OII 2018 Sperimentale

Confidential Solutions to Experimental Problem 2 Viscoelasticity of a polymer thread (J. M. Gil, J. Pinto da Cunha, R. C. Vilão, H. V. Alberto) July 22, 2018 v1.1

Confidential Experiment English (UK) SE2-1 Problem 2: Viscoelasticity of a polymer thread (10 points) Part A. Stress-relaxation measurements (1.9 points) A.1 Measurement: ${\ell }_0=42.7+2\times 0.5=43.7$ cm , A.1 ${\ell }_0=(43.7\pm 0.2)$ cm . 0.3pt A.2 A.2 $P_0=(81.11\pm 0.03)$ gf . 0.3pt A.3 The table contains the readings on the scale $P$ (Question A.3) and the force on the thread, $F(t)$, at constant strain (Question D.1). The values of $\frac{dF}{dt}$ (Question D.6) were computed numerically using equal time intervals. The function $y(t)$ is given by $y(t)=F(t)-F_0-F_1{e}^{-t/{\tau }_1}$ (Question D.10). A.3 $t$/s $P(t)$ /gf $F$/gf $\frac{dF}{dt}$/gf ${\text{s}}^{-1}$ $y(t)$ /gf 10 35.7 45.41 2.82 17 36.2 44.91 2.33 26 36.6 44.51 1.95 32 36.8 44.31 1.76 40 37.0 44.11 1.57 46 37.1 44.01 1.48 51 37.2 43.91 1.38 58 37.3 43.81 1.29 65 37.4 43.71 1.20 1.0pt

Confidential Experiment English (UK) SE2-2 A.3 $t$/s $P(t)$ /gf $F$/gf $\frac{dF}{dt}$/gf ${\text{s}}^{-1}$ $y(t)$ /gf 73 37.5 43.61 1.12 84 37.6 43.51 1.03 94 37.7 43.41 0.94 105 37.8 43.31 0.86 118 37.9 43.21 0.77 136 38.0 43.11 0.70 151 38.1 43.01 0.62 173 38.2 42.91 0.55 193 38.3 42.81 0.48 217 38.4 42.71 0.41 247 38.5 42.61 0.35 279 38.6 42.51 0.29 317 38.7 42.41 0.23 358 38.8 42.31 0.18 408 38.9 42.21 0.14 471 39.0 42.11 0.11 525 39.1 42.01 0.07 591 39.2 41.91 0.03 600 39.2 41.91 0.04 672 39.3 41.81 0.01 773 39.4 41.71 0.007 866 39.5 41.61 $-0.01$ 900 39.52 41.59 $-0.00$ 993 39.6 41.51 $-0.01$ 1124 39.7 41.41 1200 39.74 41.37 $-7.00\times 10^{-4}$ 1272 39.8 41.31 1419 39.9 41.21 1500 39.94 41.17 $-5.33\times 10^{-4}$ 1628 40.0 41.11 1800 40.06 41.05 $-4.67\times 10^{-4}$ 1869 40.1 41.01 2037 40.2 40.91 2100 40.22 40.89 $-3.83\times 10^{-4}$ 2400 40.29 40.82 0.3pt

Confidential Experiment English (UK) SE2-3 A.4 Measurement: $\ell =50.0+2\times 0.5=51.0$ cm , A.4 $\ell =(51.0\pm 0.2)$ cm . 0.3pt Part B. Measurement of the streched thread diameter (1.5 points) B.1 Two mirrors are used to maximize the distance D and consequently the distance between diffraction minima. B.1 Sketch of the method 0.6pt B.2 The total distance $D$ is the sum $D=D_1+D_2+D_3=(26.0+36.0+102.3)$ cm $=164.3$ cm $=1.643$ m . The estimated uncertainties are ${\sigma }_{D_1}={\sigma }_{D_2}={\sigma }_{D_3}\approx 0.5$ cm $\Rightarrow {\sigma }_D=\sqrt{3}\times {\sigma }_{D_1}=0.5\times \sqrt{3}=0.87$ cm . B.2 $D=(1.643\pm 0.009)$ m . 0.3pt

