Simple Harmonic Motion-should be easy

1. Dec 6, 2009

mattmannmf

A particle's position in centimeters is given by the expression x = 40 cos(45 t + 8), where t is given in seconds.

angular frequency= 45 rad/sec
period= .139 s
amplitude= 40 cm

Heres the part i cant get but seems so simple...

where is the particle at t= .05 sec?

what is its velocity?

what is its accleration?

what is the maximum speed attained by the particle?

So what i figured on doing is just plugging .05 into my original equation x = 40 cos(45 t + 8) to solve for distance....where t= .05, distance= 39.36....which came out to be wrong for some reason...(all is in cm)
then for the rest i just add a "-w" infront of the equation for velocity
and a "-w^2" for acceleration and then solve when a=0, velocity has a max or min..

any ideas?

2. Dec 6, 2009

denverdoc

First issue, the argument of the cosine is in radians--make sure to set your calculator for radians. I got something like -27.

See if that helps with the remaining questions which are modified by -w and w^2 as you note.