Question

the maximum displacement of an oscillatory motion is A=0.49m. determine the position x at which the kinetic energy of the particle is half it's elastic potential energy? (if K.E = U/2 __ x = ?)

Answer 1

Answer:0.4 m

Explanation:

Given

Maximum displacement A=0.49

The sum of kinetic and elastic potential energy is $\frac{1}{2}kA^2$

where k=spring constant

U+K.E.=$\frac{1}{2}kA^2$

when K.E.=U/2

K.E.=kinetic energy

U=Elastic potential Energy

$\rightarrow \ U+\frac{U}{2}=\frac{1}{2}KA^2\\\rightarrow \ \frac{3U}{2}=\frac{1}{2}KA^2\\\rightarrow \ U=\frac{1}{3}KA^2\\\rightarrow \ \frac{Kx^2}{2}=\frac{1}{3}KA^2\\\\x=\sqrt{\frac{2}{3}}A\\x=0.4\ m$