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If we imagine a 2d semisphere "bowl" with a ball resting at the bottom. The ball can be slightly displaced and will roll back and forth crossing the bottom of the bowl an infinite number of times over and over. I believe this motion can be approximated to be simple harmonic for small displacements from equilibrium, but it isn't completely.

What shape of bowl would be needed (it looks like $y=x^2$ doesn't work: Ball Rolling in a Parabolic Bowl) in order to have a ball placed at any point in the bowl to execute simple harmonic motion. Is this possible?

Edit: in reality I'm actually curious about a point mass "ball" with no moment of inertia for simplicity

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    $\begingroup$ I suppose what is meant is a) the harmonic motion in horizontal direction, b) the motion lies in the vertical plane passing through the lowest point in the bowl. $\endgroup$
    – Roger V.
    Nov 23, 2021 at 17:39
  • $\begingroup$ If you know a priori the solution of the brachistochrone problem then you would know that the arc length $\:s(t)\:$ is a sinusodial function of time $\:t$, see equation (b-04) in my answer here : What is the position as a function of time for a mass falling down a cycloid curve?. So you could find that a solution is the cycloid. But even in this case you have no proof that the solution is one and only one, that is that there is no other curve satisfying a SHM. $\endgroup$
    – Frobenius
    Nov 24, 2021 at 0:35
  • $\begingroup$ In my opinion the question must not be tagged "homework-and-exercises" due to reasons exposed in my previous comment. $\endgroup$
    – Frobenius
    Nov 24, 2021 at 0:43

4 Answers 4

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The shape needed is one 'loop' or 'festoon' of a cycloid (or an upside-down cycloid). Taking the bottom of a loop as origin, the loop's equation may be written as $$x=a(\theta + \sin \theta),\ \ \ \ y=a(1-\cos\theta)$$ for $-\pi<\theta<\pi$, with $\theta=0$ at the bottom of the loop. We see that the horizontal range of the loop is $-\pi a<x<\pi a$. A bead threaded on this loop will perform SHM (not just approximate SHM) from whatever point on the loop it is released. [We can, of course, envisage a particle sliding in a bowl, vertical sections of which, passing through the lowest point, are cycloidal. A rolling ball will also perform translational SHM – at $\sqrt {\frac 57}$ of the frequency derived below, if the rolling ball is homogeneous.]

We can show that the arc length from the origin is $$s=4a \sin \frac{\theta}2$$ and this gives us the neat relation $$s^2 = 8ay$$ If the amplitude of the motion, expressed as a maximum height is $y_0$ (where $y_0<2a$), and as an arc length is $s_0$ (where $s_0<4a$), we can write the energy conservation equation as $$\tfrac12m \left(\frac{ds}{dt}\right)^2+mgy=mgy_0$$ That is $$\left(\frac{ds}{dt}\right)^2+\frac g{4a} s^2=\frac g{4a} s_0^2$$ You'll find that this is satisfied by $$s=s_0 \cos (\omega t + \phi)$$ with $\omega=\sqrt {\frac g{4a}}$. No approximations and a finite amplitude!

Addendum [inspired by dialogue of comments on Eli's answer]

As an alternative to the last step, if you differentiate the energy equation, $\left(\frac{ds}{dt}\right)^2+\frac g{4a} s^2=\frac g{4a} s_0^2\ \ $ wrt time, $t$, and tidy up you get the equation of motion $$\frac{d^2 s}{dt^2} + \frac g{4a} s = 0.$$

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    $\begingroup$ Yes. It works for $-\pi<\theta<\pi$, that is $-a\pi<x<a\pi$ and $0 \leq y <2a$. $\endgroup$ Nov 23, 2021 at 7:55
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    $\begingroup$ @ Frobenius Good question! Huygens solved the so-called 'tautochrone' problem around 1659. I don't know the method he used. I find that Wiki has a good discussion of various methods under 'tautochrone curve'. The Lagrangian approach, as described there, is very accessible. $\endgroup$ Nov 24, 2021 at 8:53
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    $\begingroup$ @Frobenius Yes indeed it does. At twice the frequency of the x-motion, or of the motion along the arc. $\endgroup$ Nov 24, 2021 at 11:57
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    $\begingroup$ @Frobenius I agree with your opinion. All I meant in my comment is that the frequency of the x-motion is the same as that of the s-motion, but that the frequency of the y-motion is twice that of the s or x. $\endgroup$ Nov 24, 2021 at 12:12
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    $\begingroup$ @Frobenius Of course it would be possible to modify my answer above to show that the required curve is a cycloid. One would only need to compare my first version of the energy equation (in which the potential energy is expressed as proportional to $y$) with the known requirement for shm that the PE is proportional to $s^2$. We must therefore have $s^2=by$ in which $b$ is some constant length. The only curve, as far as I know, for which this relationship holds is the cycloid. $\endgroup$ Nov 24, 2021 at 22:49
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inspired from Mr. @ Phillip Wood

