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Experiments- Hooke's Law and Young Modulus

www.curriculum-press.co.uk Number 83

FactsheetPhysics

Factsheet 27 went through the elasticity theory required at A-level (andprobably further) in some detail.

In this Factsheet we will look at some of the experimental work linked tothe topic. We will concentrate on how best to make the practical workproduce accurate and reliable data, and the graphical work and calculationsresulting.

Hooke’s Law provides information on the properties of a specific device(spring, length of wire, etc). The Young modulus gives us a value for amaterial (steel, copper, glass, etc).

Hooke’s Law refers to a specific device; the Young modulusrefers to a material.

Hooke’s Law (revision):

In the proportional region, between O and P (the limit of proportionality):

F = ke where F is the applied force (N)e is the extension (m)k is the spring constant (Nm-1)

Practical Hint:Throughout these practicals, always do repeats and averages wherepossible, and take care with significant figures and units.

Finding the spring constant, k, of a steel spring.

m

This is a standard practical going back toGCSE level. A series of masses are carefullyadded to the mass holder, and measurementsof extension and weight are recorded in atable.

e

F

Practical Hints:1. Use only small masses. This gives you more data points. It also

makes it less likely you will exceed the limit of proportionality.2. Repeat readings with decreasing masses to ensure there is no

hysteresis effect.3. Some springs are manufactured with the coils forced so tightly together

that it takes a significant force to begin separating them. This mayaffect the starting point of the graph. Use only the straight-line sectionto find the gradient.

Example:A student performs an experiment to find the spring constant of asteel spring, obtaining these results:

Mass / g

0.05.0

10.015.020.025.030.0

Ave. extension / 10-3 cm

0.00.10.50.81.31.82.1

Find the spring constant.

Solution:Rewrite the table:

Weight/10-3N

04998

147196245294

Ave. extension / 10-3 m

0158

131821

P

e

F

0

The results are then graphed:

The spring constant, k,is the gradient (from k = ∆F / ∆e).

300

250

200

150

100

50

0e/10-3m

F/10-3N

2 6 1084 12 14 1618 20 22 24

k = gradient = (200 × 10-3 ) / ( 17 × 10-3 ) = 12 Nm-1

Notice the care that must be taken with the units, and that the beststraight line should not be started from the origin in this situation.

83. Experiments- Hooke's Law and Young Modulus Physics Factsheet

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Combinations of SpringsExperimental work is often performed to verify the rules for combiningsprings in series and parallel.

T=W

T=W

T=W

2T=

W 2

W

W

Series Parallel

Practical Hints:In the series combination, different springs can be used, but be preparedto add the inverse spring constants:

1/k = 1/k1 + 1/k

2

The most common error is forgetting to invert 1/k to find k at the end ofthe calculation.

In the parallel combination, always use identical springs. Otherwise onewill extend further than the other, causing the lower support rod to tip.

Finding “g” from SHM with an oscillating steel spring:

m

Hooke’s Law can be used in simple harmonic motion where the period ofthe mass, m, oscillating vertically from a spring depends on the springconstant.

T = 2π√π√π√π√π√(m/k)

Using k = F

= mg

e e

We find T = 2π√π√π√π√π√(e/g)

Using a range of masses, we record extension e and period of oscillation T.

T2/S2

e/m

The gradient of the graph will be 4πππππ2/g.

Practical Hints:Use small extensions for the spring (small masses) to ensure that you areoperating in the Hooke’s Law region. However remember using verysmall extensions will increase the percentage error in the measurement.Use small amplitude oscillations to stay within the Hooke’s Law region,and also to reduce the likelihood of the spring entering “swinging mode”(acting like a pendulum).

Example:With the previous set-up, the student sets the mass on the steel springinto vertical oscillation. The time for 10 oscillations is measured (andrepeated and averaged) for each mass.

Table of results:

Mass / g Ave. time for 10 osc. / s Extension, e / cm

50 3.8 3.7100 5.4 7.3150 6.7 11.1200 7.7 14.9250 8.6 18.5

Find the value of “g” from the results.

Solution:

Period2, T2 / s2 Extension, e / m

0.144 0.0370.292 0.0730.444 0.1110.596 0.1440.740 0.185

e/m0.200.100

0.20

0.40

0.60

0.80T2/s2

The gradient works out to be 4.0 s2m-1.

