Answer
(a) The amplitude is 0.055 meters.
(b) The maximum speed is 0.59 m/s
Work Step by Step
(a) The total energy during the motion is constant. We can find the amplitude of the motion as|:
$\frac{1}{2}kA^2 = \frac{1}{2}kx^2+\frac{1}{2}mv^2$
$A^2 = x^2+\frac{mv^2}{k}$
$A = \sqrt{x^2+\frac{mv^2}{k}}$
$A = \sqrt{(0.020~m)^2+\frac{(2.7~kg)(0.55~m/s)^2}{310~N/m}}$
$A = 0.055~m$
The amplitude is 0.055 meters.
(b) We can find the maximum speed as:
$v_{max} = \sqrt{\frac{k}{m}}~A$
$v_{max} = \sqrt{\frac{310~N/m}{2.7~kg}}~(0.055~m)$
$v_{max} = 0.59~m/s$
The maximum speed is 0.59 m/s.