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So I'm doing an experiment to verify the contact duration when a steel rod (Material: Mild Steel) is allowed to free fall longitudinally on a steel plate. The arrangement is real simple:

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The rod is allowed to fall longitudinally on the plate, resulting in a number of impacts. I will only be considering the first impact.

Now here's the problem I'm facing: During the first contact and off-contact, the oscilloscope reads a much higher than expected value for the contact time. Same procedure is repeated for three lengths of the rod (200mm, 350mm and 475mm), all resulting in the same anomaly. Here are the results:

enter image description here

As clear from the plot, the experimental values are similar to the theoretical values except for an offset (found to be about 121 µs). Any idea why this is happening?

P.S: The theory used is that of elastic waves produced on spherical contact where the impact results in the creation of a compressive wave that travels to the end of the rod and returns as a tensile wave which pulls the rod off contact. As such the contact time t= 2l/(wave speed). Wave speed = 5000 m/s.

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  • $\begingroup$ You dead certain there's no elastic rebound in the steel plate? $\endgroup$
    – Carl Witthoft
    Commented May 26, 2017 at 13:30
  • $\begingroup$ Although the plate is not clamped, the base is flat and doesn't vibrate on impact. It is also large enough and I believe there's no rebound. $\endgroup$
    – Aswin James
    Commented May 29, 2017 at 14:23
  • $\begingroup$ Could you test the followings : 1. change the rod and plate material. If the behavior changes, maybe you have come capacitive phenomenons happening. 2. change the room humidity. If the behavior changes, maybe you have electrical conduction through air just after the impact. 3. Check the response time of your oscilloscope. $\endgroup$
    – Aurélien Pierre
    Commented Dec 22, 2017 at 15:02
  • $\begingroup$ How distant is the edge of the sheet from the point of impact? How long would the wave traveling there and back take? $\endgroup$
    – SF.
    Commented Sep 18, 2018 at 10:25
  • $\begingroup$ I've got a couple of questions (admittedly naive) regarding the setup: - what is the mass of the steel rod compared to the steel plate? (are the comparable) - how is the plate fixed? - Did you observe the plate to bounce up after contact? All this could have an effect in prolonging the contact duration. $\endgroup$
    – NMech
    Commented Aug 5, 2020 at 18:18

2 Answers 2

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Try the experiment in a darkened room and set it up so you can see the actual contact point in profile. If the value of the sense resistor is too low, it will pass too much current and you'll create a spark upon rebound which means the setup is conducting for a certain time AFTER the rod and plate have bounced out of contact. This will make it appear like they are in contact for longer than they actually are.

I have used this same arrangement to trigger an oscilloscope upon impact but I used a much lower source voltage (1.5 volts instead of 9) and I recommend a sense resistor bigger than 1K ohm. Both of these things will minimize arcing. Turn up the vertical gain on the y-axis of the scope to catch the smaller signal, and adjust the trigger threshold accordingly.

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Above my electrical skills, but as the rod bounces up and opens the contact, is the resistor discharging through the oscilloscope giving a false time? Would a capacitor help this?

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  • $\begingroup$ If that's the case, shouldn't the oscilloscope peak reading decay over some time? I'm getting a near-perfect pulse signal. $\endgroup$
    – Aswin James
    Commented May 25, 2017 at 17:07
  • $\begingroup$ I was just thinking of the points as used in older car ignition systems where the capacitor (condensor) was used to stop arcing... The arc would continue until the air gap broke it - which would not be the same as a decay... When you say "near perfect pulse" you mean a square wave : straight up , across, then straight down? $\endgroup$
    – Solar Mike
    Commented May 25, 2017 at 17:12
  • $\begingroup$ Yeah it is a square wave $\endgroup$
    – Aswin James
    Commented May 25, 2017 at 17:19

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