. 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(x)-(C/w/A2-X2)-1/2 b) Normalize this expression to solve for C.

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shahbazali3290 shahbazali3290 · Answer 1 · 2019-12-10T01:49:40+01:00

Answer:

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