"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
so .....
d(t) = 15sin (5π/2)t
since you only had data affecting the amplitude and period I will go with the above.
testing:
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)
An object suspended from a spring is oscillating up and down. The distance from the high point to the low point is 30cm, and the object takes 4s to complete 5cycles. For the first few cycles, the distance feom the mean position, d(t) centimetres, with respect to the time, t seconds, is modelled by a sine function.
Find the amplitude, vertical shift, phase shift and k value and period. Do all calculations and then make your final equation.
Can u please help me with this pne cuz there are a lot more and i don' t understand any of them please help meeee!!!:(
3 answers
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 kØ
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.
y = a sin kØ
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.