Question 14.8: Doubling the amplitude. Suppose the spring in Fig. 14-10 is ......

(a) From Eq. 14-10a, the total energy is proportional to the square of the amplitude A, so stretching it twice as far quadruples the energy \left(2^{2}=4\right). You may protest, “I did work stretching the spring from x=0 to x=A. Don’t I do the same work stretching it from A to 2 A ?” No. The force you exert is proportional to the displacement x, so for the second displacement, from x=A to 2 A, you do more work than for the first displacement (x=0 to A). (b) From Eq. 14-10b, we can see that when the energy is quadrupled, the maximum velocity must be doubled. \left[v_{\max } \propto \sqrt{E} \propto A\right]. (c) Since the force is twice as great when we stretch the spring twice as far, the acceleration is also twice as great: a \propto F \propto x.

E=\frac{1}{2} m(0)^{2}+\frac{1}{2} k A^{2}=\frac{1}{2} k A^{2} (14-10a)

E=\frac{1}{2} m v^{2}+\frac{1}{2} k(0)^{2}=\frac{1}{2} m v_{\max }^{2}, (14-10b)