Question

A particle executes simple harmonic motion
with an amplitude of 2.69 cm.
At what positive displacement from the
midpoint of its motion does its speed equal
one half of its maximum speed?
Answer in units of cm.

Answer 1

From the conservation of energy:

$$0.5mV^2=0.5kA^2=0.5m(0.5V)^2+0.5kx^2$$

$$0.5kx^2=0.5mV^2-0.5m(0.5V)^2=(0.75)0.5mV^2=(0.75)0.5kA^2$$

$$x=\frac{\sqrt{3}}{2}A=\frac{\sqrt{3}}{2}2.69=2.33\ cm$$