency of that mode.
    You can also excite two or more specific modes of vibration by applying a force that is a sum of sine waves of the specific frequencies.
    It is difficult to do it by hand. When you 'strike a string', it is just applying an impulse force to the string, that can be represented by a sum of range of frequencies.
    But electronic drive, say, an electromagnet driven at the proper frequency can do it.
     
  4. Jan 14, 2017 #3
    Thanks Henryk.

    But how do I apply a sinusoidal force? By attaching one of the ends (the other end is fixed) of the strings to an electromechanical vibrator that pulls the string end up and down at a certain frequency? Will that automatically excite the mode pattern with the specific quantized frequency as long as the vibrator frequency is equal to that? This method is interesting since it forms and selects the entire mode by just shaking one point (the one attached to the vibrator) of the string.....

    As far as striking a string, like in a guitar, we usually place a finger on the string to create a node and plug the string somewhere else. The plucking represents an impulsive initial condition which primarily excites a specific mode but also other harmonics. The Fourier decomposition of the pulse at one point +node at another point is shown to be formed by multiple harmonics, not just one....
     
  5. Jan 14, 2017 #4

    jtbell

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    Staff: Mentor

    Yes. This is a common experiment in introductory university physics labs. As you slowly increase the frequency, you can see the different standing-wave modes appear in succession.
     
  6. Jan 14, 2017 #5
    Thanks a lot.

    What happens when the oscillator frequency does not match any of the modes' quantized frequencies? The oscillator will move but what pattern will the string assume? Where does the oscillator energy go?
     
  7. Jan 15, 2017 #6

    Baluncore

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    A travelling wave is sent in each direction along the string from the oscillator. That energy is reflected at the ends of the string and returns to repeat the transit several times. The sum of those two waves travelling in opposite directions is the standing wave.

    Standing waves only occur when the return transit time along the string is close to an integer multiple of the oscillator period. At non-integer oscillator periods, the reflected waves return slightly more out of phase on each pass. The travelling waves are not being efficiently reinforced and so suffer destructive interference.

    Energy from the oscillator enters the string and travels as a transverse wave in both directions. That energy gradually heats the string and stirs up the air. Oscillator energy flow will be determined by impedance matching of the oscillator to the string. Energy will flow into the string while part of that energy will later be reflected back from the string into the oscillator.

    This gets complicated because the impedance matching of the oscillator to the string is dependent also on where on the string the energy is being introduced. There is no point trying to excite a stationary node on the standing wave pattern as almost all that energy will be reflected back again into the oscillator. That is also true of a string driven by an oscillator with a non-integer multiple period, because the entire length of the string will then average out as a stationary node.
     
  8. Jan 16, 2017 #7

    Svein

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  9. Jan 17, 2017 #8
    Thanks everyone.

    When the string is fixed at one end and is connected to the oscillator at the other, we don't have a situation where there are two nodes at both ends since the oscillator moves up and down. So how can the oscillator truly form a standing wave if there is no node where it is connected to?

    Also, is it possible to excite two different modes simultaneously with a mechanical oscillator that oscillates up and down harmonically at a frequency that we can vary? For example, if the fundamental mode has frequency f=30Hz and the 2nd harmonic f=2*30=60Hz, how could we excite both mode with the oscillator?
     
  10. Jan 17, 2017 #9

    Baluncore

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    An oscillator cannot be used to drive a node because that is a stationary point on the standing wave and so has no amplitude. If the string does not have a node at both ends then behaviour of the system will be decided by the impedance matching of the oscillator to the string.

    If you excite the string with a harmonic waveform, both travelling waves will be that harmonic waveform, but the standing wave will be the sum of the odd harmonics of that waveform. That is because even harmonics will cancel when travelling in opposite directions.
     
  11. Jan 18, 2017 #10
    Ok, all clear.

    I read that when different instruments play the same note (same fundamental frequency) different harmonics become involved. That is called timbre. That makes each instrument distinctive from the other. Why are the excited higher harmonics for one instrument different from the excited higher harmonics of another instrument? What factors determine that? For instance, a violin and a guitar playing the same fundamental at 440Hz sound distinctively different. I assume the two strings are plucked the same way and in the same position. Why would their higher harmonic content be different?
     
  12. Jan 18, 2017 #11

    Baluncore

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    The construction of the instrument attenuates some harmonic frequencies more rapidly than others. Also, the strings are excited differently. A guitar is usually plucked, once per note like a piano. A violin string is usually excited by a continuous stroke of fibres with a dry resin coating on the bow. The friction of the resin results in a rapid repetition of a high frequency step excitation, which better maintains higher harmonics.
     
  13. Jan 19, 2017 #12
    Thanks Baluncore,

    Great answer that makes a lot of sense. You are strong on this topic.

    Another thing that is bothering me is the idea that a sound wave can be seen as a sequence of local pressure variations OR as a sequence of local displacement variations. Displacement means that the air molecules are moving away from their equilibrium position. That is clear. Pressure means the ordinary isotropic atmospheric pressure. A pressure variation seems to imply that more air molecule get pushed in the same space (hence increasing the density) or vice versa. A region of compression corresponds to an area of higher density.

    For a standing wave, a node of pressure is an antinode of displacement and vice versa. Why? At a displacement node, the air molecules don't move at all, either backward or forward. But the pressure at that point seems to vary the most from the largest to the smallest value. Is it because other air molecules get pushed towards that point, increasing the pressure and air density at the displacement node?

    For a traveling sound wave (longitudinal wave), are the pressure and displacement waves also out of phase the same as for a standing wave?
     
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