Question

A simple harmonic oscillator consists of a block of mass 2.50 kg attached to a spring of spring constant 350 N/m. When t = 0.880 s, the position and velocity of the block are x = 0.156 m and v = 2.950 m/s. (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t=0 s? (a) NumberT.294 Unitsm (b) NumberT.249 Unitsm (c) Number 3.23 Units m/s

Answer 1

0.880 Bee. biveni m= 2.soky, Ka 3 SON/m, t = V=2.95omis Y = 0.156 m, in SHm the displacement and velocita in A cos (wtto) du dt - Asiw sin (wt + o) a une both equation -W tun (wt+ o) wt +$ = tan la here wa 350 2.5 =11.83 rad/s substitute values -2.950 (11.83) (0.880) + $ "G (11.93, 601533) (I rad) - (10.41 rad) =-11.41 rad from equation .156 (-11.4157 (os (wtto) cos (11.83 x0.880 +(-11. A=0.289 m teo if equo n = A cos (2(0)+4) = (0.289) (as (-11.urrad) •116 m ean - Aw sin&w (o)+8) (0.289) (11.63) sin ( - 11.41 Tad) V= 3.13 mis Ics canned with Camscanner