Question

When a 4 kg mass is attached to a spring whose constant is 100 N/m, it comes to rest in the equilibrium position. Starting at t= 0, a force equal to f(t) = 12e-3t cos 6t is applied to the system. In the absence of damping, (a) find the position of the mass when t= 1. (b) what is the amplitude of vibrations after a very long time?

Answer 1

Since there is no damping the differential equation governing the mass spring system is

dr(t) 4 dt2 + 100.c(t) = 12e-3t cos 6t
with $x(0)=0~,~x'(0)=0$

a) The aboce differential equation can be written as

--------------------(1)

-30 de(t) +25.0(t) = 3e dt2 cos 6t

The characteristic equation for this equation is

m2 + 25 = 0 m= +52

So the complementary function is

2 (t) = ci cos(5t) + C sin(5t)

The particular solution is given by

$x_p(t)= e^{-3t}[A\cos (6t)+B\sin (6t)]$

Then, ${d^2x_p(t)\over dt^2}= e^{-3t}[9(4A-3B)\sin (6t)-9(4B+3A)\cos (6t)]$

Use in equation (1) we get

$e^{-3t}[ (36A-2B)\sin (6t)+ (-36B-2A)\cos (6t)] =3e^{-3t}\cos(6t)$

Equate coefficients we get $A=-{3\over 74}~,~B=-{9\over 37}$

So we get $x_{p}(t)=e^{-3t}\left ( -{3\over 74}\cos (6t)-{9\over 37}\sin (6t) \right )$

So the solution is

$x(t)=x_c(t)+ x_{p}(t)= c_1\cos (5t)+c_2\sin (5t)+e^{-3t}\left ( -{3\over 74}\cos (6t)-{9\over 37}\sin (6t) \right )$

$x(0)=0$ gives $c_1={3\over 74}$ also

$x'(0)=0$ gives $c_2=-{99\over 370}$

$x(t)= {3\over 74}\cos (5t)-{99\over 370}\sin (5t)+e^{-3t}\left ( -{3\over 74}\cos (6t)-{9\over 37}\sin (6t) \right )$

At $t=\pi$

$x(\pi)=- {3\over 74} +e^{-3\pi}\left ( -{3\over 74} \right )=-{3\over 74}[1+e^{-3\pi}]$

b) The amplitude after very long time

$A= \sqrt{{3^2\over 74^2}+{(99)^2\over (370)^2}}={1\over 370}\sqrt{10,026 }$