Question 14.7: Energy calculations. For the simple harmonic oscillation of ......
Energy calculations. For the simple harmonic oscillation of Example 14-5, determine (a) the total energy, (b) the kinetic and potential energies as a function of time, (c) the velocity when the mass is 0.050 \mathrm{~m} from equilibrium, (d) the kinetic and potential energies at half amplitude (x= \pm A / 2).
APPROACH We use conservation of energy for a spring-mass system, Eqs. 14-10 and 14-11.
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)
E=\frac{1}{2} m v^{2}+\frac{1}{2} k x^{2} . (14-10c)
v = \pm \sqrt{\frac{k}{m}\left(A^{2}-x^{2}\right)} (14-11a)
v= \pm v_{\max } \sqrt{1-\frac{x^{2}}{A^{2}}} (14-11b)
(a) From Example 14-5, k=19.6 \mathrm{~N} / \mathrm{m} and A=0.100 \mathrm{~m}, so the total energy E from Eq. 14-10 a is
E=\frac{1}{2} k A^{2}=\frac{1}{2}(19.6 \mathrm{~N} / \mathrm{m})(0.100 \mathrm{~m})^{2}=9.80 \times 10^{-2} \mathrm{~J}
(b) We have, from parts (f) and (g) of Example 14-5, x=-(0.100 \mathrm{~m}) \cos 8.08 t and v=(0.808 \mathrm{~m} / \mathrm{s}) \sin 8.08 t, so
\begin{aligned}& U=\frac{1}{2} k x^{2}=\frac{1}{2}(19.6 \mathrm{~N} / \mathrm{m})(0.100 \mathrm{~m})^{2} \cos ^{2} 8.08 t=\left(9.80 \times 10^{-2} \mathrm{~J}\right) \cos ^{2} 8.08 t \\& K=\frac{1}{2} m v^{2}=\frac{1}{2}(0.300 \mathrm{~kg})(0.808 \mathrm{~m} / \mathrm{s})^{2} \sin ^{2} 8.08 t=\left(9.80 \times 10^{-2} \mathrm{~J}\right) \sin ^{2} 8.08 t\end{aligned}
(c) We use Eq.14-11b and find
v=v_{\max } \sqrt{1-x^{2} / A^{2}}=(0.808 \mathrm{~m} / \mathrm{s}) \sqrt{1-\left(\frac{1}{2}\right)^{2}}=0.70 \mathrm{~m} / \mathrm{s}
(d) At x=A / 2=0.050 \mathrm{~m}, we have
\begin{aligned}& U=\frac{1}{2} k x^{2}=\frac{1}{2}(19.6 \mathrm{~N} / \mathrm{m})(0.050 \mathrm{~m})^{2}=2.5 \times 10^{-2} \mathrm{~J} \\& K=E-U=7.3 \times 10^{-2} \mathrm{~J} .\end{aligned}