Question

A 120 g block attached to a spring with spring constant 3.0 N/m oscillates horizontally on a frictionless table. Its velocity is 17 cm/s when x0 = -4.5 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

PE in spring = kx²/2
KE of mass = mv²/2
Total energy E = constant, so
kx² + mv² = 2E

Plug in the numbers in consistent units:

3x(0.045)² + 0.12x(0.17)² = 2E
2E = 0.0095

1) The max amplitude occurs when v = 0
3 x² = 0.0095
x = 5.6 cm

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

2) Force on ball at max amplitude = kx = 3 x 0.056 = 0.168 N
Acceleration = force / mass = 0.168 / 0.12 = 1.4 m/s² = 140 cm/s²

4) kx^2 + mv^2 = 2E
3x(0.029)² + 0.12 v^2 = 0.0095
v = 0.241 m/s = 24.1 cm/s