Question

The mass attached to a spring balance is 80 g. When the spring is stretched 10 cm. it requires a force of 0.2 N to hold it motionless. The mass-spring system set into motion by displacing the mass 20 cm to the right and letting it go. The period of the mass-spring system is

Answer 1

We are given information:
m = 80g = 0.080kg
x = 10cm = 0.10m
F = 0.2N
d = 20cm = 0.20m

This mass-spring system can be represented as harmonic oscilator. Harmonic oscilator is a type of oscilator where mass oscilates around equilibrium point. Maximum distance on both sides of equilibrium point remains same Period of oscilation is constant. In harmonic oscilator all outside forces are neglected.

In this problem we have mass attached to a spring so we can use spring formula:
F = -kx
where k = spring constant
We can rearrange this formula in order to calculate k. We will ignore negative sign.
k = F / x
k = 0.2 / 0.1
k = 2 N/m

Now we can use this information to calculate the period. Formula is:
$T=2 \pi \sqrt{ \frac{m}{k} } \\ \\ T=2 \pi \sqrt{ \frac{0.080}{2} } \\ \\ T=1.26s$

We can see from formula that the period does not depend on maximum distance from equilibrium point. It only depends on mass and spring constant.