Question

A 170 g block attached to a spring with spring constant 3.0 N/m oscillates horizontally on a frictionless table. Its velocity is 25 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 = 2.6 cm ?

Answer 1

here,

mass , m = 170 g = 0.17 kg

spring constant , K = 3 N/m

x0 = - 5.6 cm , v0 = 25 cm/s = 0.25 m/s

let the amplitude of motion be A

using conservation of energy

0.5 * K * A^2 = 0.5 * K * x0^2 + 0.5 * m * v0^2

3 * A^2 = 3 * 0.056^2 + 0.17 * 0.25^2

solving for A

A = 0.082 m

the amplitude of motion is 0.082 m

let the maximum speed be v

using conservation of energy

0.5 * K * A^2 = 0.5 * m * v^2

3 * 0.082^2 = 0.17 * v^2

solving for v

v = 0.34 m/s

the maximum speed is 0.34 m/s

The block's position is x = A = 0.082 m when the acceleration is maximum

let the speed be v1 when x1 = 2.6 cm = 0.026 m

0.5 * K * A^2 = 0.5 * K * x1^2 + 0.5 * m * v1^2

3 * 0.082^2 = 3 * 0.026^2 + 0.17 * v1^2

v1 = 0.33 m/s

the speed is 0.33 m/s