Amplitude of harmonic oscillation given position and velocity

  • #1
zenterix
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Homework Statement
A particle performs harmonic oscillations along the ##x##-axis about the equilibrium position ##x=0##. The oscillation frequency is ##\omega=4\text{s}^{-1}##.

At a certain moment of time the particle has a coordinate ##x_0=25\text{cm}## and its velocity is equal to ##v_{x0}=100\text{cm/s}##.

Find the coordinate ##x## and the velocity ##v_x## of the particle ##t=2.40\text{s}## after that moment.
Relevant Equations
##x(t)=a\cos{(\omega t-\phi)}##
##v(t)=-a\omega\sin{(\omega t-\phi)}##

##x(0)=a\cos{(-\phi)}=x_0##

##v(0)=-a\omega\sin{(-\phi)}=v_0##

##\implies \tan{(-\phi)}=-\frac{v_0}{\omega x_0}##

##\implies \phi=-\tan{\left (-\frac{v_0}{\omega x_0}\right )}##

The solution to this problem says that we can find ##a=\sqrt{x_0^2+(v_{x0}/\omega)^2}##

How do we find this expression?

For the given values of ##\omega, x_0##, and ##v_{x0}## we have ##\phi=-\frac{\pi}{4}## and so we can find that

##\phi=-\frac{\pi}{4}##

##x(0)=a\frac{\sqrt{2}}{2}=x_0##

##a=\frac{2x_0}{\sqrt{2}}##
 
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  • #2
The answer seems to be that we multiply the equation for ##x(0)## by ##\omega##, square it, add it to the square of the ##v(0)## equation, and solve for ##a##.

I wonder if there is a simpler way that comes about through intuition about the meaning of the equations. At least for me this was just an algebraic trick.
 
  • #3
From your expression for ##x(0)##, $$a^2=\frac {x_0^2} {cos^2(-\phi)}$$
From trig, $$cos^2(\alpha)=\frac 1 {1+tan^2(\alpha)}$$
Then, algebra...
 
Last edited:
  • #4
zenterix said:
The solution to this problem says that we can find ##a=\sqrt{x_0^2+(v_{x0}/\omega)^2}##

How do we find this expression?
It’s conventional to use upper case ##A## for amplitude. (Lower case ##a## is acceleration.) So another way to get the above formula is:

##x(t) = A \cos(\omega t - \phi)##
##v(t) = -A\omega \sin(\omega t - \phi)##
Use ##\cos^2 + \sin^2 = 1## and this will give the formula relating ##A, \omega, x## and ##v##.

If you’re allowed to use it directly, there is a handy 'standard' SHM formula for velocity as a function of displacement: ##v = \pm \omega \sqrt{A^2 -x^2}##. This can easily be rearranged to give the formula for ##A##.
 
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  • #5
If I was looking at this problem for the first time, I would do this.

Take ##t = 0## when the object is at the origin. Then, the equation of motion is:
$$x = A\sin(\omega t)$$That must be easier than messing about with a phase factor. Then, of course:$$v = \omega A\cos(\omega t)$$You are given ##\omega## and ##x_0, v_0## at some unknown time ##t_0##. That's two equations in two unknowns. Solve for ##t_0## and ##A##, then plug in ##t_0 + 2.4s##.

Even if that's not the neatest or quickest way, it's the sort of thing you should be able to do by now.

PS as a byproduct, you get the expression for ##A## when you eliminate ##t_0##.
 
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