1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    sin(12πt)+cos(12πt-π/4)

    2. Relevant equations

    T=n2T2=T1n1

    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
    sin(12πt)+cos(13πt-π/4)


    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
    sin(12πt)+sin(13πt-3π/4)

    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
    cos(π/2-12πt)+cos(13πt-π/4)
    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
    √2/2[cos(13πt)-sin(13πt)]-[cos(12πt)]



    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,

    T=n1*1/6=n2*2/13

    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:
    http://www.wolframalpha.com/input/?i=factor+sin+x+++sin+y
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding the frequency of the combined motion
Loading...