Answer
a. 0.06 m
b. 7.1
Work Step by Step
a. The bug moves in simple harmonic motion. The amplitude is half of the peak to peak motion, or 0.06 m.
b. The maximum KE is proportional to the square of the amplitude.
$$\frac{KE_f}{KE_i}=\frac{0.5kA_f^2}{0.5kA_i^2}=(\frac{0.16m}{0.06m})^2=7.1$$
The KE changed by a factor of about 7.1.
