For a particle executing simple harmonic motion, the maximum speed and acceleration are 10 cm/s and 50 cm/s ^2 , respectively. Determine the position(s) of the particle when its speed is 8 cm/s .

Step-by-step solution

Given v_ max = A = 10 cm/s and a_ max = A ^2 = 50 cm/s ^2 . Dividing the two expressions, we get: = A ^2 A = 50 10 = 5 rad/s The amplitude A is: A = 10 = 10 5 = 2 cm The velocity v of a particle in SHM at a distance x from the mean position is given by: v = A^2 - x^2 Substituting v = 8 cm/s , = 5 rad/s , and A = 2 cm : 8 = 5 2^2 - x^2 Squaring both sides: 64 = 25(4 - x^2) 25x^2 = 100 - 64 = 36 x^2 = 36 25 = 1.44 x = 1.2 cm