Confidential Experiment English (UK) SE2-4 B.3 The distance between minima, $x$, is quite small. To reduce the error, the total distance $Nx$, with $N=22$, was measured: $22x=49$ mm $\Rightarrow \bar{x}=2.227$ mm . The corresponding uncertainty is ${\sigma }_{\bar{x}}=\frac{{\sigma }_{22x}}{22}=\frac{0.25\text{mm}}{22}=0.011$ mm . B.3 $\bar{x}=(2.227\pm 0.011)$ mm . 0.3pt B.4 Using previous results, we get $d=\frac{\lambda }{\sin \theta }\simeq \frac{\lambda D}{\bar{x}}=\frac{650\times 10^{-9}\text{m}\times 1.643\text{m}}{2.227\times 10^{-3}\text{m}}=4.795\times 10^{-4}$ m $=0.480$ mm . For the uncertainties, we have $\frac{{\sigma }_d}{d}=\frac{{\sigma }_{\lambda }}{\lambda }+\frac{{\sigma }_D}{D}+\frac{{\sigma }_{\bar{x}}}{\bar{x}}=\frac{10}{650}+\frac{0.0087}{1.643}+\frac{0.011}{2.227}=0.02517\Rightarrow {\sigma }_d=0.02517\times 0.480$ mm $=0.012$ mm . B.4 $d=(0.480\pm 0.012)$ mm . 0.3pt Part C. Change to a new thread (0.3 points) C.1 Measurement: ${\ell }_0^{\prime }=31.6+2\times 0.5=32.6$ cm. C.1

${\ell }_0^{\prime }=(32.6\pm 0.2)$ cm . 0.3pt

Confidential Experiment English (UK) SE2-5 Part D. Data Analysis (5.7 points) D.1 The force on the thread was calculated as $F(t)=(P_0-P(t))$, in gram-force units. D.1 See column $F(t)$ in the table in A.3. 0.3pt D.2 D.2 0 500 1000 1500 2000 2500 40 41 42 43 44 45 46 $F$ (gf) $t$ (s) 0 500 1000 1500 2000 2500 40 41 42 43 44 45 46 $F$ (gf) $t$ (s) Left: $F(t)$ sampled at unequal time intervals. Right: $F(t)$ sampled at equal time intervals for $t>1000$ s. 0.4pt D.3 The dimensionless quantity $\epsilon$ is given by $\epsilon =\frac{\ell -{\ell }_0}{{\ell }_0}=\frac{51.0-43.7}{43.7}=0.167$ . The uncertainty in $\epsilon$, ${\sigma }_{\epsilon }$, is calculated propagating the uncertainties in the measured length, ${\sigma }_{\ell }$ and ${\sigma }_{{\ell }_0}$: $\frac{{\sigma }_{\epsilon }}{\epsilon }=\frac{{\sigma }_{(\ell -{\ell }_0)}}{\ell -{\ell }_0}+\frac{{\sigma }_{{\ell }_0}}{{\ell }_0}=\frac{\sqrt{{\sigma }_{\ell }^2+{\sigma }_{{\ell }_0}^2}}{\ell -{\ell }_0}+\frac{{\sigma }_{{\ell }_0}}{{\ell }_0}=\frac{0.2\times \sqrt{2}}{7.3}+\frac{0.2}{43.7}=0.0433$ Therefore, ${\sigma }_{\epsilon }=0.0433\times 0.167=0.0072$. D.3 $\epsilon =0.167\pm 0.007$ . 0.3pt

Confidential Experiment English (UK) SE2-6 D.4 One has $\frac{\sigma }{\epsilon }=\frac{F}{\epsilon S}$ . In this case, $S=\pi (d/2{)}^2=1.809\times 10^{-7}{\text{m}}^2$ and $\epsilon =0.167$. We also have $1$ gf $=g\times 10^{-3}$ N with $g=9.8\text{m}{\text{s}}^{-2}$. Therefore, if $F$ is in gram-force units we have $\frac{\sigma }{\epsilon }=\frac{9.8\times 10^{-3}{\text{gf}}^{-1}\text{N}}{0.167\times 1.809\times 10^{-7}{\text{m}}^2}F=(324293{\text{gf}}^{-1}\text{N}{\text{m}}^{-2})F$ , where $F$ is in gf , and $\sigma$ is in N ${\text{m}}^{-2}$. Comparing with $\frac{\sigma }{\epsilon }=\beta F$ we get $\beta =324293{\text{gf}}^{-1}\text{N}{\text{m}}^{-2}$ . Note that, if we write $F(t)=F_0+F_1{e}^{-t/{\tau }_1}+F_2{e}^{-t/{\tau }_2}+F_3{e}^{-t/{\tau }_3}+\cdots$ (1) and compare with equation $\frac{\sigma }{\epsilon }=\beta F(t)=E_0+E_1{e}^{-t/{\tau }_1}+E_2{e}^{-t/{\tau }_2}+E_3{e}^{-t/{\tau }_3}+\cdots$ (2) we conclude that $E_0=\beta F_0$ , $E_1=\beta F_1$, $E_2=\beta F_2$, etc. D.4 $\beta =3.24\times 10^5{\text{gf}}^{-1}\text{N}{\text{m}}^{-2}$ . 0.3pt D.5 For a purely elastic process, $\sigma =\epsilon E_0$ and $F=\alpha \sigma =\alpha \epsilon E_0$ . Thus, a graph of a constant function is expected. D.5

