A brick of mass 2 kg hangs from the end of a spring. When the brick is at rest,

Question

A brick of mass 2 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.

(1 pt) A brick of mass 2 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 it for the displacement s(t) of the mass from its equilibrium position (with the spring stretched 3 cm) s(t) = Note that your answer should measure t in seconds and s in centimeters.) cm

Explanation / Answer

Let s(t) be the displacement from the equilbrium position , then

m d2/dt2 (s) = m g - k ( s + 3)

noting that mg = 3 k, we get

m d2/dt2 (s) = - k s , so

d2/dt2 s = -k/m s,

Let a = sqrt(k/m), and noting that mg = 3k

then k / m = g /3 = 980/ 3 = 326.66667, so a = sqrt(326.66667) = 18.0739 sec^-1

Therefore the solution of d2/dt2 s = - (18.0739)^2 s, is

s = A cos ( 18.0739 t + phi)

The initial conditions will determine A, and phi, at t = 0 , s(0) = 4 cm , and s'(0) = 0

so 2 = A cos (phi), and 0 = - 18.0739 A sin( phi), therefore phi = 0 and A = 4, so

s(t) = 4 cos(18.0739 t) (in centimeters)

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