• Support PF! Buy your school textbooks, materials and every day products Here!

Standard deviation problem

  • #1
338
0

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):

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

The time T it takes to fall:

[tex]h=\frac{1}{2}gt^2[/tex]
[tex]T=\sqrt{\frac{2h}{g}}[/tex]

Probability a picture is taken in interval dt:

[tex]\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[/tex]

So the probability density is:

[tex]\rho(x)=\frac{1}{2\sqrt{hx}}[/tex]

Expectation Value:

[tex]\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}[/tex]

Standard deviation:

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

Look right?
 

Answers and Replies

  • #2
338
0
I'm pretty sure that's right. How about this one:

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

Find <x>, <x²> and std dev.

[tex]<x>=\int_{-\infty}^{\infty} xAe^{-\lamba(x-a)^2}dx[/tex]

[tex]<x^2>=\int_{-\infty}^{\infty} x^2Ae^{-\lamba(x-a)^2}dx[/tex]

[tex]\sigma^2=<x^2>-<x>^2[/tex]

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?
 
  • #3
Dick
Science Advisor
Homework Helper
26,258
618
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?
 

Related Threads for: Standard deviation problem

  • Last Post
Replies
0
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
2K
Replies
3
Views
533
Replies
1
Views
940
  • Last Post
Replies
6
Views
4K
Top