$F$ $t$ 0.4pt

Confidential Experiment English (UK) SE2-7 D.6 The data for $\frac{dF}{dt}$ inserted in table introduced in A.3, was computed numerically for equal time intervals. However, the graphical method is also exemplified. In the present graph, tangent lines to $F(t)$ are drawn at four different time instants (1200, 1500, 1800 and 2100 s). The slopes of those lines are a measure of $\frac{dF}{dt}$ at those instants. D.6 See in the table used in A.3, the column with $\frac{dF}{dt}$. This graph is present only if a graphical method is used. 0.5pt D.7 For a single viscoelastic process, $F=\frac{1}{\beta }(E_0+E_1{e}^{-t/{\tau }_1})=F_0+F_1{e}^{-t/{\tau }_1}$ . Therefore, D.7 $\frac{dF}{dt}=-\frac{F_1}{{\tau }_1}{e}^{-t/{\tau }_1}$ , where $F_1=\frac{E_1}{\beta }$ . 0.3pt D.8 The linearisation of the expression of $dF/dt$ is accomplished using logarithms: $\ln \left(-\frac{dF}{dt}\right)=\ln \left(\frac{F_1}{{\tau }_1}\right)-\frac{1}{{\tau }_1}t$ .

Confidential Experiment English (UK) SE2-8 The plot of $\ln (-dF/dt)$ is shown in the graph below for a case where the derivative was obtained numerically (left) and using a graphic method (right). For the left graph, the best straight line is $\ln (-dF/dt)=m_{1}t+b_1$ where $m_1=(-6.47\pm 0.62)\times 10^{-4}$ and $b_1=(-6.52\pm 0.11)$, using $t$ in seconds and the force in gram-force units. If the derivative is computed numerically for unequal time intervals, the final parameters $E_1$ and ${\tau }_1$ are similar. The best straight line for the right graph yields $m_1=(-6.00\pm 0.15)\times 10^{-4}$ and $b_1=(-6.63\pm 0.02)$ using $t$ in seconds and the force in gram-force units. Thus, using the data from the left graph, ${\tau }_1=\frac{1}{-m_1}=1546$ s and $F_1={\tau }_1{e}^{b_1}=2.28$ gf $\Rightarrow E_1=\beta F_1=7.39\times 10^5$ N ${\text{m}}^{-2}$ . For the right graph, the final parameters are ${\tau }_1=1667$ s and $E_1=7.13\times 10^5$ N ${\text{m}}^{-2}$ . D.8 ${\tau }_1=1546$ s , $E_1=7.39\times 10^5$ N ${\text{m}}^{-2}$ . 1200 1400 1600 1800 2000 2200 -8.1 -8.0 -7.9 -7.8 -7.7 -7.6 -7.5 -7.4 -7.3 -7.2

$\ln (-dF/dt)$ $t$ (s) 1200 1400 1600 1800 2000 2200 -7.9 -7.8 -7.7 -7.6 -7.5 -7.4 -7.3 -7.2

$\ln (-dF/dt)$ $t$ (s) Left: $\frac{dF}{dt}$ computed numerically using equal time intervals. Right: using data from the graph in D.6. 1.0pt D.9 For the 4 points on the left graph in D.8, we can write $F(t)=F_0+F_1{e}^{-t/{\tau }_1}\Rightarrow F_0=F(t)-F_1{e}^{-t/{\tau }_1}$ Thus, averaging $F_0$ for the 4 points of the left graph in D.8: $F_0=\left(\frac{40.32+40.31+40.34+40.30}{4}\right)=40.32$ gf Finally, $E_0=\beta F_0=324293\times 40.32$ N ${\text{m}}^{-2}$ . D.9 $E_0=1.31\times 10^7$ N ${\text{m}}^{-2}$ . 0.3pt