start with cycloidal equation

$$x=a \left( \varphi +\sin \left( \varphi \right) \right)\\ y=a \left( 1-\cos \left( \varphi \right) \right)$$ from here you obtain

$$x(y)=a \left( \arccos \left( {\frac {-y+a}{a}} \right) +\sqrt {{\frac {y \left( -y+2\,a \right) }{{a}^{2}}}} \right) $$

with the position vector

$$\mathbf R=\begin{bmatrix} \pm\,x(y) \\ y \\ \end{bmatrix}$$

the kinetic energy :

$$T=\frac m2 \mathbf{\dot{R}}\,\cdot \mathbf{\dot{R}}= \frac{m\,a}{y}\dot y^2$$

the potential energy :

$$U=-m\,g\,y$$ And the equation of motion

$$\ddot y+\frac{g}{2\,a}\,y-\frac{\dot y^2}{2\,y}=0\tag 2$$

hence

only if $\,\dot y\mapsto 0~$ you obtain SHM


edit

the solution of equation (2) with the initial conditions $~y(0)=y_0~,\dot y(0)=0~$ is

$$y(t)=\frac{y_0}{2}+\frac{y_0}{2}\,\cos(\omega\,t)\quad,\omega^2=\frac{g}{2a}$$

$\Rightarrow\quad y(t)-\frac{y_0}{2}~$ is SHM

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    $\begingroup$ (a) I agree with your equation of motion. And, as expected, you find that the frequency of vertical motion is twice that of the motion along the arc. (b) I don't understand your remark about $s=s_0 \cos(\omega t + \phi)$ not fulfilling eq 1. Note that the omega you need to use here is $\sqrt {g/4a}$. PS: It's nice to have inspired someone! $\endgroup$ Nov 23, 2021 at 18:38
  • $\begingroup$ Probably my fault i take it out $\endgroup$
    – Eli
    Nov 23, 2021 at 21:24
  • $\begingroup$ @PhilipWoodI think that „my Solution „ is straight forward, then yours? $\endgroup$
    – Eli
    Nov 23, 2021 at 21:39
  • $\begingroup$ "I think that „my Solution „ is straight forward," Yes, I agree, but I think it would be more satisfying to consider more than just the vertical motion. For example, you can get the equation of motion along the curve by the (very easy) differentiation wrt $t$ of the equation $\left(\frac{ds}{dt}\right)^2 +\frac g{4a}s^2=\frac g{4a}s_0^2$. But it's largely a matter of taste. $\endgroup$ Nov 23, 2021 at 22:20
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What shape must a bowl be to have a ball rolling in the bowl execute perfect simple harmonic motion?

starting with the position vector to the mass \begin{align*} &\mathbf{R}=\begin{bmatrix} x(y) \\ y \\ \end{bmatrix} \end{align*} where $~x(y)~$ is arbitrary function and y is the generalized coordinate

from here the kinetic , potential energy \begin{align*} &T=\frac{m}{2}\left(x'^2+1\right)\,\dot y^2\\ &U=-m\,g\,y\\ \end{align*} and the total energy

\begin{align*} &E_s= \frac{m}{2}\left(x'^2+1\right)\,\dot y^2+m\,g\,y\\ \end{align*} where $~x'=\frac{dx}{dy}$

for SHM the total energy must be (Ansatz)

\begin{align*} &E_{\text{shm}}=\frac{m}{2}\dot{y}^2+\frac k2\,y^2 \end{align*}

\begin{align*} &\text{with conservation of the energy }\\ & E_s=\frac{m}{2}\left(x'^2+1\right)\,\dot y^2+m\,g\,y=m\,g\,y_0\quad\Rightarrow\quad \dot y^2=-\frac{2\,g(y-y_0)}{x'^2+1}\\ & E_{shm}=\frac{m}{2}\dot{y}^2+\frac k2\,y^2=\frac k2\,y_0^2\quad\Rightarrow\quad \dot y^2=-\frac km\,(y^2-y_0^2) \end{align*}

