I got the following answers to the problem set. Let me know if you can't get them.

1.a. (3.91 x 10^-2 m/s)t

b. v(t)=-0.25 m/s sin(1646 rad/s)t

a(t)=-(406 m/s^2) cos (1646 rad/s)t

c. x(t)=-1.13 x 10^-5 m

2.a. k=19.6 N/m

w(omega)=8.08 rad/s

b. A=0.100 m

c. v(max)=0.81 m/s

d. a=6.53 m/s^2

e. T=0.78 s, f = 1.29 Hz

f. x(t)=-(0.1 m) cos (8.08 rad/s)t

g. dx/dt=0.757 m/s

3.a. 206.3 deg or 3.6 rad

b. A=0.11 m

c. x(t)=(0.11 m) cos (8.08 rad/s t + 3.6 rad)

If you respond to this blog by Tuesday at 2:45 pm, I'll add a point to your picture extra cedit. Just make sure it says your name and time you responded.

Subscribe to:
Post Comments (Atom)

This comment has been removed by the author.

ReplyDeleteZim,

ReplyDeleteI don't seem to be getting the correct answer for 2g. I am using the equation v(t)=-Aw(omega)sin(wt+(phase angle)). Which leads to v(t)=-(.1m*(8.08r/s))sin(8.08r/s*.15s). giving me an answer of 0.0171m/s.

Adam Bahrainwala