Hard Problem For People Who Like A Good Physics Challenge?

Question

There are two common forms for the general solution for the position of a harmonic oscillator as a function of time t:

1) x(t) = A cos(omega*t + phi) and

2) x(t) = C cos(omega*t) + S sin(omega*t)

Either of these equations is a general solution of a second-order differential equation (F = ma); hence both must have at least two--arbitrary constants--parameters that can be adjusted to fit the solution to the particular motion at hand.

1) Find analytic expressions for the arbitrary constants C and S in Equation 2 in terms of the constants A and phi in Equation 1, which are now considered as given parameters. Give your answers for the coefficients of cos(omega*t) and sin(omega*t), separated by a comma. Express your answers in terms of A and phi.

Answer 1

1)

Since both solutions are equivalent you can equate them:

A·cos(ω·t + φ) = C·cos(ω·t ) + S·sin(ω·t)

expand RHS using angle sum identity  of cosine

cos(α + β) = cos(α )·cos( β) -  sin(α)·sin(β)

=>

A·cos(ω·t)·cos(φ) - A·sin(ω·t)·sin(φ) = C·cos(ω·t ) + S·sin(ω·t)

Equivalence requires, that the coefficients of cos(ω·t ) and sin(ω·t ) are the same on both sides. Hence:

C = A·cos(φ)

S = -A·sin(φ)

2)

Use formulas derived above.

You can eliminate A by taking the quotient of s and C:

S/C = - A·sin(φ) / (A·cos(φ) )

<=>

S/C = - tan(φ)

<=>

φ = arctan(-S/C)

(because inverse tangent is an odd function)

<=>

φ = -arctan(S/C)

If you take the sum of S and C squared you can eliminate φ

using Pythagorean identity:

S² + C² = A²·sin²(φ) + A²·cos²(φ)

(sin²(φ) + cos²(φ) =1)

<=>

S² + C² = A²

=>

A = √(S² + C²)