Question

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/mN/m. At t=0t=0 the block has velocity -4.00 m/sm/s and displacement +0.200 mm.

DO 849 O ENS 12:12 PM E + - X Hoe 192. How CAR Ins In yen E Shi Exer $ 100 http WHO Tryit 192. Hel tạc ở openvellum. ollege.com/course.htmliourseld-150112928 D7 Bookmarks Adu New full iman impul de la puy de My ASU 20974450552384175N 10001 Ouh burukuks Physics 210 Summer 2020 Hochary Sign Out 1 Help Mastering Physics Course Home My Courses <Ch 14 HW Course Home Exercise 14.14 12 of 15 > Syllabus Constants Scores Correct A2.00 kg, frictionless block attached to an ideal spring with force constant 300 Nm. Att - the block has velocity 4.00 m/s and displacement 0.200 m e Text Study Area Part B Document Sharing a 102 User Settings Previous Answers Course Tools Correct - Part 6 Write an equation for the position as a tunction of time Assume (l) in meters and in seconds. VAXT ? (0) - Submit Previous Answers Best X Incorrect Try Again: 2 attempts remaining Provide Feedback Next >

Answer 1

Im=2kg smass of object. x=0 Ik=300N/m x=0.2m 3 Spring constant 311) 300 2 ..Angular frequency is w= Irad/s w=12-24745 rad's At tao, X.20.2m and velocity is V=-4 m/s If amplitude of oscilation is A then, v² w² w? (A? (A²-x²) R= 1300 (32)x(43–1027 = AP-0.04 - A210.10664 +0.04) > ²= 0.14667 > A= Jo. 14667 m > A=0.383m » 16 150

dt Equation of S.H-M is x= A costelt to velocity is v=de V=-law) sen(celt+0) At t=0, x=+0.2m and U=.-4m/s :: 0.2=0.383 cos (O+)=0.383coso >> cos$= *= cost ..^=1.02 rad -> Initial phase. - Equation of SH.m.is, x= 16.383683/12.24745++1:02Jm 0.2 0.383 0.2 0.383