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.

Answer 1

Answer:$x=\frac{x_m}{\sqrt{2}}$

Explanation:

Given

initially mass is stretched to $x_m$

Let k be the spring Constant of spring

Therefore Total Mechanical Energy is $\frac{kx_m^2}{2}$

Position at which kinetic Energy is equal to Elastic Potential Energy

$K=\frac{mv^2}{2}$

$U=\frac{kx^2}{2}$

it is given

$k=U$

thus $2U=\frac{kx_m^2}{2}$

$2\times \frac{kx^2}{2}=\frac{kx_m^2}{2}$

$2x^2=x_m^2$

$x=\frac{x_m}{\sqrt{2}}$