4πππππ2 / g = 4.0

g = 9.87 ms-2.

Practical Hints:Dynamic measurements (e.g. timing oscillations) are more difficult thanstatic measurements (e.g. measuring extension). It is essential to repeatand average, and to measure ten or twenty oscillations, not just one.

Physics Factsheet

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83. Experiments- Hooke's Law and Young Modulus

The Young modulus (revision):As mentioned, the Young modulus is a property of a material. Instead ofapplied force, we use tensile stress (the force applied per unit cross-sectional area); instead of extension, we use tensile strain (the fractionalincrease in length).

Stress = F/A (Nm-2)Strain = e/l (no units) where l is the original length.

Young modulus, E = stress/strain = (F/A) / (e/l)

Or we can write: E = (Fl) / (Ae) (Nm-2)

The Young modulus is a measure of the stiffness of a material

It may be useful to emphasise the size of the effect we may seeexperimentally with an example.

Example:Mild steel has a Young modulus E = 20 × 1010 Nm-2. A steel wire of length1.00m has a cross-sectional area of 1.0mm2. Find the extension if a force of100N is applied to the wire.

Solution:E = (Fl) / (Ae) = (100×××××1.0) / (1.0×××××10-6 ××××× 20×××××1010) = 5.0×××××10-4m = 0.50mm.

The point of this example is to illustrate the tiny extension expected inYoung modulus investigations.

This set-up, combined with careful experimentation, should maximise theaccuracy of the final result. Notice these points – they are important:

1. A micrometer is used to measure the diameter in several places on bothwires.

2. The control wire compensates for changes in temperature or “sagging”of the support frame.

3. Small loads are introduced onto both wires to make sure they arestraight before measurements are taken.

4. The spirit level and micrometer allow very small changes to be measured.

5. Readings on both loading and unloading are taken for accuracy, and tocheck for any hysteresis effect.

6. A best-straight line graph is used for increased accuracy.

contol wire Test wire

spirit levelmicrometer

test massesstraighteningmass

A graph of results will resemble this example:

E = (Fl) / (Ae)

So the experimental value for E will be found from:

E = gradient ××××× (l/A)

Exam Hint: Be prepared to discuss the ways in which the extension ofthe wire is maximised, and the steps taken to improve accuracy.Also be ready to discuss why accuracy is more difficult to achieve infinding the Young modulus than the spring constant.

Questions1. (a) State the key difference between the spring constant and the Young

modulus. (b) Why is the spring constant generally easier to determine

experimentally?2. Using weights from 0 to 10N, a steel spring exhibits extensions

increasing linearly from 0 to 20cm.(a) Find the spring constant.(b) Find the extension for a load of 25N. (Can you be certain of this?)

3. Here is a set of results for an experiment with a length of metallic wire(original length, l = 1.5m ; diameter = 2.0mm).

Tension, T / N Extension, e / mm

0 0.00100 0.19200 0.42300 0.60400 0.83500 0.99600 1.24

(a) Graph these results (with T on the y-axis).(b) Find the gradient (in Nm-1).(c) Find the cross-sectional area (in m2).(d) Calculate the Young modulus for this metal.

4. A 1.0m length of wire is stretched by 0.5mm. This causes a decreasein the cross-sectional area.(a) Estimate the % change in the stress due to this area change.(b) Is this large enough to be noticeable in the results?

Answers:1. (a) Spring constant – property of device

Young modulus – property of material.(b) Much larger extension when dealing with springs, etc.

Easier to measure extension accurately.

2. (a) k = F/e = 50Nm-1

(b) 50cm (but only if the limit of proportionality is not exceeded)3. (b) Gradient approximately 4.9 × 105Nm-1.

(c) 3.14 × 10-6m2

(d) E = gradient ××××× (l/A) = 23 x 1010Nm-2

4. (a) stress = F/A, volume = Al% change in stress of same magnitude as that in area.And % change in area of same magnitude as that in length.% change = 0.05%

(b) This is too small to be noticed.

Safety point: To maximise the extension obtained, a long and verythin wire is put under considerable tension. Goggles must be worn incase the wire snaps.

e/m

F/N

Acknowledgements:This Physics Factsheet was researched and written by Paul FreemanThe Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU