Question

An object with mass 3.9 kg is executing simple harmonic motoon, attached to a spring with spring constant 250 N/m. When the object is 0.018 m from its equilibrium position, it is moving with a speed of 0.50 m/s.

A) Calculate the amplitude of the motion

B) Calculate the maximum speed attained by the object

An object with mass 3.9 kg is executing simple harmonic motion, attached to a spring with spring constant 250 N/m. When the object is 0.018 m from its equilibrium position, it is moving with a speed of 0.50 m/s

Thank you!

Answer 1

Page No. Date Ans given: m=3.9 kg K= 250 N/m x = 0.018 m v = 0.50 m/s motton is vex) = K (A²-x²) . We know that velouty in simple harmonie (A?-X22 or taking square in both sides V(x) = K (A x²] (0.507* = K (A2-(0.01832) $ 0.950) = 250 (42= (0.01892) 9 CĂ?0.0182) = (0:50)x3.9 250 7 À 0.000324 = 0.0039 - A: 3: 00 39+Ố: 010324 ZA = 0.004224 -- A=.06499 m A +0.06499m a to.0650m So amplitude of the motion is 0.064g No 10650m.

Page No. Page No. 69 - - maximum speed attained by the object mex = AW Vmey. = 0.0650 x WIT E = 1250 = [64.10 = 8.006 ved I see So Vmax = 0.0650x8.006 max = 0.52039 m/s mon - 0.52 m/s maximum speed of the object can be 0.52 m/s-
Thank you so much