Confidential Experiment English (UK) SE2-9 D.10 The function $y(t)$ is given by $y(t)=F(t)-F_0-F_1{e}^{-t/{\tau }_1}$ , and was added in the Table introduced in A.3 using $F_0=40.32$ gf , $F_1=2.28$ gf and ${\tau }_1=1546$ s. D.10 See column $y(t)$ in the Table in A.3. 0.3pt D.11 Since $y(t)=F(t)-F_0-F_1{e}^{-t/{\tau }_1}$ , then $y(t)=F_2{e}^{-t/{\tau }_2}+F_3{e}^{-t/{\tau }_3}+\cdots ,{\tau }_2>{\tau }_3>\cdots$ At long times, when the contributions from the higher components are small enough, we expect a linear behaviour for $\ln y(t)$: $\ln y=\ln F_2-\frac{1}{{\tau }_2}t$ . In this case, the $y(t)$ data points become meaningless above 500 s. In the region 200-500 s the graph is linear and that region can be used to extract the parameters of the second component. The equation of the straight line is $\ln y_2=b_2+m_{2}t$. From the graph below, $m_2=-(5.65\pm 0.19)\times 10^{-3}\Rightarrow {\tau }_2=\frac{1}{-m_2}=177$ s $b_2=0.33\pm 0.07\Rightarrow F_2=e^{b_2}=1.39\Rightarrow E_2=\beta F_2=4.5\times 10^5$ N ${\text{m}}^{-2}$ . . D.11 $E_2=4.5\times 10^5$ N ${\text{m}}^{-2}$ , ${\tau }_2=177$ s . 0 100 200 300 400 500 600 700 800 -6 -4 -2 0

$\ln y$

Topic: Elasticity & Materials, Wave Optics, Oscillations & Waves Metodi: Stress-Strain Analysis, Interference & Diffraction Analysis, Graph Linearization, Error Propagation, Experimental Data Analysis Competenze: Experimental Data Analysis, Graph Linearization, Error Propagation Objects: String, Mirror Fonte: Testo (PDF) — p.1

Confidential Solutions to Experimental Problem 2 Viscoelasticity of a polymer thread (J. M. Gil, J. Painting by Cunha, R. C. Vilão, H. V. (Albert) July 22, 2018 v1.1

Confidential Experiments English (UK) SE2-1 Problem 2: Viscoelasticity of a polymer thread (10 points) Part A. Stress-relaxation measurements (1.9 points) A.1 Measurement: ${\ell }_0=42.7+2\times 0.5=43.7$ cm , A.1 ${\ell }_0=(43.7\pm 0.2)$ cm . 0.3pt A.2 A.2 $P_0=(81.11\pm 0.03)$ gf . 0.3pt A.3 The table contains the readings on the scale $P$ (Question A.3) and the force on the thread, $F(t)$, at The following is the list of the types of tests: The values of $\frac{dF}{dt}$ (Question D.6) were computed numerically using equal time intervals. The function $y(t)$ is given by $y(t)=F(t)-F_0-F_1{e}^{-t/{\tau }_1}$ (Question D.10). A.3 $t$/s $P(t)$ /gf $F$/gf $\frac{dF}{dt}$/gf ${\text{s}}^{-1}$ $y(t)$ /gf 10 35.7 45.41 2.82 17 36.2 44.91 2.33 26 36.6 44.51 1.95 32 36.8 44.31 1.76 40 37.0 44.11 1.57 46 37.1 44.01 1.48 51 37.2 43.91 1.38 58 37.3 43.81 1.29 65 37.4 43.71 1.20 1.0pt

