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Why does $\omega= \frac{2 \pi}{T}=\sqrt{\dfrac{k}{m}}$? If we take the differential of $a=-\frac{k}{m}x$ from $F=-kx$ by SHM definition, we get $x=A\cos{(\sqrt{\frac{k}{m}}t+\phi)}$. And we just equate $\sqrt{\frac{k}{m}}$ to a new parameter $\omega$ which is the angular frequency. This has always been defined in textbooks at the start and assumed to be true like below but how did we actually find/derive that relationship between $a=-\omega^2x$ and subsequently  $\omega $ and $\sqrt{\dfrac{k}{m}}$? Is it simply by experimental and dimensional analysis?

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  • $\begingroup$ Related: physics.stackexchange.com/questions/451080/… $\endgroup$
    – Christophe
    Commented Oct 14, 2023 at 6:26
  • $\begingroup$ @Christophe Yes, I get why it is dimensionally correct, but the post just says "Let's call it ω. You can call it with any other letter, but later on you will see that it is very related to an angular velocity, so it is a good name. Just that." $\endgroup$
    – Dian Sheng
    Commented Oct 14, 2023 at 8:14
  • $\begingroup$ your question is really just about "when does the trigonometric functions first repeat its motion" and can be answered by yourself. $\endgroup$ Commented Oct 14, 2023 at 8:58

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Given:

$$ F = m \ddot x = - kx $$

This can be simplified using division and addition:

$$ \ddot x + \frac{k}{m} x = 0 $$

For simplicity, we define $\omega^2 = \frac{k}{m}$ because we know how to solve this differential equation. If we don't know, then this step is fairly baffling, so I'll explain a little bit. We need 2 functions (because it's a second order equation) that have the property that their second derivative is the negative of themselves up to a constant. Both $\sin$ and $\cos$ fulfill this property:

$$ \frac{d}{dt} \left( \frac{d}{dt} \sin(\omega t) \right) = \frac{d}{dt} \omega \cos(\omega t) = - \omega^2 \sin(\omega t) $$

The cosine function is left to the reader. If you don't know anything about differential equations, then this analysis has been worthless and I'll start anew with some vague, hand-wavey incantations. Each of the two dots over $x$ indicate time derivatives which have the units of $t^{-1}$ meaning that $\frac{k}{m}$ must have units of $t^{-2}$. This proves nothing, but you'll be expected to have it mastered for your next exam.

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  • $\begingroup$ I understand that we replace omega for \sqrt{k/m} for the sake of convenience when solving this second-order homogenous differential equation. And by coincidence, the square root of k/m just happens to have the same unit as angular frequency but is also equal? $\endgroup$
    – Dian Sheng
    Commented Oct 14, 2023 at 8:10
  • $\begingroup$ Not coincidence. If we were working with wavelength - as in a length instead of time, we'd have a differential equation of $\frac{d^2}{x^2} \psi + \frac{\psi}{\lambda^2} = 0$ where $\lambda$ would be a shorthand for the square root of many other constants which end up together having units of length. Often there are also [unitless] numbers (i.e. $\pi$ or $2$) as part of that particular constant. It's because that's what drops out of the differential equation. $\endgroup$
    – user121330
    Commented Oct 14, 2023 at 23:54
  • $\begingroup$ sorry, $\frac{d^2}{dx^2} \psi$ $\endgroup$
    – user121330
    Commented Oct 15, 2023 at 1:18
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The question is actually not about the differential equation. You have already established that its solution has the form $x(t)=A\cos{\left(\sqrt{\frac{k}{m}}t+\phi\right)}$. Now it is just a matter of showing that the factor $\sqrt{\frac{k}{m}}$ is the same as $\frac{2\pi}{T}$ where $T$ is the period for the sine function. This is easily checked: Assuming that this is true, we get $$T=2\pi\sqrt{\frac{m}{k}}$$ so we can calculate

$$x(t+T)=A\cos{\left(\sqrt{\frac{k}{m}}(t+T)+\phi\right)}=A\cos{\left(\sqrt{\frac{k}{m}}\left(t+2\pi\sqrt{\frac{m}{k}}\right)+\phi\right)}\\=A\cos{\left(\sqrt{\frac{k}{m}}t+2\pi+\phi\right)}=A\cos{\left(\sqrt{\frac{k}{m}}t+\phi\right)}=x(t)$$

So, indeed, $T$ is the period for this sine function and thus our assumption was correct. We can then define the expression $\frac{2\pi}{T}$ to be called $\omega$ and give this new quantity some name, but the relevant part is how the expression $\sqrt{\frac{k}{m}}$ relates to the period $T$.

