Question 14.6: Spring is started with a push. Suppose the spring of Example......
Spring is started with a push. Suppose the spring of Example 14-5 is compressed 0.100 \mathrm{~m} from equilibrium \left(x_{0}=-0.100 \mathrm{~m}\right) but is given a shove to create a velocity in the +x direction of v_{0}=0.400 \mathrm{~m} / \mathrm{s}. Determine (a) the phase angle \phi,(b) the amplitude A, and (c) the displacement x as a function of time, x(t).
APPROACH We use Eq. 14-8a, at t=0, to write v_{0}=-\omega A \sin \phi, and Eq. 14-4 to write x_{0}=A \cos \phi. Combining these, we can obtain \phi. We obtain A by using Eq. 14-4 again at t=0. From Example 14-5, \omega=8.08 \mathrm{~s}^{-1}.
v =\frac{d x}{d t}=-\omega A \sin (\omega t+\phi) (14-8a)
x=A \cos (\omega t+\phi), (14-4)
(a) We combine Eqs. 14-8a and 14-4 at t=0 and solve for the tangent:
\tan \phi=\frac{\sin \phi}{\cos \phi}=\frac{\left(v_{0} /-\omega A\right)}{\left(x_{0} / A\right)}=-\frac{v_{0}}{\omega x_{0}}=-\frac{0.400 \mathrm{~m} / \mathrm{s}}{\left(8.08 \mathrm{~s}^{-1}\right)(-0.100 \mathrm{~m})}=0.495 .
A calculator gives the angle as 26.3^{\circ}, but we note from this equation that both the sine and cosine are negative, so our angle is in the third quadrant. Hence
\phi=26.3^{\circ}+180^{\circ}=206.3^{\circ}=3.60 \,\mathrm{rad} .
(b) Again using Eq. 14-4 at t=0, as given in the Approach above,
A=\frac{x_{0}}{\cos \phi}=\frac{(-0.100 \mathrm{~m})}{\cos (3.60 \mathrm{rad})}=0.112 \mathrm{~m} .
(c) x=A \cos (\omega t+\phi)=(0.112 \mathrm{~m}) \cos (8.08 t+3.60).