Question

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

Answer 1

Answer:

the equation of motion is

$x(t) =-\frac{1}{6} \cos4\sqrt{6} t$

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

$m\frac{d^2x}{dt^2} + kx=0\\\\\frac{3}{4 } \frac{d^2x}{dt^2}+72x=0$

$\frac{d^2x}{dt^2} +96x=0$

Auxiliary equation is

$m^2+96=0\\m=\sqrt{-96} \\m=\_ ^+4\sqrt{6}$

Thus the solution is,

$x(t) = c_1cos4\sqrt{6t} +c_2sin\sqrt{6t}$

$x'(t) =-4\sqrt{6c_1} sin4\sqrt{6t} +c_24\sqrt{6} cos4\sqrt{6t}$

The mass is release from rest

x'(0) = 0

$-4\sqrt{6c_1 } \sin4\sqrt{6} (0)+c_24\sqrt{6} \cos4\sqrt{6} (0)=0\\c_24\sqrt{6} =0\\\\c_2=0$

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

$x(t) =-\frac{1}{6} \cos4\sqrt{6} t$