Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The problem that got me thinking goes like this:-

Find $dp/dx$ where $p$ is the probability of finding a body at a random instant of time undergoing linear shm according to $x=a\sin(\omega t)$. Plot the probability versus displacement graph. $x$=Displacement from mean.

My work:

$$v=dx/dt=\omega \sqrt{a^2-x^2}$$

Probability of finding within $x$ and $x+dx$ is $dt/T$ where dt is the time it spends there and T$$ is the total period.
Therefore

$$dp=dt/T=\frac{dx}{\pi \sqrt{a^2-x^2}}$$

because $t=2\pi /\omega$ and the factor 2 is to account for the fact that it spends time twice in one oscillation. The answer matches the answer and also the condition that integration $-a$ to $a$ of $dp =1$.

But when i try to find p as a function of x to plot the graph I get

$$p=\frac{1}{\pi}\arcsin(x/a)+C.$$

But then I get stuck as there is no way to find $C$ (except the fact that for $C=0$ the probability at the mean position is $0$ and hence $C$ cannot equal 0) which I know of. So how can I get a restraint on $C$ to find its value and hence to properly graph it with the condition that the probability from $-a$ to $a$ be 1?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

The differential $dp(x)$ is the probability of finding the body in an interval of length $dx$ centered at $x$. The quantity $p$ you are looking for is the cumulative distribution function, $$P(x)=\int_{-\infty}^x \frac{dp}{dx}(x) dx,$$ which is the probability that the particle will be to the left of the point $x$. Since the particle cannot be to the left of $-a$ you can fix $C$ by requiring that $P(-a)=0$. This will then give $P(0)=1/2$ as expected.

It's just a matter of being precise as to exactly what you are calculating.

share|improve this answer

In your notation, $dp/dx$ is your probability density. $p(x)$ is your cumulative probability density, the probability that the particle is to the left of $x$ at a given time. Knowing this, there are at least three ways you can reason. As Emilio Pisanty pointed out in his answer, you can require that $p(-a) = 0$ (ie, that the particle cannot be to the left of $-a$. In addition to that, you can require that $p(a) = 1$ (ie, that the particle cannot be to the right of $a$), or you can require that $p(0) = 0.5$, because the system is symmetric about the origin. All of those should give you the same value for $C$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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