"distance from the high point to the low point is 30cm"
----> your a value is 15
so far: d(t) = 15 sin ( ???? t)
5 cycles takes 4 seconds
so 1 cycle takes 4/5 seconds
then 2π/k = 4/5
4k = 10π
k = 5π/2
so far: d(t) = 15 sin (5π/2)( t ± phase shift)
since you didn't say where you want the object to be at t-0 , I will assume at the mean positon of zero
d(t) = 15sin (5π/2)t
since you only had data affecting the amplitude and period I will go with the above.
since the period is .8 seconds we should have a value of -15 at the 3/4 mark
3/4 (.8) = .6
set your calculator to radians and evaluated
15 sin (5π/2)(.6)
( I get -15)
Im sorry I do not understand. How did you get 15 and the period .8 seconds how did you get the 5pie/2please explainI am so confused!!!:(
I assumed you were familiar with the basic general sine function
y = a sin kŲ
where a is the amplitude, and k is such that
2π/k is the period of the curve.
To be able to do the type of question above, you just MUST know that terminology.
Check with your textbook or your notes.
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