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Homework Help: Finding the frequency of the combined motion

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Finding the frequency of the combined motion of the following


    2. Relevant equations


    3. The attempt at a solution

    Ok so I found the periods be 1/6 for the sin function and 2/13 for the cosine
    Any hints on what to do next? I really have no clue

    the answer in 6.25/s
  2. jcsd
  3. Sep 16, 2012 #2
    Think about fundamental trig identities.
  4. Sep 16, 2012 #3
    which fundamental ones? like this one sin(θ)^2 + cos(θ)^2 = 1?
    Im not quite sure I follow you on how this could help

    Im getting my T from T=n2T2=T1n1
    to be 2s so the f is .5/s, which is wrong

    also the question is actually

    Does it have to do withe the π/4 in the cos argument?
  5. Sep 16, 2012 #4
    I was thinking of [tex] \sin(x)=\cos(\frac{\pi}{2}-x) [tex], so you can convert the cosine or sine term this way, so that you have two sine or two cosine functions. Then look to use the equation you mention. I think this is should give the right result
  6. Sep 16, 2012 #5
    If I do that I get

    But this doesn't change the angular velocity values, so it doesn't change the the period or frequency... or am I totally missing something here?
  7. Sep 16, 2012 #6
    So now I'd say try this identity, [ itex ] \cos(\phi-\th)=\cos(\phi)\cos(\th)-\sin(\phi)sin(\th) [ /itex ]. Which has its sin counterpart, [ itex ] \sin(\phi-\th)=\sin(\phi)\cos(\th)+\cos(\phi)sin(\th) [ /itex ]. I think now you'll be able to write this as one sine or cosine function with a combined frequency
  8. Sep 16, 2012 #7
    ok so

    so I changed it too
    and used that identity to get

    [cos(13πt)cos(π/4)-sin(13πt)sin(π/4)] -[cos(π/2)cos(12πt)-sin(π/2)sin(12πt)]
    which simplifies to

    i really appropriate your help, but Im just not getting it, Ill have to go into see my prof I guess, thanks for sticking with me this far
  9. Sep 16, 2012 #8
    Or simply,


    where n1 and n2 are the lowest common value multiple of the individual periods.

    Sorry for the long probably round about explanation.
  10. Sep 16, 2012 #9

    I did that part at the first and got n1=12 ans n2=13.
    this gave a T of 2sec and a f of .5/sec

    which totally makes sense to me but the back of my book says the answer is 6.25/sec
  11. Sep 16, 2012 #10
    You can use this identity:
    to get a product of two sin terms.
    It will result a typical expression encountered when you study beats between two close frequencies.
    The motion has a fast, 6.25 Hz component, modulated by a slow varying "amplitude" (0.25 Hz).
  12. Sep 17, 2012 #11
    This link shows how to derive the identity, which gives the result 6.25Hz and .25Hz

    http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/sumprod.htm [Broken]
    Last edited by a moderator: May 6, 2017
  13. Sep 17, 2012 #12
    than ks to both of you! I get it now and was able to use that steps to solve other problems
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