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=(x_m)/(√(2))`

Step-by-step explanation:

Given

initially mass is stretched to
`x_m`

Let k be the spring Constant of spring

Therefore Total Mechanical Energy is
`(kx_m^2)/(2)`

Position at which kinetic Energy is equal to Elastic Potential Energy

`K=(mv^2)/(2)`

`U=(kx^2)/(2)`

it is given

`k=U`

thus
`2U=(kx_m^2)/(2)`

`2* (kx^2)/(2)=(kx_m^2)/(2)`

`2x^2=x_m^2`

`x=(x_m)/(√(2))`