Kinematics: simple harmonic motion

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Homework Help Overview

The discussion revolves around the analysis of simple harmonic motion described by the equation x=(6.0m) cos(0.586t +0.72). Participants are addressing various aspects of the motion, including amplitude, period, velocity, acceleration, and phase at a specific time.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore methods to find the time when velocity is zero and question the use of phase constants in their calculations.
  • Some suggest using derivatives to find velocity and acceleration, while others discuss the implications of the phase constant on their calculations.
  • There are inquiries about the relationship between phase and time, with attempts to clarify how to incorporate the phase constant into their reasoning.

Discussion Status

The discussion is active, with participants sharing their attempts and questioning the validity of their approaches. Some guidance has been offered regarding the equations to use, but there is no clear consensus on the methods for solving the remaining parts of the problem.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on understanding the implications of the phase constant and its role in the equations being discussed.

mizzy
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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!
 
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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?
 

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