At a specific instant, the magnitudes of the position, velocity, and acceleration of a particle undergoing simple harmonic motion are observed to be 2 cm , 1 m/s , and 10 m/s ^2 , respectively. Determine the amplitude and the time period of the motion.

Step-by-step solution

Given x = 2 cm = 0.02 m , v = 1 m/s , and a = 10 m/s ^2 . In simple harmonic motion, the magnitude of acceleration is given by a = ^2 x . Substituting the given values: 10 = ^2 0.02 ^2 = 10 0.02 = 500 rad ^2/ s ^2 = 500 = 10 5 rad/s The time period T is given by: T = 2 = 2 10 5 2 3.14 22.36 0.28 s The velocity in SHM is given by v = A^2 - x^2 . Squaring both sides: v^2 = ^2 (A^2 - x^2) Substituting the known values: 1^2 = 500 (A^2 - 0.02^2) 1 = 500 (A^2 - 0.0004) A^2 - 0.0004 = 1 500 = 0.002 A^2 = 0.002 + 0.0004 =