# Physics Spring Coursework

Extracts from this document...

Introduction

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þÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ using logs.

The laws of logs: 1. logax + logay = loga(xy)

2. logax - logay = loga(x/y)

3. logaxn = nlogax

lnx means logex, where e is the mathematical constant of approximate value 2.718.

Experiment B

This experiment involved finding how the time period of oscillations varied with the mass added to a single spring (with spring constant k).

The laws of logs can be used to change the formula lnT= ln(pkqmr ) into a straight line graph, in the form y=mx+c.

Using Law 1, ln(pkqmr) = ln(pkq) + ln(mr)

Using Law 3, ln(pkq) + ln(mr) = ln(pkq) + rln(m)

This gives the equation lnT = rln(m)+ln(pkq)

Comparing this to y=mx+c, the gradient is r, and the y intercept is ln(pkq).

As lnT against lnm is a straight line, the original expression lnT= ln(pkqmr ) is in the correct form, and the values r and ln(pkq) can be found.

The gradient of my graph is 0.474±0.017, which is the value of r.

Using the point (-0.93, -0.23) (the point where the line of max gradient and min gradient cross) the y intercepts (ln(pkq)) of the lines can be found.

-0.23 - (0.491x-0.93) =cmax cmax= 0.227

-0.23 - (0.457x-0.93) =cmin cmin = 0.195

ln(pkq)= 0.21±0.02

Experiment C

This experiment involved finding a relationship between the spring constant and time period of oscillations (with a constant mass). To do this different arrangements of springs (each with spring constant k)

Middle

This proves that the oscillation is simple harmonic motion because the acceleration towards the rest point is negatively proportional the displacement from the rest point.

This means that w2 =k/m

w = (k/m)0.5

And as T=2p/w

T=2p(k/m)-0.5

T=2p(m/k)0.5

Using Physics by Robert Hutchings, this can be confirmed as T = 2pk-0.5m0.5.

This means: p=2p

q=-0.5

r=0.5

Units

To find the units of p, k and q I will use the correct version of the formula (taken from Physics by Robert Hutchings.

T = 2pk-0.5m0.5

As the values of p and q are powers, they can have no units.

To find the units of p, I will work out the units on both sides of the equation then rearrange to find the units of p. P will be used to represent the units of p.

s = P(Nm-1)-0.5(kg)0.5

s = P(kgm/N)0.5

s = P(kgm/kgms2)0.5

s = P(1/s2)0.5

s = P/s

s2 = P

This means that the units for p are s2.

Bibliography:

www.hookeslaw.com/hookeslaw.htm

Physics 1 - Cambridge University Press

http://www.projectalevel.co.uk/maths/logs.htm

http://www.ndt-ed.org/EducationResources/Math/Math-e.htm

http://en.wikipedia.org/wiki/Simple_harmonic_motion

Evaluation

I feel like the experiment went well, although my results do not perfectly match the theory.

Actual Values for p,q and r:

p=2p

q=-0.5

r=0.5

My values for p, q and r:

p=5.55±0.52

q = -0.468±0.019

r = 0.474±0.017

This means that none of the values I found included the actual values in their error range.

Conclusion

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