For a classical oscillator with mass m, frequency ω and amplitude a, the probability of finding the oscillator at position x is proportional to the time the oscillator spends at that position, so p(x)dx cdt a) rewrite this in terms of the velocity, v= dx/dt and use conservation of energy and the relationship between e and m, co and a for the oscillator to show: p(/w/a2-x2)-1/2 b) normalize this expression to solve for c.
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Ответ:
The proofs are shown in the explanation and Expression for C = w/π
Explanation:
Probability of finding the oscillator at position x is written as P(x) = C.dt
a.P(x).dx = C.dt = C. dt/dx
V = dx/dt
Put V in P(x), which becomes, P(x) = C/v (equation 1)
Now, if total energy, angular frequency and amplitude us E, w, A respectively then
E = ½ mw²A² (equation 2), m = mass
Again if at any time t, position and speed of the oscillator are x and v, then
E = ½ mw²x² + ½ mv² (equation 3), where ½ mw²x² is the component for the potential energy and ½ mv² is the component of kinetic energy
Using equation 2 and equation 3 we can write
½ mw²A² = ½ mw²x² + ½ mv²
V = w (A² – x²)^1/2
Use the above expression of v, into equation 1, we get
P(x) = C/w(A² - x²)^1/2 = (C/w) . (A2 – x2)^-1/2 (equation 4)
b.A classical oscillator oscillates between x = +A to x = -A. So probability of finding the oscillator between A and -A, must be 1.
So the normalization condition is
∫₋ₐᵃ P(x).dx = 1
C/w ∫₋ₐᵃ dx/(A² – x²)^1/2 = 1
2C/w ∫₀ᵃ dx/(A² – x²)^1/2 = 1, (equation 5) where (A² – x²)^1/2 is an even function of x
Let I = ∫₀ᵃ dx/(A² – x²)^1/2
Let x = AsinP, so dx = AcosP dp
When x = 0, P = 0 and x = A, P = π/2
So I = ∫(0, π/2) AcosP dp/(A² – A²sin²P)^1/2
I = ∫(0, π/2) AcosP dp/AcosP =∫(0, π/2) dp = π/2
Substitute into equation 5
2C/w x π/2 = 1
C = w/π
And P(x) = (w/π) . w . (A2 – x2)^-1/2
P(x) = (1/π) . (A2 – x2)^-1/2
Ответ:
What language is this?
Explanation:
English please!!!