Help please http://prntscr.com/1226g3
i have solve a and b i need help with part c
for "c" , we are looking for "t" just like in (b) but negative
so \[h = -1=7\cos\left({\pi\over3}t\right)\\ t={3\over\pi}\cos^{-1}(-1/7)=? \]
93.7
no. I got 1.64s
-1=7cos(3t/pi) t=1.79 secs
well + or - 1.79 but time has to be positive
for your case
and it says for the "second time". So, to that, you add the comeplete period
now, the period of oscillation for this is : \[T={3\over2}=1.5s\]
-3=7cos(3t/pi) t=2.1087
iam getting more confused for -1? -3? -5?
1) h=-1 \(t=1.5+\displaystyle{3\over\pi}\cos^{-1}(-1/7)=?\)
-5=7cos(3t/pi) 2.478
iam getting 95
change your calculator to radians
3.13
3.42 for -3
3.75 for -5
are they correct
\[ h=-1\qquad t=1.5+\displaystyle{3\over\pi}\cos^{-1}(-1/7)=3.13s\\ h=-3\qquad t=1.5+\displaystyle{3\over\pi}\cos^{-1}(-3/7)=3.42s\\ h=-5\qquad t=1.5+\displaystyle{3\over\pi}\cos^{-1}(-5/7)=3.76s \] so, yes. they are correct
also, remember that these times can be obtained when the weight is coming from extreme to rest position and from rest to extreme. These times are from mean to extreme
the values you got for part (b) are from extreme to mean. to get to 1cm, it takes more time than to go to 5cm from rest position!!!
dont use any of my answers i solve for the equation 3/pi x not pi/3 x
part b's solution is wrrong
what are your answers for part b?
no. it is correct. I am giving you the interpretation. Look at the time values.
is it 1.36 for the first one
it tell you that you reach 5cm mark before you reach the 1cm mark. i.e., you are coming from the extreme point to the rest point
1.36,1.08,0.74 ish?
ok thanks can you confirm my other question
I do not see a series there in the picture
wait
all correct here
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