Question

A 0.500-kg glider, attached to the end of an ideal spring with force constant k=450N/m, undergoes simple harmonic motion with an amplitude 0.040m. Compute a) themaximum speed of the glider; b) the speed of the glider when it is at x= -0.015m; c)the magnitude of the maximum acceleration of the glider; d)the acceleration of theglider at x= -0.015m; e) the total mechanical energy of the glider at any point in its motion.

Answer 1

a)we know that maximum velocity v _max =AωA = AMPLITUDE = 0.040mω = ANGULAR FREQUENCY = √k/mk = 450N/mm = 0.5 kgthen ω = √450/0.5 = 30 rad/secthen v_max = 0.04*30 = 1.2 m/sb) we know that velocity at distance x from the mean positionis v = ω√(A² -x²)here x =-0.015 mthen v = 30*√[(0.04)²-(-0.015)²]m/s= 1.11m/sat x = 0.015 m also v = 1.11m/sd) we know that total energy is E = (1/2)kA²=(1/2)450(0.04)² J=0.36 J