# Young's Modulus of a Metre Ruler

IB Physics Labwork

FINDING YOUNG’S MODULUS OF WOODEN METRE RULER

INTRODUCTION:

Young’s modulus is also known as tensile modulus. It is the measure of elasticity and it is a very important characteristic of a material. The formula of young’s modulus is stress by strain.

In this experiment I am going to determine the young’s modulus (E) of wood from the period of oscillation of a loaded wooden ruler. Equation relating the period of oscillation t to the overhanging length l of the ruler is;

T2=kl3/E

DESIGN:

To do this experiment I will following equipments; g-clamp,1metre wooden ruler, 50 gram of mass, tape, stop watch and materials to write. After getting all the required materials I will start to fix the materials. I will clamp the wooden ruler with the help of g-clamp. Using the tape I will attach the 50g mass at the end of the ruler.

I will start the oscillations with the length of 95cm of the ruler and then 90cm, 85cm and gradually decreasing. I will then make a angle less than 150 by the ruler and release it. With the help of stop watch I will measure the time period required to complete 20 oscillations.

I will use the same wooden ruler for doing the whole experiment so that the elasticity in my experiment does not change. I will measure the time period for 20 oscillations for every lengths and also use the 50 g mass in every length. The design of my lab experiment looks similar to the given picture.

G-clamp

Plastic scale

Wooden ruler

50g mass

DATA AND ANALYSIS:

Raw data:

length(l)| time for 20 oscillations(t) /sec|

±0.10 /cm| t1 | t2| t2| t4| tavg ± ∆tavg|

95.0| 16.00| 16.06| 16.09| 15.94| 16.02 ± 0.15|

90.0| 14.47| 14.37| 14.41| 14.50| 14.44 ± 0.13|

85.0| 13.19| 13.13| 13.11| 13.22| 13.18 ± 0.09|

80.0| 12.19| 12.09| 12.19| 12.22| 12.17 ± 0.10|

75.0| 11.06| 11.12| 11.10| 11.09| 11.09 ± 0.06|

70.0| 10.06| 10.00| 9.95| 10.09| 10.03 ± 0.14|

65.0| 9.10| 9.30| 9.22| 9.45| 9.27 ± 0.35|

To find tavg; tavg= t1+ t2+ t3+ t4

4

For 95cm, tavg= 16.00+16.069+16.09+15.94

4

= 16.02 s

To find ∆tavg; ∆tavg= tmax-tmin

For 95cm,∆tavg= 16.09-15.94 = 0.15s

Processed data:

length(l)| tavg ± ∆tavg| time period| l3 | ∆l3| T2| ∆T2 | ±0.001 /m| / sec| (T ± ∆T)/sec| /m3| /m3| /sec2| /sec2| 0.95| 16.02 ± 0.15| 0.801 ± 0.008| 0.857| 0.003| 0.642| 0.013| 0.90| 14.44 ± 0.13| 0.722 ± 0.007| 0.729| 0.002| 0.521| 0.01| 0.85| 13.18 ± 0.09| 0.659 ± 0.005| 0.614| 0.002| 0.434| 0.007| 0.80| 12.17 ± 0.10| 0.609 ± 0.005| 0.512| 0.002| 0.371| 0.006| 0.75| 11.09 ± 0.06| 0.555 ± 0.003| 0.422| 0.002| 0.308| 0.003| 0.70| 10.03 ± 0.14| 0.502 ± 0.007| 0.343| 0.001| 0.252| 0.007| 0.65| 9.27 ± 0.35| 0.464 ± 0.018| 0.275| 0.001| 0.215| 0.017|

To find ∆l3; ∆l3= 3×∆ll ×l3

For 0.95m, ∆l3=3×0.0010.95 ×0.857

= 0.003m3

To find T;T=tavg/20

For 0.95m, T=16.02/20

= 0.801s

To find ∆T; ∆T=∆tavg/20

For 0.95m, ∆t=0.15/20

= 0.008s

To find ∆T2; ∆T2=2×∆TT ×T2

For 0.95m,∆T2=2×0.0080.801 ×0.642

= 0.013s2

Now, I construct a graph of l3 vs T2. Since the uncertainty of l3 is very small, it might not be seen properly in the graph.

Gradient of best fit line(m1)= 0.60-0.260.795-0.325 = 0.34/0.47

= 0.72

Gradient of maximum worst fit line(m2)= 0.64-0.220.835-0.31 = 0.42/0.525 = 0.8

Gradient of minimum worst fit line(m3)= 0.62-0.2950.84-0.36

= 0.325/0.48 = 0.68

Uncertainty in gradient= (m2- m3)/2

= (0.8-0.68)/2

= 0.06

Gradient with uncertainty= 0.72±0.06

Now,

Comparing the equation; T2=kl3/E with...

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