Simple Harmonic Motion Clarification

  • #1
138
0

Homework Statement



I was wanting to get some clarification on some of the simple harmonic motion equations. So, say for example there is a box of mass "m" undergoing simple harmonic motion attached to a spring of spring constant K on a horizontal surface. To find where the box is, as a function of time, I would use the following equation:

x(t)=Acos(ωt+ø)

My first question is, will the "x" of this equation be the displacement from the maximum extension point, or the equilibrium point?

Second question: In regards to the phase constant "ø", if at t=0, x=A, then ø would have to be 0, correct? and this is because x(0)=Acos(ø)-->A=Acos(ø), cos(ø)=1 ∴ ø=0

Third question: I have seen in many place a formula used for the amplitude:

A = sqrt((x0)^2 + (v0/ω)^2)

I'm not sure how this is derived. I know it has to do with the velocity function: -Aωsin(ωt + φ)
It just says to square both sides and simplify, but I still don't see how that would give the amplitude equation...

V^2=(A^2)(ω^2)sin(ωt+ø)^2

(V^2)/(ω^2)(sin(ωt+ø)^2)=A^2.......

*EDIT* I believe I figured that last part out, simply by rearranging some terms.

So, I could use the equation
x(t)=Acos(ωt+ø) to get the displacement from equilibrium for ANYTHING undergoing SHM, and the associated velocity and acceleration functions to get the velocity or acceleration of ANYTHING undergoing simple harmonic motion, i.e mass on a spring etc.

Furthermore, will x(t), a(t), v(t) have different phase constants for the same graph, since they are based on different trig functions?

Thanks :)


Homework Equations



See above


The Attempt at a Solution



See above
 
Last edited:

Answers and Replies

  • #2
907
88
My first question is, will the "x" of this equation be the displacement from the maximum extension point, or the equilibrium point?
Depends on the value of phi.
Second question: In regards to the phase constant "ø", if at t=0, x=A, then ø would have to be 0, correct? and this is because x(0)=Acos(ø)-->A=Acos(ø), cos(ø)=1 ∴ ø=0
There are other values of phi also.
Furthermore, will x(t), a(t), v(t) have different phase constants for the same graph, since they are based on different trig functions?
They are just the derivatives with respect to "t" so you can easily prove youranswer.
 
  • #3
29
2
1. x is displacement from equilibrium. One way to see this must be the case is that the right side of the equation is symmetric about the equilibrium position, so the left side must be too. It couldn't be the displacement from the maximum extension point.

2. Yes. You've got it :)

3. Differentiate the cos displacement equation to get a velocity equation in sin, and use the fact that cos^2 + sin^2 = 1.

As for the phase constants: differentiate x to get v, and differentiate v to get a. You'll see the same ø in all of them, but the actual phase is different as v will be a sin, and a will be a -cos.

Hope this helps!
 
  • #4
907
88
It couldn't be the displacement from the maximum extension point.
What if ø=π/2?
 
  • #5
29
2
the value of phi is just a phase shift that determines the value of the displacement from equilibrium (x) at time t=0.
 
  • #6
907
88
Right, so what happens to x at t=0 when ø=π/2?
 
  • #7
29
2
x=0. But this just corresponds to the motion starting at the equilibrium position. For future t, x then oscillates symmetrically about x=0.
 
  • #8
907
88
You're absolutely right. I was thinking about x0 for some reason.
 
  • #9
138
0
To hit the point home, was was wondering if you could tell me if I am setting up the appropriate functions correctly for a problem:

A butcher throws a cut of beef on spring scales which oscillates about the equilibrium position with a period of T = 0.500 s. The amplitude of the vibration is A = 2.00 cm (path length 4.00 cm). Find: a. frequency b. the maximum acceleration c. the maximum velocity d. the acceleration when the displacement is 1.00 cm e. the velocity when the displacement is 1.00 cm f. the equation of motion as a function of time if the displacement is A at t = 0

For the displacement: x(t)=Acos(wt+¤)
f=2.00 Hz, w=2pi*2.00=12.6 rad/s. A=.0200m
x(0)=0, as the spring is just sitting there before the beef is put on the scale, so 0=.0200cos(¤), so ¤ would be pi/2 or any value that makes cos¤=0. So then, the displacement at any time would be modelled by this equation: x(t)=.0200cos(12.6t+pi/2), correct? I will answer the rest if I have this part correct by differentiating.

then v(t)=-.0200sin(12.6t+pi/2)*12.6

and a(t)=-.0200(12.6)(12.6)cos(12.6t+pi/2)

Jumping to part e.
Find time when x(t)=.01, 0.0100=.0200cos(12.6t+pi/2)
(Arccos(0.5)-pi/2)/12.6=t=-.04....how could it be negative time?!!!

Furthermore, I only get the correct answer using a phase constant of 3pi/2, why wont pi/2 work? The only difference I see in the graphs is that 3pi/2 phase angle starts with a positive graph, while pi/2 starts with a negative graph. Should 3pi/2 be used because the system moves down first, and we are defining down as positive?

Thanks!
 
Last edited:

Related Threads on Simple Harmonic Motion Clarification

Replies
14
Views
3K
Replies
3
Views
2K
  • Last Post
Replies
17
Views
836
  • Last Post
Replies
3
Views
746
  • Last Post
Replies
3
Views
636
  • Last Post
Replies
2
Views
655
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
1K
Top