Confidential Experiments English (UK) SE2-2 A.3 $t$/s $P(t)$ /gf $F$/gf $\frac{dF}{dt}$/gf ${\text{s}}^{-1}$ $y(t)$ /gf 73 37.5 43.61 1.12 84 37.6 43.51 1.03 94 37.7 43.41 0.94 105 37.8 43.31 0.86 118 37.9 43.21 0.77 136 38.0 43.11 0.70 151 38.1 43.01 0.62 173 38.2 42.91 0.55 193 38.3 42.81 0.48 217 38.4 42.71 0.41 247 38.5 42.61 0.35 279 38.6 42.51 0.29 317 38.7 42.41 0.23 358 38.8 42.31 0.18 408 38.9 42.21 0.14 471 39.0 42.11 0.11 525 39.1 42.01 0.07 591 39.2 41.91 0.03 600 39.2 41.91 0.04 672 39.3 41.81 0.01 773 39.4 41.71 0.007 866 39.5 41.61 $-0.01$ 900 39.52 41.59 $-0.00$ 993 39.6 41.51 $-0.01$ 1124 39.7 41.41 1200 39.74 41.37 $-7.00\times 10^{-4}$ 1272 39.8 41.31 1419 39.9 41.21 1500 39.94 41.17 $-5.33\times 10^{-4}$ 1628 40.0 41.11 1800 40.06 41.05 $-4.67\times 10^{-4}$ 1869 40.1 41.01 2037 40.2 40.91 2100 40.22 40.89 $-3.83\times 10^{-4}$ 2400 40.29 40.82 0.3pt

Confidential Experiments English (UK) SE2-3 A.4 Measurement: $\ell =50.0+2\times 0.5=51.0$ cm , A.4 $\ell =(51.0\pm 0.2)$ cm . 0.3pt Part B. Measurement of the streched thread diameter (1.5 points) B.1 Two mirrors are used to maximize the distance D and consequently the distance between diffraction The minimum. B.1 Sketch of the method 0.6pt B.2 The total distance $D$ is the sum $D=D_1+D_2+D_3=(26.0+36.0+102.3)$ cm $=164.3$ cm $=1.643$ m . The estimated uncertainties are ${\sigma }_{D_1}={\sigma }_{D_2}={\sigma }_{D_3}\approx 0.5$ cm $\Rightarrow {\sigma }_D=\sqrt{3}\times {\sigma }_{D_1}=0.5\times \sqrt{3}=0.87$ cm . B.2 $D=(1.643\pm 0.009)$ m . 0.3pt

Confidential Experiments English (UK) SE2-4 B.3 The distance between minimum, $x$, is quite small. To reduce the error, the total distance $Nx$, with $N=22$, was measured: $22x=49$ mm $\Rightarrow \bar{x}=2.227$ mm . The corresponding uncertainty is ${\sigma }_{\bar{x}}=\frac{{\sigma }_{22x}}{22}=\frac{0.25\text{mm}}{22}=0.011$ mm . B.3 $\bar{x}=(2.227\pm 0.011)$ mm . 0.3pt B.4 Using previous results, we get $d=\frac{\lambda }{\sin \theta }\simeq \frac{\lambda D}{\bar{x}}=\frac{650\times 10^{-9}\text{m}\times 1.643\text{m}}{2.227\times 10^{-3}\text{m}}=4.795\times 10^{-4}$ m $=0.480$ mm . For the uncertainties, we have $\frac{{\sigma }_d}{d}=\frac{{\sigma }_{\lambda }}{\lambda }+\frac{{\sigma }_D}{D}+\frac{{\sigma }_{\bar{x}}}{\bar{x}}=\frac{10}{650}+\frac{0.0087}{1.643}+\frac{0.011}{2.227}=0.02517\Rightarrow {\sigma }_d=0.02517\times 0.480$ mm $=0.012$ mm . B.4 $d=(0.480\pm 0.012)$ mm . 0.3pt Part C. Change to a new thread (0.3 points) C.1 Measurement: ${\ell }_0^{\prime }=31.6+2\times 0.5=32.6$ cm. C.1

