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Finding the frequency of the combined motion

  • Thread starter RockenNS42
  • Start date
  • #1
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Homework Statement



Finding the frequency of the combined motion of the following

sin(12πt)+cos(12πt-π/4)

Homework Equations



T=n2T2=T1n1

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
 

Answers and Replies

  • #2
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Think about fundamental trig identities.
 
  • #3
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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?
 
  • #4
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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
 
  • #5
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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?
 
  • #6
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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
 
  • #7
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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
 
  • #8
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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.
 
  • #9
52
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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.


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
 
  • #10
3,740
417
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).
 
  • #11
117
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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:
  • #12
52
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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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