Simple Harmonic Motion, $\omega ^2$ constant

Question

I recently came across the formula, $F=-kx$; where $F$ represents the net force vector acting on a particle undergoing SHM at an instant when the displacement vector of the particle from the mean position is x. It further stated that $k=m \omega^2$, where $m$ represents the mass of the particle undergoing SHM and $\omega ^2$ represents some constant.

My question is whether $\omega ^2$ is a universal constant? If not what is it constant for?

I would greatly appreciate if anyone could answer this question for I require some conceptual clarity beyond the textbook.

Answer 1

Writing the force as $\mathbf{F} = m\mathbf{a} = m\mathbf{\ddot{x}}$ your formula (so called equation of motion becomes $$m\mathbf{\ddot{x}} = -k\mathbf{x} \Longrightarrow \mathbf{\ddot{x}} = -\frac{k}{m}\mathbf{x}.$$

It is convenient, as @AccidentlFourierTransform pointed out in the comments, to simplify this equation by defining $\omega^{2} := \frac{k}{m}$ so that the equation now reads $$\mathbf{\ddot{x}} = -\omega^{2}\mathbf{x}.$$ The reason for this is that the solutions to this equation are periodic, $\sin$ and $\cos$ functions with argument that crucially depends not on $\omega^{2}$ but rather on $\omega$: $$\mathbf{x} = \mathbf{A} \sin(\omega t) + \mathbf{B} \cos(\omega t)$$ where $\mathbf{A}$ and $\mathbf{B}$ are constant vectors fixed by the initial conditions. Note that if we hadn't introduced $\omega^{2}$ the arguments to the functions above would involve the ugly square root $\sqrt{\frac{k}{m}}$. The new variable makes our equations look nicer and moreover has the interpretation of being the frequency of oscillation.

Since this frequency depends upon the mass of the particle undergoing oscillation, and the parameter $k$ (the stiffness of the spring, say) it is clear that $\omega$ will vary between systems but for a given system it is absolutely constant (assuming the oscillations are not violent enough to deform the spring which could change $k$ for large amplitudes).

Answer 2

Simple Harmonic Motion is a projection of a uniform circular motion with radius equal to the amplitude. You can see this video for understanding it better.

Now, imagine this figure

For uniform circular motion the radial acceleration (called centripetal acceleration) is given by $$ a = \frac{v^2}{r} \\ ~~~~~~~~~~~v = \omega^2 r ~~~~~~~~~~~~~~~~~~( \textrm{$\omega$ is the angular velocity}) \\ a = \omega^2 r $$

But as I said this is the radial acceleration (the acceleration which is pointing towards the center). Since, our aim is the projection and there is only the vertical acceleration there, therefore the vertical acceleration for our uniform circular motion is $$ a_{vertical} = - a \sin \theta \\ a_{vertical} = - w^2 r \sin \theta \\ a_{vertical} = - w^2 x $$

I have put the minus sign because this acceleration points downwards but the accretion in our SHM (in this particular case)is upwards for the moment we are concerning.

So, the vertical force in the uniform circular motion is $$ \mathbf{F} = -m \omega^2 x $$ Now, we will equate this equation of force to the force that you have written $$ -m \omega^2 x = -k x\\ k = m \omega ^2$$ .

So, you see $\omega$ is just angular velocity in our uniform circular motion and it need not be an universal constant, it depends on the velocity $v$. The $\omega$ can also be thought of as an angular frequency.

Hope it helps.

Answer 3

$\omega$ is the natural unit of time for your problem. Given $k$ and $m$, one can only produce one quantity with dimensions of inverse time, i.e. ${\rm [second] }^{-1}$, and this quantity is $\omega_k=\sqrt{k/m}$. Since it explicitly depends on $k$ and $m$, which occur in spring problems, this specific combinations is not universal.

What is closer to something universal is the form of the dimensionless equation that comes from using natural units. Starting from your equation of motion $$ \ddot x=-\frac{k}{m} x \tag{1} $$ and going to dimensionless time $$ \tau=\omega_k t\, ,\qquad \omega_k=\sqrt{\frac{k}{m}}\, ,\qquad \frac{d^2}{dt^2}=\omega^2 \frac{d^2}{d\tau^2} $$ transforms (1) into $$ \frac{d^2}{d\tau^2}x=-x \tag{2} $$

In a pendulum problem, $\omega_g=\sqrt{g/\ell}$ is the natural unit of time and, using now the dimensionless time $\tau=\omega_g\,t$, the resulting differential equation for the angle $\theta$ becomes $$ \frac{d^2}{d\tau^2}\theta=-\theta $$ which obviously has the same form as (2).

Answer 4

$\omega $ is a constant for a given system. Say you have a spring with its constant $k$ denoting its stiffness. Then $\omega = \sqrt\frac{k}{m}$ tells you the natural frequency of the oscillation.

Clearly, it cannot be a universal constant.

Answer 5

$\omega$ is NOT a universal constant. It is merely a property of the system that is being studied (e.g. a pendulum or a spring). It is a constant PROVIDED that the oscillations of the system are small.

For example, if you are dealing with an oscillating pendulum, the weight of the bob is the restoring force that causes the SHM, $\omega = \sqrt{\frac{g}{l}}$,where $g$ and $l$ are the acceleration due to gravity and the length of the pendulum respectively. This holds provided that the oscillations are small. (For large oscillations, $\omega$ depends on the amplitude of the oscillation.) Now if you time the oscillations of the same pendulum on the same spot on earth (or any other planet), expect $\omega$ to be a constant. However if you time the oscillations on another point on earth you will notice small differences due to changes in the value of $g$. Thus $\omega$ is NOT a universal constant.

Answer 6

It represents angular frequency i.e. $\frac{2\pi}{T}$ ,where T is time period of oscillation.