${\ell }_0^{\prime }=(32.6\pm 0.2)$ cm . 0.3pt

Confidential Experiments English (UK) SE2-5 The Commission shall adopt implementing acts in accordance with Article 21 of this Regulation. The data analysis is based on data analysis (5.7 points) D.1 The force on the thread was calculated as $F(t)=(P_0-P(t))$, in gram-force units. D.1 See column $F(t)$ in the table in A.3. 0.3pt D.2 D.2 0 500 1000 1500 2000 2500 40 41 42 43 44 45 46 $F$ (gf) $t$ (s) 0 500 1000 1500 2000 2500 40 41 42 43 44 45 46 $F$ (gf) $t$ (s) Left: $F(t)$ sampled at unequal time intervals. Right: $F(t)$ sampled at equal time intervals for $t>1000$ s. 0.4pt D.3 The dimensionless quantity $\epsilon$ is given by $\epsilon =\frac{\ell -{\ell }_0}{{\ell }_0}=\frac{51.0-43.7}{43.7}=0.167$ . The uncertainty in $\epsilon$, ${\sigma }_{\epsilon }$, is calculated propagating the uncertainties in the measured length, ${\sigma }_{\ell }$ and ${\sigma }_{{\ell }_0}$: $\frac{{\sigma }_{\epsilon }}{\epsilon }=\frac{{\sigma }_{(\ell -{\ell }_0)}}{\ell -{\ell }_0}+\frac{{\sigma }_{{\ell }_0}}{{\ell }_0}=\frac{\sqrt{{\sigma }_{\ell }^2+{\sigma }_{{\ell }_0}^2}}{\ell -{\ell }_0}+\frac{{\sigma }_{{\ell }_0}}{{\ell }_0}=\frac{0.2\times \sqrt{2}}{7.3}+\frac{0.2}{43.7}=0.0433$ Therefore, ${\sigma }_{\epsilon }=0.0433\times 0.167=0.0072$. D.3 $\epsilon =0.167\pm 0.007$ . 0.3pt

Confidential Experiments English (UK) SE2-6 D.4 One has $\frac{\sigma }{\epsilon }=\frac{F}{\epsilon S}$ . In this case, $S=\pi (d/2{)}^2=1.809\times 10^{-7}{\text{m}}^2$ and $\epsilon =0.167$. We also have $1$ gf $=g\times 10^{-3}$ N with $g=9.8\text{m}{\text{s}}^{-2}$. Therefore, if $F$ is in gram-force units we have $\frac{\sigma }{\epsilon }=\frac{9.8\times 10^{-3}{\text{gf}}^{-1}\text{N}}{0.167\times 1.809\times 10^{-7}{\text{m}}^2}F=(324293{\text{gf}}^{-1}\text{N}{\text{m}}^{-2})F$ , where $F$ is in gf , and $\sigma$ is in N ${\text{m}}^{-2}$. Comparing with $\frac{\sigma }{\epsilon }=\beta F$ we get $\beta =324293{\text{gf}}^{-1}\text{N}{\text{m}}^{-2}$ . Note that, if we write $F(t)=F_0+F_1{e}^{-t/{\tau }_1}+F_2{e}^{-t/{\tau }_2}+F_3{e}^{-t/{\tau }_3}+\cdots$ (1) and compare with equation $\frac{\sigma }{\epsilon }=\beta F(t)=E_0+E_1{e}^{-t/{\tau }_1}+E_2{e}^{-t/{\tau }_2}+E_3{e}^{-t/{\tau }_3}+\cdots$ (2) We conclude that $E_0=\beta F_0$, $E_1=\beta F_1$, $E_2=\beta F_2$, etc. D.4 $\beta =3.24\times 10^5{\text{gf}}^{-1}\text{N}{\text{m}}^{-2}$ . 0.3pt D.5 For a purely elastic process, $\sigma =\epsilon E_0$ and $F=\alpha \sigma =\alpha \epsilon E_0$ . Thus, a graph of a constant function is expected. D.5

$F$ $t$ 0.4pt

Confidential Experiments English (UK) SE2-7 D.6 The data for $\frac{dF}{dt}$ inserted in table introduced in A.3, was computed numerically for equal time intervals. However, the graphical method is also exemplified. In the present graph, tangent lines to $F(t)$ are drawn at four different time instants (1200, 1500, 1800 and 2100 s). The slopes of those lines are a measure of $\frac{dF}{dt}$ at those instances. D.6 See in the table used in A.3, the column with $\frac{dF}{dt}$. This graph is present only if a graphical method is used. 0.5pt D.7 For a single viscoelastic process, $F=\frac{1}{\beta }(E_0+E_1{e}^{-t/{\tau }_1})=F_0+F_1{e}^{-t/{\tau }_1}$ . Therefore, D.7 $\frac{dF}{dt}=-\frac{F_1}{{\tau }_1}{e}^{-t/{\tau }_1}$ , where $F_1=\frac{E_1}{\beta }$ . 0.3pt D.8 The linearisation of the expression of $dF/dt$ is accomplished using logarithms: $\ln \left(-\frac{dF}{dt}\right)=\ln \left(\frac{F_1}{{\tau }_1}\right)-\frac{1}{{\tau }_1}t$ .

