Question

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

(a) x=-3e^-3t sin⁡4t

(b) x=-e^-3t  (sin⁡4t+cos 4t)

(c) x=-3e^-3t  (sin⁡4t+cos⁡4t)

(d) x=-3e^-3t  cos⁡4t

The question was asked in examination.

My question is taken from Harmonic Motion and Mass topic in portion Linear Differential Equations – Second and Higher Order of Engineering Mathematics

Answer 1

The correct answer is (a) x=-3e^-3t sin⁡4t

To elaborate: x” (t) + 6x’ (t) + 25x(t) = 0 and at t=0, \(\frac{dx}{dt}\) = -12….since it is moving to the left.

m^2+6m+25=0…..A.E

\(m=\frac{-6±\sqrt{6^2-4*25}}{2} = \frac{-6±8i}{2} = -3±4i\)

complementary solution xc = e^-3t (c1  cos⁡4t + c2  sin⁡4t)

at t=0, x=0 therefore c1 = 0 –> xc = e^-3t c2 sin⁡4t

differentiating \(\frac{dx_c}{dt}\) = -3e^-3t c2 sin⁡4t + 4e^-3t  c2 cos⁡4t

at t=0 \(\frac{dx_c}{dt}\) = -12…  given R.H.S = 4c2 –> c2 = -3

xc = x = -3e^-3t sin⁡4t……as there is no particular solution.