A weight is oscillating on the end of a spring. The position of the weight relative to the point of equilibrium is given by y=1/12(cos8t-3sin8t), where y is the displacement (in meters) and t is the time (in seconds). Find the times when the weight is at the point of equilibrium (y=0) for 0<=t<=1.

Question

I have it worked out as:
0=1/12 (cos8t-3sin8t)
cos8t=3sin8t
tan8t=1/3
8t=arctan 1/3
t= 1/2 arctan 1/3

And now I'm stuck and I need three times. Can anyone please help?

Answer 1

you are ok up to
8t = arctan 1/3
make sure your calculator is set to radians, and then
8t = .32145 or 8t = π + .32145
t = .04022 or t = .43292
we know the period of tan 8t is 2π/8 or .7854
so we could add this to any previous answer to get more correct answers, as long as we don't go outside our domain of 0≤t≤1
so t = .04022, .43292, and .82562

testing the last time of .82562
I get y = appr 10^-9 , not bad

Answer 2

I redone it and I did:
0=1/12(cos8t-3sin8t)
cos8t=3sin8t
8=3tan8t
8/3=tan8t
arctan8/3=8t
1.212=8t
t=.1515

I think this one is correct. However, I'm still stuck on how I can get three times. Can someone please help?

Answer 3

your step from
cos8t=3sin8t
to
8 = 3tan8t is totally invalid, that would be pure mathemagics

test your value, it does not work

Answer 4

Alright, thank you Reiny. I understand it now!