7.3 Applications  (Page 3/13)

1. We have
   
   $A=\sqrt{c_1^2+c_2^2}=\sqrt{2^2+1^2}=\sqrt{5}$
   
   and
   
   $\text{tan}\varphi =\frac{c_1}{c_2}=\frac{2}{1}=2.$
   
   Note that both $c_1$ and $c_2$ are positive, so $\varphi$ is in the first quadrant. Thus,
   
   $\varphi \approx 1.107\text{rad,}$
   
   so we have
   
   $x(t)=2\text{cos}\left(3t\right)+\text{sin}\left(3t\right)=\sqrt{5}\text{sin}\left(3t+1.107\right).$
   
   The frequency is $\frac{\omega }{2\pi }=\frac{3}{2\pi }\approx 0.477.$ The amplitude is $\sqrt{5}.$
2. We have
   
   $A=\sqrt{c_1^2+c_2^2}=\sqrt{3^2+2^2}=\sqrt{13}$
   
   and
   
   $\text{tan}\varphi =\frac{c_1}{c_2}=\frac{3}{\mathrm{-2}}=-\frac{3}{2}.$
   
   Note that $c_1$ is positive but $c_2$ is negative, so $\varphi$ is in the fourth quadrant. Thus,
   
   $\varphi \approx -0.983\text{rad,}$
   
   so we have
   
   $\begin{array}{cc}\hfill x(t)& =3\text{cos}\left(2t\right)-2\text{sin}\left(2t\right)\hfill \\ & =\sqrt{13}\text{sin}\left(2t-0.983\right).\hfill \end{array}$
   
   The frequency is $\frac{\omega }{2\pi }=\frac{2}{2\pi }\approx 0.318.$ The amplitude is $\sqrt{13}.$

The mass stretches the spring 5 ft 4 in., or $\frac{16}{3}$ ft. Thus, $16=\left(\frac{16}{3}\right)k,$ so $k=3.$ We also have $m=\frac{16}{32}=\frac{1}{2},$ so the differential equation is

Multiplying through by 2 gives $x\text{″}+5x^{\prime }+6x=0,$ which has the general solution

Applying the initial conditions, $x(0)=0$ and $x^{\prime }(0)=\mathrm{-5},$ we get

After 10 sec the mass is at position

so it is, effectively, at the equilibrium position. We have $x^{\prime }(t)=10e^{\mathrm{-2}t}-15e^{\mathrm{-3}t},$ so after 10 sec the mass is moving at a velocity of

After only 10 sec, the mass is barely moving.