Question

After a mass weighing 10 pounds is attached to a 5-foot spring, the spring measures 7 feet. This mass is removed and replaced with another mass that weighs 8 pounds. The entire system is placed in a medium that offers a damping force that is numerically equal to the instantaneous velocity. (a) Find the equation of motion if the mass is initially released from a point foot below the equilibrium position with a downward velocity of 1 ft/s. (b) Express the equation of motion in the form given in (23). (c) Find the times at which the mass passes through the equilibrium position heading dow

Answer 1

Answer:

a) $x(t) = e^(^-^2^t^)[ cos ( 4t ) + 0.75*sin ( 4t ) ]$

b) $x(t) = 1.25*e^(^-^a^t^)sin(4t + 0.64350)$

c) t = ( 0.7957*k - 0.160875 )s  with k = 0, 1 , 2 , 3 , ....

Explanation:

Declaring variables:-

- The mass = m

- Positive damping constant = β

- Spring constant = k

- Gravitational constant, g = 32.2 ft/s^2

Given:-

- The attached initial weight, Wi = 10 lbs

- The attached second weight, Wf = 8 lbs

- Length of un-extended spring, Li = 5 ft

- The extended length, Lf = 7 ft

Find:-

(a) Find the equation of motion if the mass is initially released from a point foot below the equilibrium position with a downward velocity of 1 ft/s.

(b) Express the equation of motion in the form given in (23)

(c) Find the times at which the mass passes through the equilibrium position heading dow

Solution:-

- We will first evaluate the constants m , k and β using conditions given:

                         m = Wf / g

                         m = 8 / 32 = 0.25 slugs.

- We will use the equilibrium condition in (vertical direction) on the spring when the weight of Wi = 10 lbs was attached onto the spring. The spring restoring force (Fs) acts up while weight attached combats it by pointing downward.

                         Fs - Wi = 0

- From Hooke's Law we have:

                         Fs = Wi = k*ΔL   .......  ΔL : Extension of spring

                         k = W / ΔL = W / ( Lf - Li )

                         k = 10 / ( 7 - 5 )

                         k = 5 lb/ft

- Since we have insufficient information about the damping constant of the medium we will assume it as unit. β = 1.

- Now, we will consider the dynamic motion of the spring attached with mass with weight Wf damped in a medium with constant β. We will use Newton's second equation of motion for the spring.

                        F_net = m*a

Where, F_net : Net force acting on the attached mass

            a : Acceleration of the block

- There are two forces acting on the spring ( Damping force - D and restoring force of spring (ks) ). We will consider a displacement of mass in vertical direction as (x).

                       - D - Fs = m*a

- Using hooke's law and damping force (D) is proportional to velocity of attached mass. We have:

                       - β*dx/dt - k*x = m*( d^2 x / dt^2)

 - Plug in the constants:

                          $\frac{d^2x}{dt^2} + 4*\frac{dx}{dt} + 20*x = 0$

- Now solve the derived ODE. The Auxiliary equation for the above ODE is:  

                         s^2 + 4s + 20 = 0              

- Solve the quadratic and evaluate roots.

                        s = -2 +/- 4i ....... (Complex Roots)

- The complementary solution (yc) for complex roots of the auxiliary is:

                        $xc (t) = e^(^-^2^t^)[ A*cos ( 4t ) + B*sin ( 4t ) ]$

- Use the given initial conditions and evaluate constants A and B :

                       x ( 0 ) = 1 ft , x ' (0) = 1 ft/s

                       $xc (0) = e^(^0^)[ A*cos ( 0 ) + B*sin ( 0 ) ] = 1* [ A + 0 ] = 1\\\\A = 1\\\\xc'(t) = -2*e^(^-^2^t^)[ A*cos ( 4t ) + B*sin ( 4t ) ] + e^(^-^2^t^)[ -4A*sin ( 4t ) + 4B*cos ( 4t ) ]\\\\xc'(0) = -2*e^(^0^)[ cos ( 0 ) + B*sin ( 0 ) ] + e^(^0^)[ -4sin ( 0 ) + 4B*cos ( 0 ) ]\\\\ 1 = -2*[ 1 + 0 ] + 1*[ 0 + 4B]\\\\B = 3 / 4 = 0.75$

- The complementary solution becomes:

                         $xc (t) = e^(^-^2^t^)[ cos ( 4t ) + 0.75*sin ( 4t ) ]$

- Since there is no excitation force acting on the system( Homogenous ). The particular solution does not exist and the general solution to equation of motion is:

                         $x(t) = e^(^-^2^t^)[ cos ( 4t ) + 0.75*sin ( 4t ) ]$

- An alternative form of the displacement (x) - Time (t) motion is:

                        $x(t) = Pe^(^-^a^t^)sin(w_dt + theta)$

Where,

                        P = √ ( A^2 + B^2)   .... Amplitude

                        theta (θ) = arctan ( B / A )

                        Auxiliary roots = ( a +/- bi) ..... a = -2 , b = 4  

                        wd = w*√ 1 - ρ^2 = w = b = 4 .......... (β = 1 => ρ = 0)

- Evaluate P and theta (θ):

                        P = √ ( 1^2 + 0.75^2) = 1.25 m

                        theta (θ) = arctan ( 3 / 4 ) = 0.64350 rads

- The alternative form of Equation (23) is:

                        $x(t) = 1.25*e^(^-^a^t^)sin(4t + 0.64350)$

c)

- To determine the range of times (t) when the mass passes the the equilibrium position can be evaluated by setting x(t) form (23) equal to 0.

                       $x(t) = 1.25*e^(^-^a^t^)sin(4t + 0.64350) = 0\\\\e^(^-^a^t^) \neq 0 , sin(4t + 0.64350) = 0\\\\4t + 0.64350 = 0 , \pi , 2\pi , 3\pi = k\pi\ \ .......\ \ (k = 0, 1 , 2 , 3 , ...)\\\\t = \frac{k\pi - 0.64350}{4} = k*0.78571429 - 0.160875$

- The domain for time (t) is as follows:

                       t = ( 0.7957*k - 0.160875 )s  with k = 0, 1 , 2 , 3 , ....