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Is this how i would calculate the energy stored in an object on a wave?

  1. Apr 15, 2009 #1
    So say you have an object of mass "m" on a water wave that happens to be a perfect sine wave. Just to reassure you this is NOT a homework question. So here's my idea so tell me if its correct. We know that Fg = m x g x delta"h". We know that the hight traveled by the object on the wave is going to be 2 x the amplitude of the wave. So can i say that the force felt by the object will be the mass of the object x g x (2 x the amplitude)? I mean not the force at a specific point because the object on the wave is not in free fall but the overall difference of potential energy and therefore energy should be the same as it would be if the object were in free fall but it is felt over a longer period of time. So with this said i thought well the maximum force is at the peak or troph of the wave and seeing as in a sine wave there is one peak and one troph per cycle the calculation for the amount of times the peak and troph is passed should be (0.5 x lambda)/v so therefore if the object was on the wave for 3600s, we could calculate how many times a peak or troph was passed. So to sum up my idea i say that if the mass of the object on the wave was 5kg, lambda was 10s, v was 2m/s, the amplitude of the wave was 1.5m, and the object was on the wave for 3600s, I would wright the equation as (5 x 9.81 x 2(1.5)) x 3600) divided by ((0.5 x 10)/2) and find that the total energy that the mass felt was 21186N. I see room for a few problems here but i wanted your opinion before i started from scratch again.
  2. jcsd
  3. Apr 15, 2009 #2
    I'm afraid this is very difficult to follow. You've never actually said what you're trying to find. Some things I've noticed though:

    -The force of gravity IS NOT mgh this is the (approximate) GRAVITATIONAL POTENTIAL ENERGY
    -N = newton's is a unit of force, not energy.
    -In a conservative system the work (i.e. energy) is independent of path so the question "The total energy the mass felt" is kinda a nebulous concept. If after a billion crests the mass is right back where it started (with zero velocity) then the net WORK was zero regardless of how many crests went by.
  4. Apr 15, 2009 #3
    well i know that... bust say you could store the work so it wasn't lost by the canceling built into the wave. i.e. regardless of the vertical motion of the mass be it up or down, the work is stored. By the way sorry im a high school student equipped with grade 11 and half of grade 12 physics so please bear with me :P
  5. Apr 15, 2009 #4
    You can't "store the work". That would violate the conservation of energy.
  6. Apr 15, 2009 #5
    what if i built a machine that could add both the up and down motion of a mass on a wave together? This isn't just hocus pocus and theories i have actually built it and it does exactly what i mentioned... even if the waves were just the ones in my bathtub lol but i have proved the concept and now i want to calculate the energy it can derive from a water wave (in theory seeing as im not an idiot and know that water waves are not like sine waves at all... usually compound waves)
  7. Apr 15, 2009 #6
    but its super cool :D im patenting it this week so i will give you a description of the machine when thats all done :)
  8. Apr 16, 2009 #7
    Green: I did not follow your post either...

    "So here's my idea so tell me if its correct. We know that Fg = m x g x delta"h". We know that the hight traveled by the object on the wave is going to be 2 x the amplitude of the wave..."
    Can you explain in words what you are thinking here....
    the "g"s drop out and F = m x delta h makes no sense to me...
    how can height be twice the amplitude of the wave? This is not true for a single sinusoidal cycle....did you mean distance up then back down? when returned to sea level all potential energy returns to zero........

    however, you may gain some insights and references here:

    In particular, note the section SCIENCE OF WAVES....

    I'd suggest you search online as waves have been modeled extensively for many years.

    I've boated all my life and I can verify one thing: the power in waves is quite unbelievable.
    Last edited: Apr 16, 2009
  9. Apr 16, 2009 #8
    This is an example of an asymmetrical force. The downward force is just mg (the weight of the object), and the potential energy is mgh. But if the object displaces a mass M of water, and the object's mass is m, then the upward buoyant force is about (M-m)g, and the upward potential energy is (M-m)gh. If you wanted to generate electricity, I would use the upward bouyant force to operate a ratchet of some sort.
  10. Apr 16, 2009 #9
    my machine uses both the upward force of the wave and the downward force of gravity after the wave and combines the two in one unidirectional motion. The idea is to get as much energy from the wave as possible.
    Last edited: Apr 16, 2009
  11. Apr 16, 2009 #10
    The rate of change of the wave goes from + to - or vise versa in the middle of the wave. The amplitude is the distance from the middle of the wave in both the up and down direction so the max displacement of an object on a sine wave is gonna be 2 x the amplitude i.e. it goes from zero to the peak which is the amplitude's hight from zero then back to zero which is the amplitude in the other direction then it goes back to zero so the total energy in that cycle is going to be related to the total height traveled by the object BUT lets say every 1/4 of the cycle represents an energy interval. The object goes up the amplitude, down the amplitude, down the amplitude, the up the amplitude so i can say the total vertical distance traveled is 2 x the amplitude twice in every cycle.... seems to make sense. The object is still in motion after reaching the zero point... the force stored in it's rate of change reduces but doesn't just hit zero...
  12. Apr 16, 2009 #11
    oh and i meant to say that the max potential energy was at the peak and trough
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