Lab 8 (2)

February 2, 2018 | Author: Brandon Sookdeo | Category: Young's Modulus, Stress (Mechanics), Oscillation, Physical Sciences, Science
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Name: Jivan Raghoo Date: 23/03/12 Title: Young’s Modulus Aim: 1. To investigate how the period of oscillations of a loaded steel blade varies with the length of the blade. 2. To determine a value for the Young Modulus of steel using the results of the experiment. Materials & Apparatus:      

Block of wood G-clamp 50g masses (2) Blade Ruler Micrometer screw gauge

Diagram:

Name: Jivan Raghoo

Figure 1 showing setup of apparatus. Theory: This experiment shows simple harmonic motion which a body experiences when its acceleration is directed towards a fixed point in its path and is directly proportional to its distance from that point. In this case the blade describes simple harmonic motion about its equilibrium position, oscillating back and forth until its energy is dispersed and it returns to rest. The longer the steel blade is the longer the period (time for one complete oscillation) and the equation that expresses this: T =k d n Where, T is the period of oscillation d is the distance from the centre of the masses to the edge of the blocks k and n are constants

Name: Jivan Raghoo The Young’s modulus (E) of a material is the ratio of the tensile stress to the tensile strain the material experiences provided that the stress is not so high that the limit of proportionality has been exceeded. Stress (σ) is defined as the Force per unit area and strain ∈ is the ratio of extension produced and the original length of the material. A theoretical treatment of the oscillator suggests that k=



16 π 2 M Eb t 3

Where: E= Young modulus M= the mass attached to end of the blade k= constant from oscillation equation b= width of blade t= thickness of blade Method: 1

The apparatus was setup as shown in the diagram.

2

The distance, d, from the center of the mass to the edge of the blocks of wood was recorded.

3

The end of the blade was displaced from its equilibrium position and released so that small oscillations occurred in a vertical plane.

4

The time taken for 20 oscillations to occur was recorded twice and the period (T) was determined.

5

The procedure was repeated 5 additional times with different values for d, ranging from 0.13m to 0.25m being used. All results were recorded.

Name: Jivan Raghoo 6

The width of the blade, b, and the thickness of the blade, t, were both measured using a micrometer screw gauge and recorded.

Variables: Manipulated: length of blade Responding: period of oscillation Constant: mass attached to end of blade Results: Table 1 showing values of time, period, distance between center of masses and the edges of the blocks, lg T and lg d. d/m

Time for 20

Time for 20

Average

T/s

lg T

lg d

0.24 0.22 0.20 0.18 0.16 0.14

oscillations1/s 11.84 10.28 8.91 7.58 6.78 5.81

oscillations2/s 11.44 10.37 8.90 7.72 6.59 5.37

time/s 11.64 10.33 8.91 7.65 6.69 5.59

0.58 0.52 0.45 0.38 0.33 0.28

-0.24 -0.28 -0.35 -0.42 -0.48 -0.55

-0.62 -0.66 -0.70 -0.74 -0.80 -0.85

Table 2 showing values obtained for the width and thickness of the blade. Width of blade (b) /m Thickness of blade (t) /m

Treatment of results: Gradient =

y 2− y 1 x 2−x 1

0.0126 0.00065

Name: Jivan Raghoo −0.2125−(−0.5775) = −0.60−(−0.87) = 1.35 T =k d

n

Take log of both sides: lgT =lgk d n lgT =lgk +lgd

n

lgT =nlgd +lgk y=mx +c

Therefore, using graph, Gradient = n ∴ n=1. 35 y-intercept = lgk Where n =1.35, and using pt. (-0.75, -0.415) −0.415=1.35 (−0.75 )+ lgk 0.598=lgk

∴ k=3.96 To find E we use the formula,

Name: Jivan Raghoo k=



16 π 2 M 3 Eb t



16 π 2 ( 0.1) 3.96= E ( 0.0126 ) (0.00065)3 16 π 2 ( 0.1) 15.6816= E(3.460275 x 10−12) E= 2.91 x 1011 Pa Error Calculations: T =k d

n

δk δT nδd = + k T d δk 0.005 1.35(0.0005) = + 3.9 6 0.58 0.24 δk=± 0.0 453

From equation: k=



k 2=

16 π 2 M 3 Eb t

16 π 2 M Ebt 3

δE δb 3 δt 2δk = + + E b t k

Name: Jivan Raghoo δE δb 3 δt 2δk = + + E b t k

δE=E (

δb 3 δt 2 δk + + ) b t k

δE=2.91 x 1011 (

δE=1.95 ×1010

0.0005 3 ( 1 ×10 ) 2(0.0453) + + ) 0.012 6 0.0006 5 3.9 6 −6

Pa

Precautions: 1. The stop-watch was started after three full oscillations in order to reduce error due to reaction time. 2. When measuring d, the blade was held horizontally to get accurate measurements. Sources of error: 1. Due to the quick movements of the blade, it was possible to make errors when taking values for time taken for 20 oscillations. Conclusion: 1. The period of the oscillation was found to decrease as the length of the blade decreased. 2. The Young modulus of steel was found to be

2.91 x 1011 ±1.95 ×10 10 Pa.

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