Question

A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 10 times the instantaneous velocity. Determine the equations of motion if the following is true. (a) the mass is initially released from rest from a point 1 meter below the equilibrium position

Answer 1

Answer:

x(t) = 7/4(e^-7t) - ¾(e^-3t)

Explanation:

Given

Let m represents the mass attached to the spring. m = 1kg

Represent the spring constant with k; k = 21N/m

Represent the positive damping constant with β; β = 10

From Newton second law for the system, we have

m.d²x/dt² = -kx - β.dx/dt --- divided through by m

d²x/dt² = -(k/m)x - (β/m)dx/dt

Suppose the system is in equilibrium

d²x/dt² + (k/m)x + (β/m)dx/dh = 0

Substitute in the values of m,k and β.

This gives

d²x/dt² + (21/1)x + (10/1)dx/dt = 0

d²x/dt² + 21x + 10dx/dt = 0

The auxiliary equation is

m² + 21m + 10 = 0

And the solution is

m = -7 and m = -3.

The general solution is then

x(t) = c1e^-7t + c2e^-3t

Given that

x(0) = 1 and x'(0) = 0

c1 + c2 = 1

-3c1 - 7c2 = 0

c2 = -¾

c1 = 7/4

x(t) = 7/4(e^-7t) - ¾(e^-3t)

Answer 2

Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = $-\beta\frac{dx}{dt} = \frac{-10dx}{dt}$

$\beta = 10$

By Newtons second law ,

The diffrential equation of motion with damping is given by

$m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}$

substitute the value of m =1kg, k = 21N/M, and $\beta = 10$

$1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}$

$\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0$

suppose the equation of the form $x =e^m^t$,

and the auxilliary equation is given by

$m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3$

The general solution for the above differential equation is

$x(t) =C_1e^{-3t}+C_2e^{-7t}$

Derivate with respect to t

$x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}$

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$