The Effects of the Extension of a Spring on the Time it Takes a Weight to Oscillate

The Effects of the Extension of a Spring on the Time it Takes a Weight to Oscillate

Introduction

I am investigating the relationship between the extension of the

spring and the effect it has on the time it takes for the wait on the

spring to oscillate.

Scientific Knowledge

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Spring at rest

Spring extended

As the spring is extended the spring stores potential kinetic energy.

So the larger X is, the more energy is stored. To work out the energy

we must work out the amount of work done first:

Work Done = Force x Distance

When the mass is released the potential energy of the spring is

converted into kinetic energy of the mass which is at a maximum when

it passes through the mid-point of the oscillation which is the point

where the spring is not extended at all.

So the work done by the spring is equal to the mass times the

acceleration of the mass times the distance. This gives the energy

released by the spring:

Work Done = mass x acceleration x distance

At the centre point Kinetic energy is equal to Potential energy. To

work out the kinetic energy:

K.E = 1/2 mv2

This is the energy gained by the mass after releasing it on the

extended spring. So therefore:

1/2 mv2 = maX

½ mass x velocity2 = mass x acceleration x extension (distance)

The velocity value is the velocity at the mid point which is where the

mass final comes to rest after oscillating. The formula can be

simplified to:

v2 = 2aX

Velocity is distance/time and acceleration is force/mass therefore

with substituting the formula is:

(distance/time)2 = 2 x (Force/Mass) x X

Which is equal to:

(time2 / distance2) = mass / 2 x force x X

The values on the top half of the formula are the constants in this

experiment. The values on the bottom are the things which change