Question

A 150 g toy is undergoing SHM on the end of a horizontal spring with force constant of 300.0 N/m. When the object is 1.20 cm from its equilibrium position, it is observed to have a speed of 0.300 m/s. Find the total energy of the object at any point in its motion, the amplitude of the motion, and the maximum speed obtained by the object during its motion.

Answer 1

Mass M = 0.150 kg

k = 300 N/m

Displacement from equilibrium x = 0.0120 m

Speed v = 0.300 m/s

a)Total energy E = kinetic energy + potential energy

= (1/2)Mv^2 + (1/2)kx^2

= (1/2) * 0.150 * 0.300^2 + (1/2) * 300 * 0.0120^2

= 0.00675 + 0.0216

= 0.02835 J

Ans: 0.02835 J

b) Let amplitude = A

E = (1/2)kA^2

Or A = sqrt(2E/k) = sqrt(2 * 0.02835/300)

= 0.0137 m

Ans: 0.0137 m

c) Let maximum speed = vmax

E = (1/2)M * vmax^2

vmax = sqrt(2E/M) = sqrt(2 * 0.02835/0.150) = 0.615 m/s

Ans: 0.615 m/s