Question

A brick of mass 4 kg hangs from the end of a spring. When the brick is at rest, the spring is stretched by 3 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/s2. 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 3 cm). s(t) = cm(Note that your answer should measure t in seconds and s in centimeters.)

Answer 1

Answer:

Step-by-step explanation:

In this system we have the force of the spring and the gravitational force. The equation that describes that is

$F_{s}+F{g}=ma\\k(y_0+y)-mg=ma$

where y0 is the equilibrium position when the string is free and y0+y is the new equilibrium position when the object is hanged of the string. By replacing by derivatives we have

$ky_0+ky-mg=ma\\mg+ky-mg=ky=ma\\\\ky=m\frac{d^2y}{dt^2}\\\\my''+ky=0$

the solution for this differential equation is (by using the characterisic polynomial)

$y(x)=Acos(kt)+Bsin(kt)\\k=\omega^2m$

hope this helps!!

Answer 2

Answer:

A) d²x/dt² = -326.67x

At time t = 0,the spring is stretched to 5cm and and released.

Thus, the initial velocity is zero. Thus, the initial condition is;

x(0) = -5 and x'(0) = 0

B) The solution to the equation is;

x(t) = -5cos 18.07t

Step-by-step explanation:

A) Due to stretching, the force due to gravity and that due to stretching of spring will be the same. Thus;

mg = -kx

m = 4kg ; x = -3cm and g=980 cm/s²

Thus,

k = mg/x = 4x980/3 = 1306.67

The motion of mass attached to a spring is given as;

d²x/dt² = -ω²x

Where, ω² = k/m = 1306.67/4 = 326.67

Thus, the differential equation for the problem would be;

d²x/dt² = -326.67x

At time t = 0,the spring is stretched to 5cm and and released.

Thus, the initial velocity is zero. Thus, the initial condition is;

x(0) = -5 and x'(0) = 0

B) The general solution would be given by;

x(t) = C1cos ωt + C2sinωt

From earlier, ω² = 326.67

So, ω = √326.67 = 18.07

Thus,

x(t) = C1cos 18.07t + C2sin 18.07t

Now, applying initial conditions ;

At x(0) = -5;

-5 = C1cos 0 + C2sin 0

C1 = -5

x'(t) = -18.07C1Sin18.07t + 18.07C2Cos18.08t

Thus, at x'(0) = 0

0 = -18.07C1Sin 0 + 18.07C2Cos0

18.07C2 = 0

So,C2 = 0

Thus,the solution to the equation is;

x(t) = -5cos 18.07t