Question

A frictionless spring with a 4-kg mass can be held stretched 0.2 meters beyond its natural length by a force of 30 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1 m/sec, find the position of the mass after t seconds.

Answer 1

Answer:

The position for mass is $x(t)=36.75sin(37.5t)$

Explanation:

Let x(t) donate the position of mass at time t,Then x satisfies the differential equation

$m\frac{d^2x}{dt^2} +kx=0\\here\\m=4kg\\k=30N/0.2m\\thus\\w^2=k/m=30N/0.2m*4kg=37.5$

The general solution is

$(x)t=C_{1}cos(37 .5t)+C_{2}sin(37.5t)$

It follows

$x'(t)=-37.5C_{1}sin(37.5t)+37.5C_{2}cos(37.5t)\\now\\x(0)=0\\gives\\C_{1}=0\\ and\\x'(0)=1m/s\\gives\\C_{2} =36.75$

thus the position of mass is

$x(t)=36.75sin(37.5t)$