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Recall that cosine is a periodic function with period $2\pi$ $$ \cos (\theta) = \cos(\theta + 2\pi n) $$ for any integer $n$.

Applying it to your example with $n=1$ \begin{eqnarray} \cos\left(\sqrt{\frac{k}{m}}t + \phi\right) &=& \cos\left(\sqrt{\frac{k}{m}}t + \phi + 2\pi\right) \\ &=& \cos\left(\sqrt{\frac{k}{m}}\left(t + T\right) + \phi\right) \end{eqnarray} where $$ T = 2\pi \sqrt{\frac{m}{k}} $$ In other words, we can reinterpret the periodicity of the cosine under a translation of $2\pi$ radians of its argument, as a periodicity of the motion of the harmonic oscillator in time by one period $T$, as defined where $T$ is defined as above.

It is then customary to define the angular frequency as $$ \omega = \frac{2\pi}{T} $$ which in this case means $$ \omega = \sqrt{\frac{k}{m}}. $$

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Of course it simply by experimental and dimensional analysis. To be honest, it is not clear what do you want to see as the answer. But if you would like to understand (in some sense) why it is useful to define $\omega^2=k/m$, the following logic seems reasonable.

You start from the equation $$m\ddot{x}=-kx,\tag{1}$$ which is simply the 2nd Newton law and dot denotes derivative w.r.t "proper time" $t$. Now consider that you have a clock, which measures time. You observe the dynamics and measure time using the clock. It means that you effectively deal with timensionless time $\tau=t/T$, where $T$ is clock time. Now we rewrtie the equation of motion in terms of $\tau$, $$\frac{m}{T^2}\frac{d^2x}{d\tau^2}=-k x.\tag{2}$$ Now assume that you have a lot of clocks with different clock times. You ask yourself: what is the best clock time $T$ to observe dynamics? Dividing both sides of $(2)$, by $m$ you find that $$\frac{1}{T^2}\frac{d^2x}{d\tau^2}=-\frac{k}{m}x.$$ The derivative in left hand side now does not know about time. Looking on left hand side and right hand side, you conclude that $T^2=m/k$, or, equivalently $\omega^2=k/m$.

So, you should choose the clock which measures time in units $T=\sqrt{m/k}$ to observe dynamics easily.

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For uniform circular motion the angular velocity $\omega$ is constant. So it makes sense to define a period $$T=\frac{1}{\nu}=\frac{2\pi}{\omega}$$ which means the time took to complete a complete cycle. Frequency $\nu$ means how many cycles are completed in a second.

Now when you have a simple harmonic motion, you can describe the motion of the object as if it is in an uniform circular motion. It is mathematically equivallent to describe it this way. So in that case, angular velocity is the same as angular frequency. The time it takes for the object to do a complete oscillation (to come back to the initial position) is equivalent to the time elasped to go around the circle and get to the initial position.

For simple harmonic oscillations $\omega$ is a constant which tells you how many times per second a full cycle is completed. To complete a full cycle means to go around the circle the come back to the original position. This means 360$^{\circ}=2\pi$ radians. So if you have $\omega=2\pi\ rad/s$ it means that a full cycle is completed in a second.

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One way to do it is to invent imaginary extra motions to simplify the analysis. This is not the usual way, which focuses on getting good equations of motion to model the thing first, then solving them, then identifying the $2\pi$-periodicity of cosine to recognize that this parameter is the “angular velocity.” But I think you will not be happy unless you can directly see where the $2\pi$ comes from and this alternate way might help?

