A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 2 inches above the equilibrium position. Find the equation of motion. (Use g

the equation of motion is

Explanation:

Given that,

The weight attached to the spring is 24pounds

Acceleration due to gravity is 32ft/s²

Assume x is the string length, 4inches

convert the length inches to to feet = 4/12 = 1/3feet

From Hookes law , we calculate the spring constant k

k = W / x

k = 24 / (1/3)

k = 24 / 0.33

k = 72lb/ft

If the mass is displace from its equilibrium position by amount x

the differential equation is

Auxiliary equation is

Thus the solution is,

The mass is release from rest

x'(0) = 0

Therefore

x(t) = c₁ cos4 √6t

x(0) = -2inches

c₁ cos4 √6(0) = 2/12feet

c₁= 1/6feet

There fore, the equation of motion is

massxgravity x heght=weightxheight= pe

4900=490xh

h=10