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I am performing an experiment where a steel ball of diameter 7.5 mm is dropped by hand on an aluminised kapton foil. Four piezoelectric sensors are placed 55 mm away from the edges at a 45 degree angle and they measure the change in strain of the material from the drop and convert it to electrical signals, which can be measured using a picoscope. The set-up can be seen in the image above.

I do not understand if I am measuring the longitudinal waves, transverse waves or a mix of both here. How do I find which wave I am measuring?

My thoughts were that since the drop is perpendicular to the foil, I am measuring the transverse waves, but I am not sure. Any help in understanding would be appreciated.

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  • $\begingroup$ This could be also surface waves or even bending waves with rather poorly defined notion of polarization, since you are not really dealing with a bulk material. $\endgroup$
    – Roger Vadim
    Jul 21, 2021 at 15:45

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Both types of waves will be produced in the foil. You may want to try your transducers in a situation where you know that the stresses are only longitudinal.

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  • $\begingroup$ Why should I be measuring only the longitudinal waves ? I was comparing the drops on foil to ripples in water and thought they would be producing transverse waves and this is what I should be measuring. $\endgroup$
    – lqope54
    Jul 22, 2021 at 8:42
  • $\begingroup$ I was suggesting that you determine the extent that your transponders would respond to longitudinal waves. $\endgroup$
    – R.W. Bird
    Jul 22, 2021 at 14:05
  • $\begingroup$ Any wave propagates out from the ball at the speed of sound through the membrane. Transverse ripples will only occur if the frequency of vibration is so high that its wavelength along the membrane is shorter than the membrane size. Here, the oscillations of the ball are far too slow and the membrane approximates a minimal surface bounded by its edge and the ball. $\endgroup$
    – Guy Inchbald
    Jul 22, 2021 at 14:10
  • $\begingroup$ So because of the low impact energy, we are most probably measuring the longitudinal waves? For clarification, by membrane you mean the foil surface or the sensor surface ? $\endgroup$
    – lqope54
    Jul 26, 2021 at 11:22
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Strain is longitudinal within the foil. You are measuring that component of the oscillating system which comprises longitudinal oscillation of the foil.

Geometry correlates this with the vertical oscillations of the foil-and-ball-and-gauges, but that correlation is not linear.

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  • $\begingroup$ So even though the foil oscillates vertically due to the drop, it is a longitudinal oscillation because it is a component of it ? $\endgroup$
    – lqope54
    Jul 22, 2021 at 8:29
  • $\begingroup$ The energy of motion alternates between the vertical deflection and the nominally horizontal strain of the membrane. These are the two modes of energy storage, components if you will, of the motion. $\endgroup$
    – Guy Inchbald
    Jul 22, 2021 at 10:30
  • $\begingroup$ I still did not get it. What I understand from your explanation is that the horizontal strain response is going to be lesser than the longitudinal ? could you direct me towards some literature which talks about the non-linear relationship that you mentioned in your answer? $\endgroup$
    – lqope54
    Jul 22, 2021 at 12:48
  • $\begingroup$ The nonlinearity is in the elementary geometry of right-angle triangles; you can approach it either through Pythagoras or trigonometry. This will give you the exact relationship between membrane strain and vertical deflection. For example when the ball is halfway between its top and bottom positions, the membrane is stretched less than half its maximum. Similarly, the strain (distance stretched) will always be less than the vertical deflection. $\endgroup$
    – Guy Inchbald
    Jul 22, 2021 at 14:05
  • $\begingroup$ Actually, right angled triangles are an approximation, but it will be valid for all but the grossest oscillations or if high accuracy is required. The membrane forms a curved minimal surface (minimising the energy of stretching) and requires sophisticated maths to model exactly. $\endgroup$
    – Guy Inchbald
    Jul 22, 2021 at 14:13

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