## University Physics with Modern Physics (14th Edition)

(a) $A = 0.980~m$ (b) $\phi = 1.57~rad$ (c) $x(t) = (0.980)~cos(12.2~t+1.57)$
(a) If the spring is neither stretched nor compressed, then all the energy in the system is in the form of kinetic energy. We can find the amplitude. $\frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$ $A = \sqrt{\frac{m}{k}}~v_{max}$ $A = \sqrt{\frac{2.00~kg}{300~N/m}}~(12.0~m/s)$ $A = 0.980~m$ (b) If the spring is neither stretched nor compressed at $t=0$, we can say that $x = 0$ and the object is moving in the negative direction. Therefore, the basic cos-curve is shifted to the left an angle of $\frac{\pi}{2}$. Therefore, the phase angle $\phi = \frac{\pi}{2}~rad = 1.57~rad$. (c) We can find the angular frequency. $\omega = \sqrt{\frac{k}{m}}$ $\omega = \sqrt{\frac{300~N/m}{2.00~kg}}$ $\omega = 12.2~rad/s$ We can write the equation for the position. $x(t) = A~cos(\omega~t+\phi_0)$ $x(t) = (0.980)~cos(12.2~t+1.57)$