Question

A 0.600 kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.0 cm. (Assume the position of the object is at the origin at t = 0.) (a) Calculate the maximum value (magnitude) of its speed and acceleration. (b) Calculate the speed and acceleration when the object is 5.00 cm from the equilibrium position. (c) Calculate the time interval required for the object to move from x = 0 to x = 5.00 cm. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.

Answer 1

The equation of the position in simple harmonic motion is Asin(at + φ) x -24 sin a(11)sin | | ปุ่0.600 If t-0, then x-0 0-(11)sin 0.600 (0)-φ 0-sin(φ) If the position x=5.00 cm 5.00=(11)sinu yo 600 5.00 0.600 3.65t=sin- t= 0.129 s