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How many people needed to excite the lift cable to the 15th harmonic?

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Homework Statement



A lift cable has mass per unit length 4kg/m. The lift has mass of 920kg (assume that the tension in the cable is provided solely by the weight it supports). When the lift stops at the 3rd floor, the distance from the top of the lift to the top of the cable is 4.6m. Assuming that people weigh (on average) 80kg and act as vibrational sources at 100Hz (from the motion of their feet), how many people would need to enter the lift to excite the 15th harmonic in the lift cable (to the nearest person)?

Homework Equations



v=√T/μ

fn = (n/2L)(√T/μ)

The Attempt at a Solution



T= (920+80N)×9.8 where N=number of people inside the lift

fn =(n/19.2)(√(9016+784N))
 

Answers and Replies

  • #2
NascentOxygen
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Hi A9876, welcome to Physics Forums.

I'd interpret this to mean that 100Hz equates to the 15th harmonic of the cable's natural frequency.
 
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One person is all you need, if that person has some instrumentation feedback and an understanding of what he is doing.

But this is a very simplified mode of an elevator. In real life he would have much damping in the system so that excitation would be small enough to require the instrumentation to know what was going on.
 
  • #4
gneill
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One person is all you need, if that person has some instrumentation feedback and an understanding of what he is doing.

But this is a very simplified mode of an elevator. In real life he would have much damping in the system so that excitation would be small enough to require the instrumentation to know what was going on.
If the person is restricted to applying a continuous 100Hz stimulus, I don't see what advantage instrumentation will give him.
 
  • #5
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That is not the natural frequency of the human body. So if he is to maintain 100 Hz, then he needs some sort of feedback to know he is doing that. If the amplitude of motion of the lift was great enough, then that would provide sufficient feed back. But the 15th harmomic motion will be too small for him to feel. In a real elevator, it will be much greater than 100 Hz, too.
 
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  • #6
gneill
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That is not the natural frequency of the human body. So if he is to maintain 100 Hz, then he needs some sort of feedback to know he is doing that. If the amplitude of motion of the lift was great enough, then that would provide sufficient feed back. But the 15th harmomic motion will be too small for him to feel. In a real elevator, it will be much greater than 100 Hz, too.
The problem is posed in the nature of a physics thought experiment (gedanken); all you can do is take the problem statement at face value: "Assuming that people weigh (on average) 80kg and act as vibrational sources at 100Hz (from the motion of their feet)..."

So, 100 Hz it is, no matter what technology or contrivances you might need to imagine in order to have it work out that way so long as they don't interfere with the underlying physics of the system as presented.
 
  • #7
NascentOxygen
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One person is all you need, if that person has some instrumentation feedback and an understanding of what he is doing.
For one person to fit the criterion he or she would be so morbidly obese that they may be unable to deliver the necessary "tap dance" excitation. http://img854.imageshack.us/img854/9793/tazm.gif [Broken]
 
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  • #8
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Hi A9876, welcome to Physics Forums.

I'd interpret this to mean that 100Hz equates to the 15th harmonic of the cable's natural frequency.
Thankyou but the question seems to imply that each person acts as a vibrational source at 100Hz so wouldn't the frequency change with the number of people in the lift?
 
  • #9
gneill
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Thankyou but the question seems to imply that each person acts as a vibrational source at 100Hz so wouldn't the frequency change with the number of people in the lift?
100 Hz is fixed. What other frequency is important here?
 
  • #10
NascentOxygen
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Thankyou but the question seems to imply that each person acts as a vibrational source at 100Hz so wouldn't the frequency change with the number of people in the lift?
No. They all shuffle in and out, and each set of feet generates consistent vibrations. People don't usually shuffle 5 times faster if the lift contains 5 people!
 
  • #11
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I'm sorry, but I don't seem to understand the physics under discussion here. It does not seem to match my real world experience, in which I've excited many structures to find many modes, but usually not beyond the tenth. By then the natural frequencies are so high that nobody cares what they are, beyond the 100 khz range.

In a wire rope supported system, a single person can with little effort excite the first one or two modes, but after that mechanical means are required because the frequencies get too high. But the higher they get, the smaller the size of the mechanical shakers that is required. If a customer wanted me to test to the 15th mode, the shakers would weigh about two pounds, less than one kg.

If someone could help me expand my understanding of physics so as to explain then behavior of structures in the real world, the I would appreciate that.
 
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  • #12
gneill
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I'm sorry, but I don't seem to understand the physics under discussion here. It does not seem to match my real world experience, in which I've excited many structures to find many modes, but usually not beyond the tenth. By then the natural frequencies are so high that nobody cares what they are, beyond the 100 khz range.

In a wire rope supported system, a single person can with little effort excite the first one or two modes, but after that mechanical means are required because the frequencies get too high. But the higher they get, the smaller the size of the mechanical shakers that is required. If a customer wanted me to test to the 15th mode, the shakers would weigh about two pounds, less than one kg.

If someone could help me expand my understanding of physics so as to explain then behavior of structures in the real world, the I would appreciate that.
The THEORETICAL physics under discussion here: Standing wave harmonics of an ideal cable under tension.

In particular, cable tension is provided by a weight load which causes the 15th harmonic of the natural frequency of the cable to be 100Hz. The object is to determine what that weight load must be, and hence the number of 60kg units of mass that should be added to the initial 920kg load.

There's no 'real life' practicalities to consider here; it's strictly a theoretical Gedanken with ideal components.
 
  • #13
NascentOxygen
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It's barely worth discussing whether "15th" may be a typing mistake. But on noticing OP's error in typing 19.2 when meaning to type 9.2, it has crossed my mind ... https://www.physicsforums.com/images/icons/icon11.gif [Broken]
 
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  • #14
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It's barely worth discussing whether "15th" may be a typing mistake. But on noticing OP's error in typing 19.2 when meaning to type 9.2, it has crossed my mind ... https://www.physicsforums.com/images/icons/icon11.gif [Broken]
19.2 wasn't an error, I just simplifed the equation

fn =(n/2L)(√T/μ) = (n/(4.6×2))(√(920+80N)g/4) = (n/9.2)(0.5√(9016+784N)) = (n/(19.2))(√(9016+784N))

I hope that clears everything up
 
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  • #15
NascentOxygen
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I hope that clears everything up
Have you a final answer for the number of persons?
 
  • #16
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Have you a final answer for the number of persons?
I'm not sure if I should round to 9 or 10 people. I did the following calculation:

f15 = 100Hz = (15/19.2)(√(9016+784N)) => N=(1282-9016)/784 = 9.39
 

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