write an equation of a sine function with amplitude 3, period 3pi/2 and phase shift pi/4

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write an equation of a sine function with amplitude 3, period 3pi/2 and phase shift pi/4

Mathematics
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is it y=3sin(3x/2-pi/4)?
3sin((4/3)t+(pi/4))
what about -3sin(4x/3-pi/4)?

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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...
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.
so my answer is wrong?
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)
those are my choices @AERONIK
you got me wrong friend, your choices are definitely correct....
wait i'm confused
\[\huge y=a \sin(bx-c)+d\]\[a=amplitude\]\[\frac{2\pi}{b}=period\]\[c=phase \; shift\]
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
\[\large y=3\sin\left(\frac{4}{3}x-\frac{\pi}{4}\right)\]
Mmmmm I'm not sure if I did that correctly, was I suppose to factor the 4/3 into the pi/4? I forget...
thats not an option :(
Hmm
wait yes i think so
Yah I think the b is suppose to be like this... \[\huge y=a \sin (b(x-c))+d\]
Which IS one of your options, if you distribute the 4/3 to the pi/4 term.
Hmm how did you arrive at your answer? :o
i think its b. y = 3 sin (4x/3 - π/3)
Hmmm I think so too :O

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