Question

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

Answer 1

is it y=3sin(3x/2-pi/4)?

Answer 2

3sin((4/3)t+(pi/4))

Answer 3

what about -3sin(4x/3-pi/4)?

Answer 4

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...

Answer 5

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.

Answer 6

so my answer is wrong?

Answer 7

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)

Answer 8

those are my choices @AERONIK

Answer 9

you got me wrong friend, your choices are definitely correct....

Answer 10

wait i'm confused

Answer 11

\[\huge y=a \sin(bx-c)+d\]\[a=amplitude\]\[\frac{2\pi}{b}=period\]\[c=phase \; shift\]

Answer 12

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

Answer 13

\[\large y=3\sin\left(\frac{4}{3}x-\frac{\pi}{4}\right)\]

Answer 14

Mmmmm I'm not sure if I did that correctly, was I suppose to factor the 4/3 into the pi/4? I forget...

Answer 15

thats not an option :(

Answer 16

Hmm

Answer 17

wait yes i think so

Answer 18

Yah I think the b is suppose to be like this...
\[\huge y=a \sin (b(x-c))+d\]

Answer 19

Which IS one of your options, if you distribute the 4/3 to the pi/4 term.

Answer 20

Hmm how did you arrive at your answer? :o

Answer 21

i think its b. y = 3 sin (4x/3 - π/3)

Answer 22

Hmmm I think so too :O