Question

QUESTION 1 a) Consider a flexible spring is suspended vertically from a rigid support and a mass m is attached to its free end. s and x are the amount of elongation and displacement respectively. The two parameters involved are constant of proportionality, k and acceleration of gravity, g. Derive the differential equation of free undamped motion using the appropriate physical laws. {Hint: Hooke's law and Newton's Second Law.} (5 marks)

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Answer 1

wwwmga (equilibriurs position) claim. To find the differential equation of free undamped motion Hooks lov! The restoring force of the spring propotional to the displacement and acts as oppoor fe direction from the displacement so therefor restoring force is given by f = - kx where K-constant x displacement. Newton end law of motion. The force acting on body is equal mass of body times accettation acting on it. lie, fama where m-mass a. accration to the Free undamped mohon: O no intesferoce @ no friction 6 no air distarbances. By hooks las restoring force kr.oh system is and by newtons and law of motion the force an system is equal to the f = mg - gravitational acch mg Fama a is acdartion ading and, mass & By hooks low total restoring force on system is equal to fs = -K (s+x) by Newton second low of motion, the sum of the forces on system is equal to the time accederation. total force on system is eq equal to the restoring force + gravitational force and this total force 16 16 equal to mass Hmes acceptation Q = -K (S+x) + mg Page No

But at becomes mg-ks therefore ean 0 ma = -ks-kor + ks. ma= -ke > ma+Kx=0 But a=dy de Vis Velocity and v= dy → a=d2x md²x tkx=0 d² +Kx=0 11 et² where to displacement *- spolna proportionality constant m-mass of object attached. dt?