Answer to Question #183814 in Mechanics | Relativity for Charissa Hall

Question

A particle executes simple harmonic motion with an amplitude of 1.57 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:

$$0.5mV^2=0.5kA^2=0.5m(0.5V)^2+0.5kx^2$$ $$0.5kx^2=0.5mV^2-0.5m(0.5V)^2=(0.75)0.5mV^2=(0.75)0.5kA^2$$ $$x=\frac{\sqrt{3}}{2}A=\frac{\sqrt{3}}{2}1.57= 1.36\ cm$$