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Homework Help: Brief explantation as to why x(max)= v(max)/ang freq?

  1. Nov 16, 2015 #1

    I don't need help with my homework problem as I figured out the answers. But I'm still really confused on why it happened to work. I might be over thinking something that is soo simple but I've read over my chapter a few time on Simple Harmonic Motion and can't seem to find an explanation. So I come to you all to seek knowledge.
    As the title states, why is it that the max displacement is equal to the max velocity over the angular frequency?

    Thank you all kindly for any and all help,

  2. jcsd
  3. Nov 16, 2015 #2


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    Hello Shawn, :welcome:

    You should really quote some relevant equations from your chapter, because now I have no idea if I can use the derivative of a sine to show you, or if I have to resort to more elementary ways to demonstrate this...
    Can you follow the wiki lemma ?
  4. Nov 17, 2015 #3
    Ok so I know that x(t)=xmaxcos(ωt+Φ), that v(t)=-ωxmaxsin(ωt+Φ), and that ω=2π/T or 2πf or √(k/m). *(Note: xmax is also known as A, for amplitude.)
    My specific homework problem ask me to find the angular frequency and the max displacement give the mass of particle, period of oscillation, and the velocitymax.
    Ill just go ahead and post those values even though I don't need the answer, I just want to understand why xdisplacement=velocitymax/ω.
    Mass= 4.6E-20

    I worked out ω=(2π/3.7E-5) ⇒ 1.7E5 rad/s
    xmax=(4.4E3/1.7E5) ⇒ 0.026m.
    I'm just having trouble understanding why that work (for xmax).
    It might be something simple I missed during my lecture or something that is from a previous chapter I might have overlook or missed, but I can't seem to find an explination in my current chapter.

    Hope this helps you understand my question to better explain to me.

    Thank you,

  5. Nov 17, 2015 #4


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    Sine and cosine have maximum absolute value 1, so all that's left is a factor ##\omega##. There isn't much more to explain about it.

    Unless you count: v has to do with the rate of change of x: ##\ v = {\Delta x \over \Delta t}\ ## (in the limit ##\ \Delta t\downarrow 0##), so the faster x oscillates, the bigger v .

    Another thing you can do is make a few drawings of sine waves: at x = 0 the rate of change in x (a.k.a. v) is maximum. Convince yourself that for a double amplitude in x, v doubles, and for the same amplitude but a double frequency, v doubles too.

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