Question

A block with mass 2 kg is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period of 0.50 s. The surface is frictionless. The amplitude of the oscillation is 0.1 m. (a) What is the spring constant of the spring? (b) What is the total mechanical energy of the system (the spring and block system)? (c) What is the maximum speed of the block? (d) What is the speed of the block when the displacement of the block is 0.05 m (in other words, the distance between the block and the equilibrium position is 0.05 m, or the block is half way between the equilibrium and the maximum displacement)?

Answer 1

What we know: m = 2 kg, a = 0.1, T = 0.50

k = mw^2 , w = 1/T = 1/0.50 = 2

k = 2 * 2^2 = 8

(a) The spring constant is 8.

(b) Total mechanical energy is 1/2 mw^2 A^2

= 1/2 (8) (0.1)^2 = 4* 0.01 = 0.04

0.04 J

(c) The maximum speed is when kinetic energy equals mechanical energy

0.04 J = 1/2 mv^2

v = sqrt(0.08/2)

(d) 1/2 mv^2 + 1/2 kx^2 = 1/2 kA^2

mv^2 + kx^2 = kA^2

mv^2 = k(A^2-x^2)

v^2 = k(A^2 - x^2) / m

v^2 = 8 * (0.1^2 - 0.05^2) / 2

v= 1.7 * 10^-1 m/s