Kinematics: simple harmonic motion

In summary, the motion of a particle is given by x=(6.0m) cos(0.586t +0.72). The amplitude is 6.0m and the period is 10.7s. To find the time when velocity is equal to 0, find dx/dt and equate it to zero. The maximum acceleration is -Aomega^2 and the phase at time t = 1.38s can be found by multiplying omega and t. When solving for the time when sin(x) = 0, the equation is (0.586t + 0.72) = pi. To find the phase
  • #1
mizzy
217
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!
 
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  • #2
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
to find the phase at t = 1.38s, can't we just just multiply omega and time? Theta = omega * t.

??
 
  • #4
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
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
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
 
  • #7
how did you get that equation?
 
  • #8
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:
 
  • #9
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?
 

FAQ: Kinematics: simple harmonic motion

What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium and is directed towards the equilibrium point. This results in a sinusoidal oscillation.

What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x = A*sin(ωt + φ), where x is the displacement from equilibrium, A is the amplitude, ω is the angular frequency, and φ is the phase constant.

What is the difference between simple harmonic motion and periodic motion?

Simple harmonic motion is a type of periodic motion, but not all periodic motion is simple harmonic. Simple harmonic motion has a constant period and is characterized by a sinusoidal shape, while periodic motion can have varying periods and shapes.

What is the relationship between frequency and period in simple harmonic motion?

The frequency of simple harmonic motion is inversely proportional to the period. This means that as the frequency increases, the period decreases, and vice versa.

How is the energy of a simple harmonic oscillator related to its amplitude?

In a simple harmonic oscillator, the total energy is proportional to the square of the amplitude. This means that as the amplitude increases, the energy of the oscillator also increases.

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