Amplitude of harmonic oscillation given position and velocity

[zenterix]

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

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

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

$a=\frac{2x_0}{\sqrt{2}}$

 

[zenterix]

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.

 

[Hill]

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...

 

[Steve4Physics]

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 $A$.

 

[PeroK]

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 $t_0$.