Question

A 100 g block attached to a spring with spring constant 2.7 N/m oscillates horizontally on a frictionless table. Its velocity is 22 cm/s when x0 = -5.6 cm . What is the amplitude of oscillation? What is the block's maximum acceleration? What is the block's position when the acceleration is maximum? What is the speed of the block when x1 = 2.9 cm ?

Answer 1

You could do this from your equation, or you could use conservation of energy.

PE in spring = kx^2/2

KE of mass = mv^2/2

Total energy E = constant, so

kx^2 + mv^2 = 2E

Plug in the numbers in consistent units:

2.7x0.056^2 + 0.1x0.22^2 = 2E

2E = 0.0133072

1) The max amplitude occurs when v = 0

2.7 x^2 = 0.0133072

x = 0.070m = 7.0cm

3) Position when the acceleration is the maximum is at the max amplitude

2) Force on ball at max amplitude = kx = 2.7 x 0.070 = 0.19 N

Acceleration = force / mass = 0.19 / 0.1 = 1.9 m/s^2 = 190 cm/s^2

4) kx^2 + mv^2 = 2E

2.7x0.029^2 + 0.1 v^2 = 0.0133072

v = 0.332 m/s = 33.2 cm/s