A particle moves along the x-axis according to the law \(\frac{{{d^2}x}}{{d{t^2}}} + 6\frac{{dx}}{{dt}} + 25x = 0.\) If the particle is started at x = 0 with an initial velocity of 12 ft/s to the left, determine x in terms of t.

Concept:

Laplace transform:

\(L~[x''(t)]~=~s^2~×~L~[x(t)]~-~s~×~x(0)~-~x'(0)\)

\(L~[x'(t)]~=~s~×~L~[x(t)]~-~x(0)\)

L [Sin at] = \(\frac{a}{s^2~+~a^2}\)

Calculation:

Given:

\(\frac{{{d^2}x}}{{d{t^2}}} + 6\frac{{dx}}{{dt}} + 25x = 0.\)................................. (1)

At t = 0, x = 0;  \(\frac{dx}{dt}~=~-12\), at t = 0

i.e. x (0) = 0, x' (0) = -12 are the initial conditions

Now, taking laplace transform on both sides of (1)

∴ L [x'' (t)] + 6 × L [x' (t)] + 25 × L [x (t)] = L (0)

⇒ { \(s^2~×~L~[x(t)]~-~s~×~x(0)~-~x'(0)\) } + 6 × {\(s~×~L~[x(t)]~-~x(0)\)} + 25 × L [x (t)] = 0

Using initial conditions:

\((s^2 ~+~6s~+~25)~\times~L~[x(t)]\) = -12

⇒ L [x (t)] = \(\frac{-12}{(s^2 ~+~6s~+~25)}\)

⇒ x (t) = \(-12~\times~L^{-1} [\frac{1}{(s^2 ~+~6s~+~25)}]\)

⇒ x (t) = \(-12~\times~L^{-1} [\frac{1}{(s~+~3)^2 ~+~4^2}]\)

⇒ x (t) = \(-12~\times~e^{-3t}~\times~L^{-1} [\frac{1}{s^2 ~+~4^2}]\)

⇒ x (t) = \(-12~\times~e^{-3t}~\times~ [\frac{\sin 4t}{4}]\)

⇒ x (t) = \(-3~\times~e^{-3t}~\times~\sin 4t\)