SHM question (1 Viewer)

[dwarven]

x = 4 cos (3t+ pi)

what is the distance of the particle when the velocity is 6m/s

my working out:

-12 sin (3t+pi) = 6
sin (3t+pi) = -1/2
3t+pi = -sin^-1(1/2)

what do i take the RHS as??
like 7pi/6 and 11pi/6 or -pi/6

thnx =]

 

[vds700]

x = 4 cos (3t+ pi)

what is the distance of the particle when the velocity is 6m/s

my working out:

-12 sin (3t+pi) = 6
sin (3t+pi) = -1/2
3t+pi = -sin^-1(1/2)

what do i take the RHS as??
like 7pi/6 and 11pi/6 or -pi/6

thnx =]

i would use 7pi/6, 11pi/6 will work as well. But u cant use -pi/6, coz you'll get a negative value for t which is impossible.

so sub in t = pi/18 into x and see if u get the answer.

 

[samwell]

if x=4cos(3t +pi)
v=-12sin(3t +pi) v=6 t=?
-0.5=sin(3t+pi)
arcsin(-0.5)=3t+pi(remember time can only be >0 so consider the angle which will be greater than pi[consider 3rd quadrant])
(7pi/6-pi)/3=t
t=pi/18
x=4cos(7pi/6)=2sqr root 3