# Homework Help: Kinematics: simple harmonic motion

1. Aug 18, 2010

### mizzy

1. The problem statement, all variables and given/known data
The motion of a particle is given by x=(6.0m) cos(0.586t +0.72)
a) find amplitude
b) find period
c) find the first time for t>0 when v=0
d) find the max acceleration
e) find the phase at time t = 1.38s

2. Relevant equations
T = 2pie/omega

v = -A omega sin omega*t

a = -A omega^2 cos omega*t

3. The attempt at a solution
I got the first 2.
amplitude = 6.0m

period = 2pie/omega = 10.7s

i'm not sure about the next 3:
for c) do i use the second equation above and make v = 0 and just solve for t?

for d) do i use the third equation? but what do i use for t?

e) Phase is looking for the angle, right?

THANKS!

2. Aug 18, 2010

### rl.bhat

c)To find time when v = 0, find dx/dt and equate it to zero.

d) max. acceleration = aω

e) find x when t = 1.38 s. Then phase φ = 2πx/λ

3. Aug 20, 2010

### mizzy

to find the phase at t = 1.38s, can't we just just multiply omega and time? Theta = omega * t.

??

4. Aug 20, 2010

### mizzy

I'm stuck. Do I have to put into account the phase constant of 0.72??

For part c i have the equation, v = -3.516 sin(0.586t + 0.72). I set this equal to 0, but i can't figure out how to solve for t??

5. Aug 20, 2010

### novop

About the phase: yes, you certainly have to take into account the initial phase of 0.72 when t=0.

A hint for part c: when is sin(x)=0?

6. Aug 20, 2010

### rl.bhat

In one period (T) change of phase is 2π. Ιn t seconds the change of phase is
φ = 2πt/T + φo

7. Aug 20, 2010

### mizzy

how did you get that equation?

8. Aug 20, 2010

### mizzy

for part c, can i do it this way?

v = -A*omega*sin omega*t + delta
0 = -A*omega*sin omega*t + delta
-delta/ -A*omega = sin omega*t

but i still can't get t...

9. Aug 20, 2010

### novop

Ok... sin(x) = 0 when x=0 or some multiple of pi.

If v = -3.516 sin(0.586t + 0.72) = 0 then
(0.586t + 0.72) = pi

Easy to solve now, I hope.

10. Aug 20, 2010

### mizzy

for part e)

in the text it says, theta(phase) = omega * t

so could i use this and solve for the phase in radians? what i did:
theta = omega*t
= 2 pie f *t
= 2 pie (1/T) * t

where T is found in part b and t is given, t = 1.38s.

can i do that??

11. Aug 21, 2010

### mizzy

sorry, this might be a stupid question...why pi?