Question

A mass m is tied to an ideal spring with force constant k and rests on a frictionless surface. The mass moves along the x axis. Assume that x=0 corresponds to the relaxed position of the spring. The mass is pulled out to a position xm and released. Derive an expression for the positions at which the kinetic energy of the mass is equal to the elastic potential energy of the spring.

Express your answer in terms of Xm.

Answer 1

Initial potential energy = total mechanical energy

$(1/2) kX^2 = E$

at position x,

Kinetic energy , $K=(1/2) mv^2$

Potential energy , $U=(1/2) kx^2$

By conservation of energy

$K +U =E$

$K +U =(1/2)kX^2$

For K = U

$U+U =(1/2)kX^2$

$2U =(1/2)kX^2$

$2(1/2) kx^2 =(1/2)kX^2$

$x^2 =X^2/2$

$x =X/\sqrt{2}$ m and $x = -X/\sqrt{2}$ m