anonymous
  • anonymous
write an equation of a sine function with amplitude 3, period 3pi/2 and phase shift pi/4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
is it y=3sin(3x/2-pi/4)?
anonymous
  • anonymous
3sin((4/3)t+(pi/4))
anonymous
  • anonymous
what about -3sin(4x/3-pi/4)?

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More answers

anonymous
  • anonymous
general equation of a simple harmonic oscillator is Asin(wt). where A is the amplitude and w is the angular frequency.. amd 2pi/w is the time period...
anonymous
  • anonymous
the equation you are giving should be a function of time and not distance... and by adding a minus sign you are just changing its phase by pi.
anonymous
  • anonymous
so my answer is wrong?
anonymous
  • anonymous
a. y = -3 sin (3x/2 - 3π/8) b. y = 3 sin (4x/3 - π/3) c. y = -3 sin (4x/3 - π/4) d. y = 3 sin (3x/2 - π/4)
anonymous
  • anonymous
those are my choices @AERONIK
anonymous
  • anonymous
you got me wrong friend, your choices are definitely correct....
anonymous
  • anonymous
wait i'm confused
zepdrix
  • zepdrix
\[\huge y=a \sin(bx-c)+d\]\[a=amplitude\]\[\frac{2\pi}{b}=period\]\[c=phase \; shift\]
zepdrix
  • zepdrix
Let's figure out our b term. \[\large \frac{2\pi}{b}=\frac{3\pi}{2}\]\[\large b=\frac{4}{3}\] I hope I calculated that correctly hehe
zepdrix
  • zepdrix
\[\large y=3\sin\left(\frac{4}{3}x-\frac{\pi}{4}\right)\]
zepdrix
  • zepdrix
Mmmmm I'm not sure if I did that correctly, was I suppose to factor the 4/3 into the pi/4? I forget...
anonymous
  • anonymous
thats not an option :(
zepdrix
  • zepdrix
Hmm
anonymous
  • anonymous
wait yes i think so
zepdrix
  • zepdrix
Yah I think the b is suppose to be like this... \[\huge y=a \sin (b(x-c))+d\]
zepdrix
  • zepdrix
Which IS one of your options, if you distribute the 4/3 to the pi/4 term.
zepdrix
  • zepdrix
Hmm how did you arrive at your answer? :o
anonymous
  • anonymous
i think its b. y = 3 sin (4x/3 - π/3)
zepdrix
  • zepdrix
Hmmm I think so too :O

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