and solving for $~x'^2~$ with $~y_0=0$ \begin{align*} &\left(\frac{dx}{dy}\right)^2=\frac{2\,a}{y}-1\quad, a=\frac{\,m\,g}{k} \end{align*}

this differential equation fulfilled the cycloid equation

\begin{align*} &x=a\,(\varphi +\sin(\varphi))\\ &y=a\,(1 -\cos(\varphi))\\ \end{align*}

hence the shape that create SHM is cycloid


edit

for $~y_0\ne 0~$ you obtain

$$\left( {\frac {d}{dy}}x \left( y \right) \right) ^{2}=-{\frac {y}{y+ {\it y0}}}+2\,{\frac {mg}{ \left( y+{\it y0} \right) k}}-{\frac {{\it y0}}{y+{\it y0}}} \tag A$$

this differential equation fulfilled also the cycloid equation. solving Eq. (A) you obtain $~x=x(y)~$ the solution of the equation of motion is now

$$y(t)=y_0\,\cos\left(\omega\,t\right)\\ \omega^2=\frac km=\frac ga$$

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  • $\begingroup$ If I may say so, this has the makings of an excellent answer. I followed it up to $E_\text{shm}=\frac m2 \dot y^2 +mgy^2$, but couldn't understand this equation. For one thing the dimensions of the last term are not those of energy. $\endgroup$ Nov 24, 2021 at 18:26
  • $\begingroup$ Thank you this was not correct I think I have now the general solution, I was wondering why you "took" , the cycloid for your solution ? $\endgroup$
    – Eli
    Nov 24, 2021 at 19:43
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    $\begingroup$ I had read, a long time ago, that Huygens, around 360 years ago, showed that the cycloid was a 'tautochrome'' so the periodic time for a particle sliding in a cycloidal bowl was independent of amplitude. I was delighted to discover that this was so easy to show, using the slick mathematical formalism that was not available in Huygens's day. $\endgroup$ Nov 24, 2021 at 20:29
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    $\begingroup$ I've got the gist of your answer now. Not quite sure, though, why you can put $y_0 =0$. $\endgroup$ Nov 24, 2021 at 23:40
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    $\begingroup$ But $y_0$ is the amplitude of the $y$–motion, is it not? It can't, sensibly, be zero. Anyway, your edit has dealt with the problem. $\endgroup$ Nov 25, 2021 at 15:15
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Simple Harmonic Motion takes place along a straight line, so the bowl shape leads approximately to SHM only if the displacement is very small.

The acceleration down a slope depends on the angle $\theta$ with the horizontal ($x$ axis), it's $g\sin\theta$. For small angles that's equal to $g\tan\theta$ as both sin and cos approximate to $\theta$ for small angles.

For SHM the acceleration must be proportional to $x$

so the acceleration is $$g\sin\theta = g\tan\theta = g\frac{dy}{dx} = kx$$

Where $k$ is a constant of proportionality

Solving this gives $$y = \frac{kx^2}{2g}$$

and the ball will do SHM in such a bowl, but only for small displacements.

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  • $\begingroup$ this comment I got with my solution which is down vote "OP is asking for the shape of the bowl. Your answer assumes a spherical bowl. This shape doesn't cause the ball to execute perfect SHM for all displacements." $\endgroup$
    – Eli
    Nov 22, 2021 at 23:01
  • $\begingroup$ @ Eli This answer doesn't assume a spherical bowl. It's a $y=x^2$ shape. Also it was mentioned at the beginning of the answer that solution for large displacements isn't possible as SHM must be along a straight line. $\endgroup$ Nov 22, 2021 at 23:04
  • $\begingroup$ The ball will do SHM in many different bowl shapes for small displacements... The OP is asking for a bowl shape that causes SHM for all displacements. $\endgroup$
    – hft
    Nov 22, 2021 at 23:04
  • $\begingroup$ for example, if the bowl is spherical, small displacements also undergo SHM $\endgroup$
    – hft
    Nov 22, 2021 at 23:05
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    $\begingroup$ @John Hunter We both know that the most popular answers aren't always the best, but I cannot comment on answers to this particular question! We must agree to differ on what counts as SHM: I'm much more liberal than you are on this. For me, SHM means simply "varying sinusoidally with time". I think you'll find that this is a very common usage. $\endgroup$ Nov 23, 2021 at 19:58

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