m a = m d^2x/dt^2 = -k x
.66 d^2x/dt^2 = -23.7 x
let x = A sin (wt-phi)
where w = 2 pi f
then v = dx/dt= A w cos(wt-phi)
and a = d^2x/dt^2 = -Aw^2 sin (wt-phi)
= - w^2 x
so
-m w^2 x = - k x
snd
w^2 = k/m
w = sqrt(k/m) = sqrt(23.76/.66)
= 6 radians/s
so in the end
x = A sin (6 t - phi)
v = 6 A cos(6 t- phi)
BUT the max v = 15 at t = 0
cos is max of 1 at 6t-phi = 0
so - phi = 0 so phi = 0
so
x = A sin (6 t)
v = 6 A cos (6 t)
when t = 0, cos (6t) = 1
so
15 = 6 A
A = 15/6
so now
x = (15/6) sin 6 t
and
v = 15 cos 6 t
OK? I think you can take it from there
like(1/2) k x^2 = (1/2) m v^2
A particle with a mass of 0.660 kg is attached to a horizontal spring with a force constant of 23.76 N/m. At the moment t = 0, the particle has its maximum speed of 15 m/s and is moving to the left. (Assume that the positive direction is to the right.)
(a) Determine the particle's equation of motion, specifying its position as a function of time. (Use the following as necessary: t.)
x = _______
(b) Where in the motion is the potential energy three times the kinetic energy?
± ______ m
(c) Find the minimum time interval required for the particle to move from x = 0 to x = 1.00 m. (Be sure to enter the minimum time and not the total time elapsed.)
_______ s
(d) Find the length of a simple pendulum with the same period.
________ m
1 answer