A spring has natural length 075 m and a 5 kg mass A force of

Solution

From Hooke’s Law, the force required to stretch the spring is
k(0.25) = 25
so k = 100 (Divide the 25 by .25)
Using this value of the spring constant , together with m= 5
we use the Equation
m(d²x/dt²⁾ + Kx=0

Plug in all values

5(d²x/dt²) + 100 = 0

This is a second-order linear differential equation. Its auxiliary equation is mr² + k =0  with roots r= +/- wi, where w=(k/m). Thus, the general solution is x(t)= c1cos(w)t + c2 sin(wt)

In our problem this would mean the auxiliary equation is 5r² + 100 = 0 with roots r = ±25 i, so the general solution to the differential equation is x(t) = c₁cos(25 t) + c₂ sin(25 t).We are given that x(0) = 0.35 c₁ = 0.35 and x\'(0) = 0 25 c₂ = 0 c₂ = 0, so the position of the mass after t seconds is x(t) = 0.35 cos(25 t).