Explanation Formula used: The governing equation of motion is m d 2 y d t 2 + δ d y d t + k y = 0 . The roots of the quadratic equation a x 2 + b x + c = 0 is calculated as x = − b ± b 2 − 4 a c 2 a . The Hooks Law is as F = k x . Theorem used: If r 1 = α + j β and r 1 = α − j β are the two complex roots of the auxiliary equation a r 2 + b r + c = 0 , then y = e α x ( c 1 cos β x + c 2 sin β x ) is the general solution of a y ″ + b y ′ + c y = 0 . Calculation: The mass of the spring is 0.820 kg with initial stretch( x ) is 0.250 m and the mass is pulled down is 0.150 m below the equilibrium position and then released. From the Hooks Law, F = k x . Where k is spring constant, x is displacement of spring and F is force applied on the spring. Substitute the value of m , g and x in F = k x . F = k x m g = k x 0.820 × 9.8 = k ( 0.250 ) k = 32.144 So, the differential equation governing the problem as follows: m d 2 y d t 2 + δ d y d t + k y = 0 . If there is no damping means δ = 0 . Substitute the value of m, δ and k in the above equation. 0.820 D 2 x + 32.144 x = 0 The auxiliary equation is: 0.820 m 2 + 32.144 = 0 m 2 = − 32.144 0.82 m 2 = 39.20 m = ± 6.26 j So, the general solution to the above differential equation is x = e 0 t ( c 1 sin 6

11th Edition

ISBN: 9780134508290

Author: Evans

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 31.10, Problem 12E

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The equation of motion of the mass if there is no damping.

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Chapter 31.10, Problem 11E

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