What is the standard deviation for a rock dropped off a cliff?

[Bill Foster]

Homework Statement

Problem 1.2 from the book Introduction to Quantum Mechanics (2e) by Griffiths:

Suppose a rock is dropped off a cliff of height h. As it

falls, a million photos are snapped, at random intervals. On each picture the distance the rock has fallen is measured.

a) Find the standard deviation of the distribution.

The Attempt at a Solution

Starting with the equation of motion (assuming it falls in the positive x direction):

x(t)=x_0+v_0(t)+\frac{1}{2}at^2=\frac{1}{2}gt^2

The time T it takes to fall:

h=\frac{1}{2}gt^2
T=\sqrt{\frac{2h}{g}}

Probability a picture is taken in interval dt:

\frac{dt}{T}=dt \sqrt{\frac{g}{2h}}=\frac{dx}{gt}\sqrt{\frac{g}{2h}}=\frac{1}{2h}\frac{1}{\sqrt{gt}}dx=\frac{1}{2\sqrt{hx}}dx

So the probability density is:

\rho(x)=\frac{1}{2\sqrt{hx}}

Expectation Value:

\mu=\int_0^h x \rho(x)dx=\int_0^h x \frac{1}{2\sqrt{hx}}dx=\frac{1}{2\sqrt{h}}\frac{2}{3}x^\frac{3}{2}=\frac{h}{3}

Standard deviation:

\sigma^2=\int_0^h (x-\mu)^2 \rho(x)dx
\sigma^2=\int_0^h (x-\frac{h}{3})^2 \frac{1}{2\sqrt{hx}}dx=\frac{4}{45}h^2
\sigma=\frac{2h}{\sqrt{45}}

Look right?

 

[Bill Foster]

I'm pretty sure that's right. How about this one:

Consider the Gaussian distribution \rho(x)=Ae^{-\lamba(x-a)^2} where A, α, and λ are positive real constants.

Find <x>, <x²> and std dev.

&lt;x&gt;=\int_{-\infty}^{\infty} xAe^{-\lamba(x-a)^2}dx

&lt;x^2&gt;=\int_{-\infty}^{\infty} x^2Ae^{-\lamba(x-a)^2}dx

\sigma^2=&lt;x^2&gt;-&lt;x&gt;^2

I tried using integration by parts to evaluate the integrals, but I didn't get anywhere. I also tried looking them up on a table of integrals - also no luck.

I'm guessing <x>=a because a is the peak in a Gaussian distribution. But how do I show it?

 

[Dick]

For <x> just change variables to t=(x-a). If A is properly normalized, you should find <x>=a after you cancel the antisymmetric part of the integral. For <x^2> there's a standard trick to find this. You know a formula for the integral of exp(-k*x^2) from x=-infinity to infinity, right? Call it I(k). Then the integral of x^2*exp(-x^2) is closely related to d/dk(I(k)) evaluated at k=1. Isn't it?