The position of a mass on a spring (relative to equilibrium) at time t <= 0 is
x(t) = 2 cos((pi)t) where x is in centimeters and t is in seconds. Answer the ques-
tions. Include units!
(a) What is the initial position of the mass?
(b) What is the initial velocity of the mass?
(c) What is the initial acceleration of the mass?
(d) Does the mass initially move towards the wall or away from it?
(e) At what point does the mass rst turn around?

- anonymous

- chestercat

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- goformit100

post the question in physics section Not here @GrizzlyChicken

- anonymous

its in my calc book, isn't this math?

- anonymous

a) put t = 0 and solve for x
b) find x' and put t = 0 and solve for x'
c) find x'' and put y = 0 and solve for x''

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

it's both physics and math lol

- anonymous

*typo fix* sorry
c) find x'' and put t=0

- anonymous

wait, its x(t)=2cos((pi)(t))

- anonymous

sorry

- anonymous

still~ same way of solving :P

- anonymous

damn, theres another error, i'm gonna look it all over again, for some reason copy and pasting changed everything

- anonymous

at time t is greater or equal to 0. its right now

- anonymous

so if i put 0 for t, its 2cos(0)

- anonymous

yup, that's the initial position

- anonymous

is cos(0)=1?

- anonymous

yes

- anonymous

then for b) derivative of 2 is 0 so its all 0

- anonymous

maybe

- anonymous

no no, find the derivative of the function x

- anonymous

so the derivative of 2cos(pi*0)?

- anonymous

which uses the product rule (0)(cos...)+(2)(-sin(0))

- anonymous

and sin(0)=0

- anonymous

so the velocity is 0

- anonymous

wops sorry i'm here
x = 2 cos (pi t)
x' = -2 pi sin (pi t)

- anonymous

do you know how to derivative?

- anonymous

yea, did you use the chain rule?

- anonymous

is that why pi is with -2?

- anonymous

no chain rule.
derivative of cos is -sin. pi 's brought out when derivative.

- anonymous

|dw:1349943924669:dw|

- anonymous

derivative again for the acceleration equation

- anonymous

-2cos(pi*t)

- anonymous

wait

- anonymous

-2pi*cos(pi*t)

- anonymous

almost correct, but now pi is brought one more time so it's
-2 pi^2 cos (pi t)

- anonymous

ohhh

- anonymous

x = 2 cos (pi t) x0 = 2
x' = v = -2 pi sin (pi t) v0 = 0
x'' = a = -2 pi^2 cos (pi t) a0 = -2pi^2
and idk how to do d and e

- anonymous

does the mass move toward the wall because the acceleration is negative?

- anonymous

oh, thats alright. you've helped me enough.Thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.