4
$\begingroup$

There is a question on the number of times a body comes to a place in a simple harmonic motion. Have a look:

Enter image description here

I thought that the answer was B because in each oscillation a body is at the extremes only once, while for other positions, it comes two times. But the correct answer is C which I don't know why.

How come this is true?

$\endgroup$
5
  • $\begingroup$ @Nihar Karve if the body starts from origin towards +x and then to -x then in this process it had crossed the positions between origin and +x twice once going and once while returning.. $\endgroup$
    – Ankit
    Commented Jan 30, 2021 at 5:45
  • 1
    $\begingroup$ Did you take into account the variation in speed as a function of angle? (In my previous comment what I meant to say was: the argument of only reaching the end once per oscillation has no real implications, since displacements arbitrarily close to $\pm x_0$ are reached twice). Since the graph should be symmetric, try solving it for "half an oscillation" from $+x_0$ to $-x_0$. $\endgroup$ Commented Jan 30, 2021 at 6:13
  • 2
    $\begingroup$ It moves very quickly near the equilibrium points (all potential energy converted to kinetic energy), and it moves slowly at the extremes. Therefore, you are more likely to see it at the extremes because it spends more time (since its velocity is slower) $\endgroup$
    – KingLogic
    Commented Jan 30, 2021 at 19:01
  • 1
    $\begingroup$ The title as a whole is incomprehensible. Can you fix it? $\endgroup$ Commented Jan 30, 2021 at 19:10
  • $\begingroup$ @A student where did you get this question from? $\endgroup$ Commented Feb 7, 2021 at 20:45

3 Answers 3

9
$\begingroup$

What the question is asking you to find is the classical probability distribution for the harmonic oscillator. I'll show you how to do it mathematically further down, but let's start off by guessing what the solution would look like. First of all, let's interpret the question differently: imagine you're walking into a shop that sells pendulum clocks, each oscillating independently of the others. The instant you walk in, you take a picture of all the different pendulums that are in the shop. Now, you ask yourself the following question: If you divided the total amplitude of all the pendulums into discrete bins, and you looked at every individual pendulum, what fraction of the total number of pendulums would you find in any particular bin.

This is the same as the probability that a pendulum will be found in any particular location. Let's see how we can answer this intuitively: where is the pendulum most likely to be found? Well, you already know the answer to that, if you've ever done an experiment with a simple pendulum! From which point in its trajectory do you start counting oscillations? Well, the point furthest away from the centre, of course! Why do you start counting when the pendulum reaches its amplitude? Because that that's when it is slowest, and therefore it spends the most time at the amplitude, meaning that the relative error you would make in identifying that it has reached the end of its cycle would be the least. If you tried to start counting the oscillations from the mean position (where the pendulum is the fastest) you'd have a lot of trouble measuring the time period accurately, since the pendulum barely spends any time at the mean position, since it's going by so fast!

Below is a great long-exposure photograph taken from a simple paper on the subject (Understanding probabilistic interpretations of physical systems). As you can see, the trail is denser at the edges since the pendulum spends more time there than at the centre:

                                                enter image description here

Just by using this analysis, you should see that the answer to the question has to be $(C)$, since it's the only distribution where the probability of detection is minimum at the centre, and maximum at the edges.

If you'd like to derive this mathematically, this is how you'd do it:

Let's start off by agreeing that the probability of finding the pendulum in a small region $\Delta x$ is proportional to the time that the pendulum spends in $\Delta x$, which we can call $\Delta t$. If we assume $\Delta x$ to be very small, we can say that the velocity of the pendulum is roughly constant in the interval $\Delta x$, and so $$\Delta t \approx \Bigg|\frac{\Delta x}{v}\Bigg|.$$

Now, consider the oscillation of the pendulum over one full time period $T$: in this time, it covers all the locations twice (once "to" and once "fro"). So, the fraction of time it spends in any "bin" $\Delta x$ is: $$P(x, x+\Delta x) =2 \frac{\Delta t}{T} = \Bigg|\frac{2}{T \cdot v(x)}\Bigg| \Delta x.$$

