Question

A mass of 0.5 kg is attached to a horizontal spring with a spring constant of 8.0 N/m. The spring is stretched and released and it is found that the velocity of the mass is -0.204 m/s and the acceleration of the mass is +0.907 m/s2 at time t= 0.14 s. + 1. Write an expression for the position of the mass as a function of time. 2. What is the position of the mass and which direction is it moving at t=0?- 5. Sketch a graph of the position of the mass as a function of time.

Answer 1

ball w brojem por mm ml streebed coral? Given Force m = 0.5kg ke sam on mass m i f = akon - man & Kno. at ata 1. 2. - dom +KMEO of a won to where wenskim The solution of this differential en is given by a = A sin(wita) where A mand displacement from equilibrchem poc? ore amplitude. At t=0, it is released from extrome point on at an it will move towards me an posto . Cas shown in figur At 1-0 mA 3 A = A & sin ( wxo + 0) 2) sing= 1 = 12 Thus n = Asin (wtt W2 = A cos ut

WS Velocity at line t v acce feration at time t am a do All - Aweet - Awo cocent stwo w = 5 km = = 4 when t= 0.14, Ur 201m's - Ax Yg einluxo.14) F-1204 -) A .204 - 204 = 5.2m 0.097 on untretched roc? At 4:0 o n = 5.2m n = 5.2m t.e . e from from point 'o ( pol. of mass we t graph,