Answer
See answers.
Work Step by Step
The equation of motion is given as $x=0.25sin(4.70t)=A\;sin\omega t$.
a. The amplitude is the maximum x value, or 0.25 meters.
b. The frequency $f = \frac{\omega}{2 \pi}=\frac{4.70\;rad/s}{2 \pi}=0.748\;Hz$
c. The period $T =\frac{2 \pi}{\omega}=\frac{2 \pi}{4.70\;rad/s}=1.34\;s$
d. The total energy is equal to the kinetic energy at maximum speed.
$$E_T=\frac{1}{2}mv_{max}^2=\frac{1}{2}m(\omega\;A)^2=\frac{1}{2}(0.650\;kg)((4.70\;rad/s)\;0.25\;m)^2\approx 0.45\;J$$
e. First find the potential energy when x = 15 cm.
$$PE=\frac{1}{2}kx^2=\frac{1}{2}m\omega^2x^2=\frac{1}{2}(0.650\;kg)(4.70\;rad/s)^2(0.15\;m)^2= 0.1615\;J$$
The kinetic energy is the total energy minus the potential energy.
$$KE=0.4487\;J-0.1615\;J\approx 0.29\;J$$