Consider the following probability distribution corresponding to a particle located between pint x=0 and x=a:
P(x)dx= Csin^2 [pi*x/a]dx
a. determine the normalization constant, C.
b. determine <x>.
c. determine <x^2>.
d. determine the variance.
2 Answers
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a. according to this integral table: http://www.math.osu.edu/~nevai.1/H16x/DOCUMENTS/In...
the integral from 0 to a will be C [(pi/2) - (1/4)sin[2pi]]. Or just C*pi/2.
so C*(pi/2) = 1 (this is the normalization step)
so C = 2/pi
b, c, d: i dont know what you mean by the <>. and i forget how to find variance/ too much work sorry!
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If ok ( particularly of lambda) is 0 y´´=0 so y´=m and y=mx+n so m could be 0 and y=n (any consistent) If ok is helpful call it w^2 the answer is y=Acos(wx-phi) y´=-Aw sin(wx-phi) y´(0) =0 so sin (-phi)=0 and phi =0 y´(pi))=-Aw sin(w*pi)=0 so wpi=n*pi so w= n( an integer)