Question

A toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/m . When the toy is 0.140 m from its equilibrium position, it is observed to have a speed of 3 m/s and a total energy of 5.2 J . A) find the mass of the toy. B) Find the amplitude of the motion C) Find the maximum speed attained by the object in motion.

Answer 1

Energy E = 5.2 J

k = 300 N/m

Displacement from equilibrium x = 0.140 m

Speed v = 3 m/s

a)Total energy E = kinetic energy + potential energy

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

5.2 = (1/2) * m * 3^2 + (1/2) * 300 * 0.140^2

5.2 = 4.5m + 2.94

=> m = 0.50 kg

Ans: 0.50 kg

b) Let amplitude = A

E = (1/2)kA^2

Or A = sqrt(2E/k) = sqrt(2 * 5.2 /300)

= 0.186 m

Ans: 0.186 m

c) Let maximum speed = vmax

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

vmax = sqrt(2E/M) = sqrt(2 * 5.2/0.50) = 4.55 m/s

Ans: 4.55 m/s