Kinematics: simple harmonic motion

AI Thread Summary
The discussion revolves around solving a simple harmonic motion problem defined by the equation x=(6.0m) cos(0.586t +0.72). The amplitude is correctly identified as 6.0m, and the period calculated as approximately 10.7 seconds. For finding when velocity equals zero, participants suggest setting the velocity equation to zero and solving for t, with the sine function's properties guiding the solution. The maximum acceleration can be determined using the formula a = Aω², while the phase at a specific time requires incorporating the initial phase constant into the calculations. Participants express confusion about the calculations but clarify that the phase can be derived from the relationship between angular frequency, time, and the initial phase.
mizzy
Messages
217
Reaction score
0

Homework Statement


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


Homework Equations


T = 2pie/omega

v = -A omega sin omega*t

a = -A omega^2 cos omega*t


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?

Can someone please help me with the last 3?

THANKS!
 
Physics news on Phys.org
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
 
to find the phase at t = 1.38s, can't we just just multiply omega and time? Theta = omega * t.

??
 
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??
 
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?
 
mizzy said:
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??

In one period (T) change of phase is 2π. Ιn t seconds the change of phase is
φ = 2πt/T + φo
 
how did you get that equation?
 
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...:confused:
 
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
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
sorry, this might be a stupid question...why pi?
 
Back
Top