Answer
See answers.
Work Step by Step
a. At equilibrium, the speed is at a maximum.
$$v_{max}=\omega A = 2\pi fA = 2 \pi (2.2\;Hz)(0.15m)\approx 2.1\;m/s$$
b. Equation 11-5b tells us the velocity at that position.
$$v=v_{max}\sqrt{1-\frac{x^2}{A^2}}$$
Evaluating, we find 1.5 m/s.
c. The total energy is given by the kinetic energy when the speed is highest:
$$E_{total}=\frac{1}{2}mv^2_{max}$$
$$=\frac{1}{2}(0.25\;kg) (2.073\;m/s)^2\approx0.54\;J$$
d. The object has a maximum displacement at t=0. The position is described by a cosine function.
$$x = (0.15\;m)cos(2\pi(2.2\;Hz)t)$$
$$x = (0.15\;m)cos(4.4\pi t)$$