Therefore, the probability density of finding the particle in some interval $\Delta x$ is given by: $$\rho(x) = \frac{2}{T \,\, |v(x)|}.$$

I'll leave it to you to write the velocity as a function of position and show that $$\rho(x) = \frac{1}{\pi \sqrt{x_0^2 - x^2}}.$$

Choosing $x_0 = 1$, you can plot this function out, and it looks like you'd expect:

                                                enter image description here

$\endgroup$
6
  • $\begingroup$ Thats a great answer! Do you see any geometric interpretation for why $\rho(x) =\frac{1}{\pi} \frac{d}{dx} \text{sin}^{-1}(\frac{x}{x_o})$ ? $\endgroup$ Commented Jan 30, 2021 at 7:20
  • 1
    $\begingroup$ Hm, good question. Not offhand, no. I understand it as follows: I've shown that $$\text{d}t = \frac{\text{d}x}{v(x)} = \frac{\text{d}x}{\omega_0 \sqrt{x_0^2 - x^2}}.$$ (You can get the relation for $v(x)$ by using conservation of energy, for example.) Integrating from $0$ to some time $t$, you can show that $$\omega_0 t = \sin^{-1}\left(\frac{x}{x_0}\right).$$ Which is just just a restatement of $x = x_0 \sin(\omega_0 t).$ Not sure this is the sort of motivation you wanted, but it does sort of justify the sine function: it appears because the position goes as $\sim \sin(\omega_0 t + \phi)$. $\endgroup$
    – Philip
    Commented Jan 30, 2021 at 7:58
  • $\begingroup$ @Philip can you please explain your example further which you mentioned in the last lines of the first para ? $\endgroup$
    – Ankit
    Commented Jan 30, 2021 at 9:14
  • $\begingroup$ Sure, what exactly are you finding confusing? You walk into a shop with a bunch of pendulums all oscillating independently. You take each pendulum, and measure its position (relative to its own mean). In general, all the pendulums won't be at exactly the same position (since they're oscillating independently). Now, plot a graph of "Number of times a pendulum was found at some position" versus "position". (Of course, to do this practically you'd have to divide the distance into bins i.e. "3 cm to 4 cm", "4 cm to 5 cm" etc.) The resulting distribution is what the question is asking for. $\endgroup$
    – Philip
    Commented Jan 30, 2021 at 9:41
  • $\begingroup$ You can do this in a different way: suppose you just had one simple pendulum (or simple harmonic oscillator), and you took a million photos of it while it was oscillating. Then, just as before, you make little bins. Now, you ask yourself the same question as before, how many photos show the oscillator "in" some distance interval $\Delta x$? That's the same distribution. In fact, this is what the figure in my answer is showing, it's just superposing all the images (on a black background), so that the brighter locations are where the oscillator is more likely to be found. $\endgroup$
    – Philip
    Commented Jan 30, 2021 at 9:46
6
$\begingroup$

This is basically because the particle has a slow speed at the ends $( \ at \ \pm x_0)$ and hence a greater chance of being captured there.

Let us look at the equation of motion of the particle:

$x=x_0 sin(\omega t) $

Enter image description here The $X$ axis shows time and the $Y$ axis shows displacement of particle $(x)$.

I have marked points on the curve with a fixed time interval between them. Each of these dots represent a snapshot of the oscillator. We can see from the figure that at the turning points: $ x = x_0$ and $x=-x_0$, points cluster together, giving a greater chance of seeing the particle there.

This can be mathematically shown using density of states, but I feel for multiple choice questions, intuition is what is necessary.

$\endgroup$
2
  • $\begingroup$ not just slower speed at the ends, but it also stops at the ends, so the amount of time it spends there (hence probability to be caught on camera) is longer than during its fast motion in the central area. $\endgroup$
    – Alex P
    Commented Jan 30, 2021 at 9:51
  • 1
    $\begingroup$ yes exactly. And that explains why the mathematical approach as in @Philip answers goes to infinity at the ends. But unlike the mathematical approach, step sizes in real life do not tend to zero and hence gives only finite probability $\endgroup$ Commented Jan 31, 2021 at 17:15
2
$\begingroup$

At the ends, the oscillator slows down. So the probability of finding it at ends is high.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.