Question

PLEASE DON'T ANSWER PREVIOUSLY ASKED QUESTION, TRIED DELETING IT BUT THAT DIDN'T WORK. (Question about: Consider two masses, both with mass M, attached to a spring with spring constant k. They slide along angled rails, and the angle between the rails is theta...)There is no friction: the masses slide freely along the rails. Assume that the masses move together so that the spring remains parallel to its equilibrium position. The masses are initially moving upwards such that the spring is being stretched past its equilibrium length. Describe what happens next, by using Newton's second law to identify the equation of motion, solving that equation to determine the motion of this system, and then describing your result in words. (Do not consider gravity in this question, consider the masses and the spring lying flat on a horizontal surface) PLEASE DON'T ANSWER THIS QUESTION AND DELETE IT INSTEAD.

Answer 1

the angle between spring and rail  $\Phi$ =  $(180-\Theta )/2$ =   $90-\Theta /2$ (angle in a isosceles triangle)

for the displacement of x along the rail the expansion in spring y is

$y=2x cos(90-\theta/2)$

  $y=2x sin(\theta/2)$  

서

external force upon mass due to spring along the rail =  $ky sin(\theta/2)$

  $F=k(2xsin(\theta/2))*sin(\theta/2)$

  $Ma=2kxsin(\theta/2)^{2}$

$a=[(2ksin(\theta/2)^{2})/M]*x$

$a=\gamma^{2} *x$

where $[\gamma^{2} =(2k(sin\Theta/2)^{2})/M]$

by simple harmonic motion we know that the solution is $x(t)=Asin(\omega t+\Lambda ) ....where -->\Lambda=phase..difference -->\omega =\gamma$

hence equation of motion of masses is given by

$x(t)=A*sin([((2k/M)^{1/2})sin(\Theta /2)]t+\Lambda )$