1 \begin{problem*}{12.18}
2 A $2.00\U{kg}$ object is attached to a spring and placed on a
3 horizontal, smooth surface. A horizontal force of $20.0\U{N}$ is
4 required to hold the object at rest when it is pulled $0.200\U{m}$
5 from its equilibrium position (the origin of the $x$ axis). The
6 object is now released from rest with an initial position of
7 $x_i=0.200\U{m}$, and it subsequently undergoes simple harmonic
8 oscillations. Find \Part{a} the force constant of the
9 spring, \Part{b} the frequency of oscillations, and \Part{c} the
10 maximum speed of the object. Where does this maximum speed
11 occur? \Part{d} Find the maximum acceleration of the object. Where
12 does it occur? \Part{e} Find the total energy of the oscillating
13 system. Find \Part{f} the speed and \Part{g} the acceleration of the
14 object when its position is equal to one third of the maximum value.
25 Surface surf = Surface(pFrom=(-.7u,-a/2), pTo=(2u,-a/2));
26 Block b = Block(center=(0,0), width=a);
27 Spring spring = Spring(pFrom=(b.center+(a/2,0)), pTo=(2u,0));
28 Vector Fspring = Force(b.center, mag=2a, dir=0, L="$F_s$");
29 Vector Fexternal = Force(b.center, mag=2a, dir=180, L="$F_e$");
39 The external force $F_e$ must exactly balance the spring force $F_s$
40 to hold the object at rest on a frictionless surface, so
42 |F_s| &= k|x| = |F_e| \\
43 k &= \p|{\frac{F_e}{x}}| = \ans{100\U{N/m}} \;.
47 The frequency of oscillation is then
49 f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{k}{m}}
50 = \ans{1.13\U{Hz}} \;,
54 \omega = 2 \pi f = 7.07\U{rad/s} \;.
58 The maximum speed of the object is (see 12.5)
60 v_\text{max} = \omega A = 7.07\U{rad/s} \cdot 0.200\U{m}
61 = \ans{1.41\U{m/s}} \;,
63 which occurs at $x=0$, when all the oscillation energy is kinetic.
66 The maximum acceleration of the object is (see 12.5)
68 a_\text{max} = \omega^2 A = (7.07\U{rad/s})^2 \cdot 0.200\U{m}
69 = \ans{10\U{m/s$^2$}} \;,
71 which occurs at $x=-A=-0.200\U{m}$. If you are only interested in the
72 peaks in the \emph{magnitude} of the acceleration, they occur for
76 Lets find the energy in the initial situation, right after the object
77 was released. It's at rest, so its kinetic energy is $0$ and all the
78 enegy is spring-potential energy
80 E = \frac{1}{2} k A^2 = \ans{2.00\U{J}}
86 E &= \frac{1}{2} k \p({\frac{A}{3}})^2 + \frac{1}{2} m v^2
87 = \frac{E}{9} + \frac{1}{2} m v^2 \\
88 v &= \pm\sqrt{\frac{2}{m} \cdot \frac{8}{9}E} = \ans{\pm1.33\U{m/s}} \;.
92 This is very similar to Problem 12.15 \Part{c}.
95 a &= \frac{k}{m}x = \pm \frac{kA}{3m} = \ans{\pm3.33\U{m/s$^2$}} \;.