The imaginary pendulum

Look at a full pendulum’s motion in 3D: it moves along the bottom of a hemisphere, $$ z = R - \sqrt{R^2 - x^2 - y^2}$$ by the 3D Pythagorean theorem. Taking a differential assuming $x^2+y^2\ll R^2$ gives $$ z = \frac1{2R}\big(x^2+y^2\big),\tag{1}$$ the potential of gravity is $U=mgz$ and because this is a sum $f(x)+g(y)$ the two directions decouple completely, motion in the $x$-direction is independent from motion in the $y$-direction. So a pendulum with a gravity and radius carefully tuned to $k=mg/R$ can model a spring’s motion, where $U=\frac12 kx^2$, and projecting out only the $x(t)$ component will track the spring’s motion perfectly fine no matter what weird stuff is happening in the $y$-direction. (The condition that $R\gg x$ needed for eq (1) here may require us to imagine $g > g_\text{Earth}$ for large $k$ here, this limitation can be fixed with more advanced dimensional analysis though.)

If you let me invent an imaginary $y(t)$ to analyze the spring’s motion, then I choose the one which makes the imaginary pendulum go around in a perfect circle! This motion is observed to be perfectly rotationally symmetric and even. And the distance traveled in one period is one circumference $2\pi A$, so the speed is the constant $2\pi A/T$, where $T$ is the period.

Even though the pendulum is an imaginary analysis tool and its y-axis was invented for pure convenience on our part, it has pushed back on us. It says the maximum speed that we see $|v_\text{max}|$ of the spring must be related to the period by $|v_\text{max}|=2\pi/T.$ It also tells us that the motion of this $\frac12 k x^2$ potential is sinusoidal. Finally, since the velocity is also traveling in a uniform circle in velocity space, 90° out-of-phase with the position, the acceleration) (velocity of the velocity) is also constant in magnitude, $$a={2\pi\over T}\left({2\pi A\over T}\right),\tag{2}$$and 180° out of phase with the position, so it always points inward $\mathbf F = ({-k x},{- ky})$. So combining eq (2) with Newton’s laws, the imaginary pendulum still tells us that $$ -m~{2\pi\over T}\left({2\pi x\over T}\right) = {- k x},\\ {2\pi\over T}=\sqrt{k\over m}.$$

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The standard differential equation is parametrized with $\omega$ in order to bring it into a solvable form

$$ m \ddot{x} + k x = 0$$

becomes

$$ \ddot{x} + \omega^2 x = 0$$

when we substitute $\omega^2 = \frac{k}{m}$. At this point, there is no significance to $\omega$ other than having two parameters $k$ and $m$ our equation has only one $\omega$.

Now we do some differential gymnastics and take ${\rm d}v =a {\rm d}t$, multiply by velocity on both sides $v {\rm d}v = a v {\rm d}t$ and note that $v {\rm d}t = {\rm d}x$ in order to arrive at

$$ \int v {\rm d}v = \int a {\rm d}x + C$$

$$ \int \dot{x} {\rm d}\dot{x} = \int \ddot{x} {\rm d} x + C$$

$$ \tfrac{1}{2} \dot{x}^2 = -\tfrac{1}{2} \omega^2 x^2 + C$$

We can pick zero time when $x(0)=0$ and initial velocity is $\dot{x}(0) = v_0$ to find the constant $C$ and solve for $\dot{x}$ in terms of $x$

$$ \dot{x} = \pm \sqrt{v_0^2 - \omega^2 x^2 } $$

Now introduce another parameter for $d=\tfrac{v_0}{\omega}$ to re-write the above as

$$ \dot{x} = \pm \omega \sqrt{ d^2 - x^2} $$

Now use ${\rm d}x = v {\rm d}t$ to write

$$ t = \int \frac{1}{\dot{x}} {\rm d}x + C$$

or

$$ t = \tfrac{1}{\omega} \int \frac{1}{\sqrt{d^2-x^2}} {\rm d}x + C$$

which is a solvable integral (look it up on a table)

$$ t = \tfrac{1}{\omega}\; \sin^{-1} \left( \frac{x}{d} \right) + C$$

And since $x(0)=0$ the factor $C=0$. Most importantly the coefficient $\omega$ is outside the integral. Solving the above in terms of displacement you have

$$ x(t) = d \sin(\omega t) = \tfrac{v_0}{\omega} \sin ( \omega t) $$

which is a solution to the SHM problem. As you can see $\omega$ sets the frequency of oscillation because it factors out if the integral and ends up in $\omega t = \int \ldots$ form

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