Confidential Experiments English (UK) SE2-8 The plot of $\ln (-dF/dt)$ is shown in the graph below for a case where the derivative was obtained numerically (left) and using a graphic method (right). For the left graph, the best straight line is $\ln (-dF/dt)=m_{1}t+b_1$ where $m_1=(-6.47\pm 0.62)\times 10^{-4}$ and $b_1=(-6.52\pm 0.11)$, using $t$ in seconds and the force in gram-force units. If the derivative is computed numerically for unequal time intervals, the final parameters $E_1$ and ${\tau }_1$ are similar. The best straight line for the right graph yields $m_1=(-6.00\pm 0.15)\times 10^{-4}$ and $b_1=(-6.63\pm 0.02)$ using $t$ in seconds and the force in gram-force units. Thus, using the data from the left graph, ${\tau }_1=\frac{1}{-m_1}=1546$ s and $F_1={\tau }_1{e}^{b_1}=2.28$ gf $\Rightarrow E_1=\beta F_1=7.39\times 10^5$ N ${\text{m}}^{-2}$ . For the right graph, the final parameters are ${\tau }_1=1667$ s and $E_1=7.13\times 10^5$ N ${\text{m}}^{-2}$ . D.8 ${\tau }_1=1546$ s , $E_1=7.39\times 10^5$ N ${\text{m}}^{-2}$ . 1200 1400 1600 1800 2000 2200 -8.1 -8.0 -7.9 -7.8 -7.7 -7.6 -7.5 -7.4 -7.3 -7.2

$\ln (-dF/dt)$ $t$ (s) 1200 1400 1600 1800 2000 2200 -7.9 -7.8 -7.7 -7.6 -7.5 -7.4 -7.3 -7.2

$\ln (-dF/dt)$ $t$ (s) Left: $\frac{dF}{dt}$ computed numerically using equal time intervals. Right: using data from the graph in D.6. 1.0pt D.9 For the 4 points on the left graph in D.8, we can write $F(t)=F_0+F_1{e}^{-t/{\tau }_1}\Rightarrow F_0=F(t)-F_1{e}^{-t/{\tau }_1}$ Thus, averaging $F_0$ for the 4 points of the left graph in D.8: $F_0=\left(\frac{40.32+40.31+40.34+40.30}{4}\right)=40.32$ gf Finally, $E_0=\beta F_0=324293\times 40.32$ N ${\text{m}}^{-2}$ . D.9 $E_0=1.31\times 10^7$ N ${\text{m}}^{-2}$ . 0.3pt

Confidential Experiments English (UK) SE2-9 D.10 The function $y(t)$ is given by $y(t)=F(t)-F_0-F_1{e}^{-t/{\tau }_1}$ , and was added in the Table introduced in A.3 using $F_0=40.32$ gf , $F_1=2.28$ gf and ${\tau }_1=1546$ s. D.10 See column $y(t)$ in the Table in A.3. 0.3pt D.11 Since $y(t)=F(t)-F_0-F_1{e}^{-t/{\tau }_1}$ , Then $y(t)=F_2{e}^{-t/{\tau }_2}+F_3{e}^{-t/{\tau }_3}+\cdots ,{\tau }_2>{\tau }_3>\cdots$ At long times, when the contributions from the higher components are small enough, we expect a linear behaviour for $\ln y(t)$: $\ln y=\ln F_2-\frac{1}{{\tau }_2}t$ . In this case, the $y(t)$ data points become meaningless above 500 s. In the region 200-500 s the graph is linear and that region can be used to extract the parameters of the second component. The equation of the straight line is $\ln y_2=b_2+m_{2}t$. From the graph below, $m_2=-(5.65\pm 0.19)\times 10^{-3}\Rightarrow {\tau }_2=\frac{1}{-m_2}=177$ s $b_2=0.33\pm 0.07\Rightarrow F_2=e^{b_2}=1.39\Rightarrow E_2=\beta F_2=4.5\times 10^5$ N ${\text{m}}^{-2}$ . . D.11 $E_2=4.5\times 10^5$ N ${\text{m}}^{-2}$ , ${\tau }_2=177$ s . 0 100 200 300 400 500 600 700 800 -6 -4 -2 0

$\ln y$

Topic: Elasticity & Materials, Wave Optics, Oscillations & Waves Metodi: Stress-Strain Analysis, Interference & Diffraction Analysis, Graph Linearization, Error Propagation, Experimental Data Analysis Competenze: Experimental Data Analysis, Graph Linearization, Error Propagation Objects: String, Mirror Fonte: Testo (PDF) — p.1