Question

A 145 g block connected to a light spring with a force constant of k = 5 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 3 cm from equilibrium and released from rest. Find the period of its motion. (Recall that the period, T, and frequency, f, are inverses of each other.) s Determine the maximum acceleration of the block.

Answer 1

Answer:

T = 1.07 s

a = 1.034 m/s2

Explanation:

Metric unit conversion

m = 145 g = 0.145 kg

x = 3 cm = 0.03 m

Suppose this is a simple harmonic motion, then its period T can be calculated using the following equation

$T = 2\pi\sqrt{\frac{m}{k}}$

$T = 2\pi\sqrt{\frac{0.145}{5}} = 1.07 s$

The maximum acceleration would occurs when spring is at maximum stretching length, aka 0.03m

The spring force at that point would be

$F_s = kx = 5*0.03 = 0.15 N$

According to Newton's 2nd law, the acceleration at this point would be

$a = F_s/m = 0.15 / 0.145 = 1.034 m/s^2$