you can write it as x=-.98cos(12.2t+PI/2)
That is the same as x=-.98sin(12.2t)
This is using the trig identity given
cos(wt+PI/2)=-sinwt
a 2 kg frictionless block is attached to an ideal spring with force constant 300N/m. At t=0, the spring is neither stretched or compressed and the block is moving in the negative direction at 12 m/s. find (a) amplitude, (B)phase angle (c) write an equation for the position as a function of time.
I got a and b. For the amplitude, it is 0.98 m.
For the phase angle, I got 90 degrees or pi/2
C) I know the velocity equation is
v(t)=-wAsin(wt+phaseangle)
x(t)=Acos(wt +phaseangle)
My answer was x(t)=0.98cos(12.23t+pi/2)
The way I got the angular acceleration was using w=sqt(k/m)
HOWEVER in the book, the answer was
cos (ωt+(π/2))=−sin ωt, so x=(−0.98 m) sin((12.2 rad/s)t).
WHY IS IT SINE AND NOT COSINE?
1 answer
1 answer