Simple Harmonic Motion Clarification

1. Apr 14, 2014

ThomasMagnus

1. The problem statement, all variables and given/known data

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 :)

2. Relevant equations

See above

3. The attempt at a solution

See above

Last edited: Apr 14, 2014
2. Apr 14, 2014

paisiello2

Depends on the value of phi.
There are other values of phi also.
They are just the derivatives with respect to "t" so you can easily prove youranswer.

3. Apr 14, 2014

Naz93

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. Apr 14, 2014

paisiello2

What if ø=π/2?

5. Apr 14, 2014

Naz93

the value of phi is just a phase shift that determines the value of the displacement from equilibrium (x) at time t=0.

6. Apr 14, 2014

paisiello2

Right, so what happens to x at t=0 when ø=π/2?

7. Apr 14, 2014

Naz93

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. Apr 14, 2014

paisiello2

You're absolutely right. I was thinking about x0 for some reason.

9. Apr 14, 2014

ThomasMagnus

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+¤)
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: Apr 14, 2014