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Homework Help: Harmonics Question; prove that v max of a mass on a spring is given by 2 pi f A

  1. Feb 12, 2008 #1
    The question is exactly this:

    Prove that the maximum speed (Vmax) of a mass on a a spring is given by 2(pi)fA.

    where f = frequency and A =Amplitude

    We are given such formulas as:
    f=[1 / 2(pi)]square_root (k /m) or
    f=[1 / 2(pi)]square_root (a /-x)
    the formula for total energy in a system (1/2*m*vsquared + 1/2*k*xsquared)

    When I try to do it, I try to rearrange the formulas and get stuck in a giant maze of variables and such.

    This question was asked before on this forum as part of another question but was never answered
  2. jcsd
  3. Feb 12, 2008 #2
    Use energy equations!
    What does k/m equals to?
    Go back to the basics.
  4. Feb 13, 2008 #3
    that doesnt really help me. Maybe mentioning which ones I should use would help, because I start to use some but they dont seem to get me where I want. It's like math class all over again with that darn unit with rearranging cos(x), sin(x), and tan(x), i hated that too.
  5. Feb 13, 2008 #4
    nvm, i figured it out
  6. Jul 16, 2008 #5
    can anyone elaborate on how he figured it out???
  7. Jul 17, 2008 #6


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    For SHM

    [tex]v= \pm \omega \sqrt{A^2-x^2}[/tex]

    x is the displacement from the equilibrium position.At +A, you have max pe and 0 ke.
    At -A you have max pe and 0 ke. At the equilibrium position you have 0pe and max ke (which means?)
  8. Jul 17, 2008 #7
    im not really following because ive never seen that equation before; i understand the amplitude part where at equilibrium the mass has kinetic energy while before or after the equilibrium the mass will have potential energy.
  9. Jul 17, 2008 #8


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    The basic SHM eq'n: [itex]a=- \omega^2 x[/itex]

    [tex]a=\frac{dv}{dt}=v \frac{dv}{dx}[/tex]

    so that

    [tex]v \frac{dv}{dx}= - \omega^2 x[/tex]

    and solve that, remembering that when x=A,v=0 and you'll arrive at the equation.
  10. Jul 17, 2008 #9
    oh okay so i read up on SHM and circular motion on the internet (not in my text book) and it says that: Note that the in the SHM displacement equation is known as the angular frequency. It is related to the frequency (f) of the motion, and inversely related to the period (T):

    [tex] \omega = 2 \pi f[/tex]
    [tex]v= \pm \omega \sqrt{A^2-x^2}[/tex]
    [tex]v= 2 \pi f \sqrt{A^2-x^2}[/tex]

    and since velocity is at a maximum that means that displacement is 0 so x = 0 and then:

    [tex] v= 2 \pi f \sqrt{A^2-0}[/tex]
    [tex] v= 2 \pi f A^2 [/tex]

    right? :confused:
    I just dont really understand what [tex] \omega [/tex] stands for?
  11. Jul 17, 2008 #10


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    [itex]\omega[/itex] is the angular frequency i.e. the number of oscillations in [itex]2 \pi[/itex] seconds
  12. Jul 17, 2008 #11
    oh yea i forgot to take that off, but thank you for all your help!
  13. May 9, 2010 #12
    But if we want an answer without straying from the text (assuming you are using ILC's SPH4U funded by TVO), then we can go to Lesson 6, page 12.

    Right at the top, it gives the equation T=2(pi)r/v
    *If you know how to graph trig functions, or understand them even in the least, you know that the (r)adius is the same as the (A)mplitude

    Thus: T= 2(pi)A/v
    Isolating for v, we get: v=2(pi)A/T
    But since T=1/f;

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