Question

The position of a SHO as a function of time is given by x=4.0cos( 5π/4 t + π/6) where t is in seconds and x in meters.

(A) Find the velocity at = 2.1

(B) Find the acceleration at = 2.1

Answer 1

x(t)=4*cos(wt +pi/6 ), where w= 5*pi/4
so,
velocity = v(t) = dx(t)/dt
or,
v(t) = -4*w*sin(wt+pi/6)
and a(t) = acceleration= dv(t)/dt
or,
a(t) = -4*w*w*cos(w*t+pi/6)
at t= 2.1,
v(t=2.1s)= -9.562 m/s
and
a(t=2.1) = 48.938 m/s^2

Answer 2

x(t)=4*cos(wt +pi/6 ), where w= 5*pi/4
so,
velocity = v(t) = dx(t)/dt

velocity= d/dt(4*cos(wt +pi/6 ))

d/dt(wt+pi/6) =d/dt(wt) =wsince pi/6 is a constant d/dt (pi/6)=0

d/dt (cosθ)=-sinθ
or,
v(t) = -4*w*sin(wt+pi/6)
and a(t) = acceleration= dv(t)/dt
or,
a(t) = -4*w*w*cos(w*t+pi/6)
at t= 2.1, w=5*pi/4v(t) = -4*5(pi/4)*sin(5pi/4 *2.1+pi/6)
v(t=2.1s)= -9.562 m/s
and
a(t=2.1) = 48.938 m/s^2