Question

A 927 g mass is connected to a light spring of force constant 5 N/m that is free to oscillate on a horizontal, frictionless track. The mass is displaced 8 cm from the equilibrium point and released from rest. 5 N/m 927 g 8 cm x = 0 x Find the period of the motion. Answer in units of s.

Answer 1

Answer:

2.7 s

Explanation:

The period of the motion of a massless loaded spring is given by

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

where m is the mass of the load and k is the force or spring constant.

Using values in the question and converting to appropriate units,

$T = 2\pi\sqrt{\dfrac{927\times10^{-3}\text{ kg}}{5 \text{ N/m}}}= 2.7\text{ s}$

Answer 2

Explanation:

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