# The displacement of a particle at t = 0.350 s is given by the expression x = (3.00 m) cos(6.00xt), where x is in meters and is in seconds (a) Determine the frequency and period of the motion Hz s (b) Determine the amplitude of the motion m (c) Determine the displacement of the particle at t = 0.350 s. m (d) At what time after t=0 does it first reach equillibrium? (e) At what time after t=0 does it reach zero velocity for the first time? (f)What total distance does it travel during the first period?

The displacement of a particle at t = 0.350 s is given by the expression x = (3.00 m) \cos(6.00xt), where x is in meters and is in seconds (a) Determine the frequency and period of the motion Hz s (b) Determine the amplitude of the motion m (c) Determine the displacement of the particle at t = 0.350 s. m (d) At what time after t=0 does it first reach equillibrium? (e) At what time after t=0 does it reach zero velocity for the first time? (f)What total distance does it travel during the first period?
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Kyle Delacruz
Expression of particle comping with the standard equation
$X=A\mathrm{cos}\left(2\pi ft+\varphi \right)$
we have
a) $2\pi ft=600\pi t$

Time period of motion $T=\frac{1}{f}$

b) Amplitude of motion A=3.00 m
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Gisselle Hodges
Ans (c) at next t=0.350 s
$x=\left(3.00\right)\mathrm{cos}\left(6.00\pi x\left(0.35+0.35\right)\right)$
x=2.427 m
x=2.43 m
(d) For the equilibrium
$x=0⇒\mathrm{cos}6\pi \left(0.35+t\right)=0$
$⇒6\pi \left(0.35+t\right)=\frac{5\pi }{2}$
$t=\frac{5×1}{12}-0.35$
t=0.0666 second