Question

Problem 42P: Chapter: CH9 - Problem: 42P At time t = 0, a forced harmonic oscillator occupies position (0) = 0.1 mand has a velocity x(0) 0. The mass of the oscillator is m = 10 kg, and the stiffness of the spring is k-1000 N/m. Calculate the motion of the system if the forcing function is AO - FO sin wor, with F0 - 10 N and wo - 200 rad/s.

Answer 1

the motion of the system is given by:

I = A cos(w.t + o)
where $A=\frac{F_o/m}{\sqrt{{(\omega_o^2-\omega^2)^2+\left(\frac{b\omega}{m} \right)^2}}}$

How the valous of the angular frequency are diferents, then b=0, therefore:

$A=\frac{F_o}{m(\omega_o^2-\omega^2)}$

how the x(0)=0.1 m and v(0)=0 m/s, this implies that $\phi=0$ thus we have:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{1000 N/m}{10 kg}}=10 rad/s$, thus

, therefore:

$x(t)=0.1\cos(10t) m$

$v(t)=-\omega A\sin(\omega t)=-\sin(10t))m/s$

$a(t)=-A\omega^2\cos(\omega t)=-10\cos(10 t)$

$F(t)=10\sin(200t)$