Answered: A block of mass= 1 is sitting on a…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

A block of mass= 1 is sitting on a table and is attached to a spring of strength k=5 so that it can slide horizontally on the table. the coefficient of linear friction between the block and the table is b=2, and an external force of F(t)= 13cos(3t) acts on it. Find the general solution to this differential equation, and determine if the spring-mass system is over-damped, critically damped, or underdamped.

Expert Solution

Step 1

The differential equation governing the spring-mass system is

$m \frac{d {}^{2}x}{d t^{2}} + b \frac{d x}{d t} + k x = F (t)$

where, $F (t)$ is the external force

$b$ is the coefficient of friction

$k$ is the spring constant

$m$ is the mass

Here Mass, $m = 1$

Spring constant, $k = 5$

Coefficient of friction, $b = 2$

External force , $F (t) = 13 \cos (3 t)$

Step 2

The non-homogeneous differential equation governing the given spring-mass system is

$\frac{d {}^{2}x}{d t^{2}} + 2 \frac{d x}{d t} + 5 x = 13 \cos (3 t)$

Put $D^{2} = \frac{d^{2}}{d t^{2}}, D = \frac{d}{d t}$

Then the differential equation becomes, $(D^{2} + 2 D + 5) x = 13 \cos (3 t)$

Then the associated characteristic equation is $λ^{2} + 2 λ + 5 = 0$

$λ = - 1 \pm i 2$

Since the roots are complex conjugate, the spring mass system is underdamped.

The complementary function has the form $C . F = e^{- t} (c_{1} \cos (2 t) + c_{2} \sin (2 t)) . . . . . . . . . . . . . . . (1)$

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