Must the amplitude of a driving force increase so as to match the increase in the amplitude of the standing wave it generates? If not, how can standing waves of increasing amplitude form if the amplitude of the waves generated by the source does not match the amplitude of the waves prevailing in the system?

  • 3
    $\begingroup$ You never pushed a swing? Do you have to move your hand over the amplitude of the swing's motion? $\endgroup$ – nasu Feb 23 '17 at 20:04
  • $\begingroup$ I see your point for the case of a single particle oscillator. But what about for more complex systems in which the driving force acts continuously? For example, someone shaking a rope or forcing air into a pipe? $\endgroup$ – Vinny Feb 23 '17 at 20:16
  • 1
    $\begingroup$ @Vinny same thing. $\endgroup$ – DanielSank Feb 23 '17 at 23:09

Your 2nd sentence and comment suggest that you are thinking about a standing wave on a string, with the driving force being provided by a mechanical oscillator at one end, like a hand shaking a rope. You are perhaps confusing the amplitude of the oscillator (measured in metres) with the amplitude of the force (measured in Newtons).

In this case there is an unnecessary complication. The oscillator is performing the multiple functions of (i) keeping the string taut, (ii) providing a restraint from which the wave is reflected, while also (iii) doing work on the string so as to input energy. As the amplitude of the waves increase, and the accelerations of points on the string increase, then the reaction forces on the restraints at both ends of the string will also increase - but not in proportion to the amplitude.

We need to consider a simpler case, in which the oscillator is not doing several jobs at the same time, only providing a periodic push.

The driving force could be provided without any mechanical contact. For example, parallel electrified plates above and below the string providing a force on a small, insulated, charged section of the string. Then the force on the charge does not depend on the position, velocity or acceleration of the charge, only on the uniform electric field between the plates. This field with constant amplitude would only need to oscillate at the natural frequency of the string in order to create or increase the amplitude of standing waves.

As with all harmonic oscillators which are driven at resonance (ie at the natural frequency of the system), the amplitude of oscillations continues to increase although the amplitude of the driving force is constant, until damping removes energy from the system as fast as the driving force is putting it in.

See Standing waves due to two counter-propagating travelling waves of different amplitude and also How are standing waves a result of constructive and destructive interferences? which includes a demo in Brionius' answer.

  • $\begingroup$ Thank you very much for the clear response. I have a follow up question regarding resonating systems experiencing driving forces at one end (i.e. shaken ropes, kundt's tubes, etc.). As wave amplitude increases, do these systems exhibit traveling waves or standing waves? $\endgroup$ – Vinny Feb 25 '17 at 14:53
  • $\begingroup$ A standing wave is the superposition of two waves travelling in opposite directions. The driving force creates waves which travel along the rope or tube to the right, say, are reflected from the end, travel to the left, and interfere with other waves travelling to the right. $\endgroup$ – sammy gerbil Feb 25 '17 at 17:27
  • $\begingroup$ Understood. But, mustn't interfering waves that form standing waves possess the same amplitude? If so, how would standing waves appear when a driving force continuously generates larger amplitude waves (even though the amplitude of the driving force need not increase proportionally, as you clearly stated above) that encounter smaller amplitude waves? $\endgroup$ – Vinny Feb 25 '17 at 17:41
  • $\begingroup$ Thank you for the links. It appears that, under the conditions I describe, differing amounts of mixing of traveling/standing waves occur. $\endgroup$ – Vinny Feb 25 '17 at 18:39

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.