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A brick of mass 8 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 4 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)= 4 cos(sqrt(980/3)t) cm

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Answer:

s(t) = 0.04cos(sqrt(980/3))

Step-by-step explanation:

Find spring constant. Use balance force between weight of brick and spring elastic force when brick at rest:

Weight of brick = force of spring

mg = kx

Mass brick, m = 8kg

Gravity constant, g = 9.8 m/s2

Spring elongation,x = 0.03m

Hence, spring constant, k= mg/x

= 8*9.8/0.03

= 7840/3

Since there is no other external forces, spring acts in simple harmonic motion

-kx = ma

a = -kx/m

note that a is the acceleration which is the double derivative of distance over time. Hence

d2y/dx2 = -kx/m

d2y/dx2 + kx/m = 0

Note that this equation is similar to simple harmonic motion:

d2y/dx2 + (w2)x = 0

Comparing these two equations we found:

w2 = k/m

Using the values obtained earlier:

w2 = (7840/3)/8 = 980/3

w = sqrt(980/3) = 18.07 rad/s

Since the movement of the spring will be sinusoidal, similar to the movement of pendulum, we use the general equation for oscillating motion:

s(t) = A sin (wt + c)

Note that in initial condition when t=0,

displacement of spring s = A = 0.04m

Hence 0.04 = 0.04 sin(0+c)

sin c = 1

c = π/2 - hence indicating that the motion is π/2 further than the equation.

Using trigonometry identity, we know that cos(theta) = sin(π/2 + theta)

So, we can change sin(wt+π/2) to cos(wt)

Updating the equation, we'll get

s(t) = 0.04 cos(wt).

w = sqrt(980/3)

Finally,

s(t) = 0.04cos(sqrt(980/3)) (in m)

Or

s(t) = 4cos(sqrt(980/3)) (in cm)

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