Question

Nick Camelot · Accepted Answer

To solve the equation of motion for a damped oscillator with the given initial conditions, we can use the trial function:

x (t) = C_{1} e^{- λ_{1} t} + C_{2} e^{λ_{2} t}

where

C_{1}

and

C_{2}

are constants to be determined, and

λ_{1}

and

λ_{2}

are parameters that we need to find.
We start by taking the derivatives of

x (t)

with respect to time:

\frac{d x}{d t} = - λ_{1} C_{1} e^{- λ_{1} t} + λ_{2} C_{2} e^{λ_{2} t}

\frac{d^{2} x}{d t^{2}} = λ_{1}^{2} C_{1} e^{- λ_{1} t} + λ_{2}^{2} C_{2} e^{λ_{2} t}

Now, substitute these derivatives into the equation of motion:

λ_{1}^{2} C_{1} e^{- λ_{1} t} + λ_{2}^{2} C_{2} e^{λ_{2} t} + \frac{b}{m} (- λ_{1} C_{1} e^{- λ_{1} t} + λ_{2} C_{2} e^{λ_{2} t}) + \frac{k}{m} (C_{1} e^{- λ_{1} t} + C_{2} e^{λ_{2} t}) = 0

Simplifying the equation, we have:

(λ_{1}^{2} - \frac{b}{m} λ_{1} + \frac{k}{m}) C_{1} e^{- λ_{1} t} + (λ_{2}^{2} + \frac{b}{m} λ_{2} + \frac{k}{m}) C_{2} e^{λ_{2} t} = 0

For this equation to hold for all values of

t

, the coefficients of the exponential terms must be zero:

λ_{1}^{2} - \frac{b}{m} λ_{1} + \frac{k}{m} = 0

λ_{2}^{2} + \frac{b}{m} λ_{2} + \frac{k}{m} = 0

These are quadratic equations in

λ_{1}

and

λ_{2}

. We can solve these equations to find the values of

λ_{1}

and

λ_{2}

. The solutions will depend on the specific values of

b

,

m

, and

k

in the equation of motion.
Once we find the values of

λ_{1}

and

λ_{2}

, we can substitute them back into the trial function

x (t)

and use the initial conditions

x (0) = A

and

\frac{d x}{d t} (0) = 0

to determine the constants

C_{1}

and

C_{2}

.

Answered question