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

Step-by-step explanation:

Given

Maximum displacement A=0.49

The sum of kinetic and elastic potential energy is
`(1)/(2)kA^2`

where k=spring constant

U+K.E.=
`(1)/(2)kA^2`

when K.E.=U/2

K.E.=kinetic energy

U=Elastic potential Energy

`\rightarrow \ U+(U)/(2)=(1)/(2)KA^2\\\rightarrow \ (3U)/(2)=(1)/(2)KA^2\\\rightarrow \ U=(1)/(3)KA^2\\\rightarrow \ (Kx^2)/(2)=(1)/(3)KA^2\\\\x=\sqrt{(2)/(3)}A\\x=0.4\ m`