Question

A mass of 4kg stretches a spring 40cm. Suppose the mass is displaced an additional 12cm in the positive (downward) direction and then released. Suppose that the damping constant is 3 N⋅s/m and assume g=9.8m/s2 is the gravitational acceleration. (a) Set up a differential equation that describes this system. Let x t

Answer 1

Answer:

A differential equation is $4x''+3x'+98x=0$.

Explanation:

Given that,

Mass = 4 kg

Stretch string = 40 cm

Additional distance = 12 cm

Damping constant = 3 N-s/m

Let xx to denote the displacement, in meters, of the mass from its equilibrium position, and give your answer in terms of x,x′,x′′ .

We need to calculate the spring constant k

The net force in y direction at equilibrium position

$F_{y}=0$

$mg-kx=0$

Put the value into the formula

$4\times9.8-k\times40\times10^{-2}=0$

$k=\dfrac{4\times9.8}{40\times10^{-2}}$

$k=98\ N/m$

The initial displacement from equilibrium

$x(0)=12\ cm$

The initial velocity is

$v(0)=0$

We need to set up a differential equation

The net force in y direction is zero at equilibrium position .

$\Sum F_{y}=0$

$mx''+cx'+kx=0$

Put the value into the equation

$4x''+3x'+98x=0$

Hence, A differential equation is $4x''+3x'+98x=0$.