Question

A 160 gg block hangs from a spring with spring constant 7.0 N/m . At t=0s the block is 20 cm below the equilibrium point and moving upward with a speed of 125 cm/s .

What is the block's oscillation frequency?

What is the block's distance from equilibrium when the speed is 39 cm/s ?

What is the block's distance from equilibrium at t=6.0s

Show work please.

Answer 1

m= 1609 K-7Nm at t=0; xo= 20cm ; Vo = 125 cmo using, a= A Cos (t+S); Putto 20 cm = A COOS - Now ,v= - A ce Suillettd) But to 125 m/= - A co find шу 04, 0 125 f = - 20 ale di d Tong (2) 125 - 20 o using kamwe = 6.6143 Tand put in eq (2) 125=-20x 6.6143 Tand= – 0.945 S=(-43.387 Put in 2on - A cos(-43.38) A=127.517 cm

е) к = m G 7 = 0.16 + 27 8² f=11.053 Hz coscillation frequency) V – A 6 si (6t+8) 39 ans =- 27.517 6.6143 Sun (at - 43.38) - 0.214279 = Sui (Gut - 43.38) - 0.216 redion = tot - 43.38. 43.164= 6.6143t t=6.52586 see At this time, X= A coletis) = 27.517 Cos (6.6143) 6.52586) -43.380) = /26.88 cm Put t= 6 see in 2. = A Cos (cott) 1.- (27.517) Cos (16.614346) - 43.28) = – 23.42 cm distance from equilibsium point (C)