A brick of mass 6 kg hangs from the end of a spring When the

A brick of mass 6 kg hangs from the end of a spring. When the brick is at rest, the spring is stretched by 9 cm. The spring is then stretched an additional 2 cm and released. Assume there is no air resistance. Note that the acceleration due to gravity, g, is g = 980 cm/s^2. Set up a differential equation with initial conditions describing the motion and solve it for the displacement s(t) of the mass from its equilibrium position (with the spring stretched 9 cm). (Note that your answer should measure t in seconds and s in centimeters.)

Solution

initial condition

u(t) = 9 initial displacement

u\'(t) = 0 initial velocity

the equation is :

mu\'\' +yu\' +ku =0

F= kx => mg = kx

=> k = mg / x = 6*900/9 =600

hence equation is : 6u\'\'+ y*0 +600 u = 0

=> 6r^2 +6r = 0

=> 6r(r+100) =0 =>r=0 , r=-100

hence u(t) = Asin (-100)t +Bcos(-100t)

plugging in initial conditions

at t =0 , u(t) = 9

9 = 0+B =>B= 9,

A =0

Final equation is

u(t) = 9cos(-100)t