ZiIP I rok 2008/2009		16.12.08
Nr 5	Wyznaczanie przyśpieszenia ziemskiego za pomocą wahadła prostego oraz logarytmicznego dekrementu tłumienia wahadła fizycznego.

3. Obliczenia

Obliczenie przyspieszenia ziemskiego

$$g = \frac{4\pi^{2}L}{T^{2}}$$

$$g_{1} = \frac{4{(3,14)}^{2} \bullet 0,618}{{(1,51)}^{2}} = 10,68\frac{m}{s^{2}}$$

$$g_{2} = \frac{4{(3,14)}^{2} \bullet 0,638}{{(1,56)}^{2}} = 10,29\frac{m}{s^{2}}$$

$$g_{3} = \frac{4{(3,14)}^{2} \bullet 0,552}{{(1,47)}^{2}} = 10,05\frac{m}{s^{2}}$$

$$g_{4} = \frac{4{(3,14)}^{2} \bullet 0,79}{{(1,74)}^{2}} = 10,25\frac{m}{s^{2}}$$

Obliczenie niepewności standardowej dla u(T)

$$u\left( T \right) = \sqrt{\frac{\left( \frac{_{d}t}{30} \right) + \left( \frac{_{e}t}{30} \right)}{3}} = \sqrt{\frac{\left( \frac{0,2}{30} \right) + \left( \frac{0,2}{30} \right)}{3}} = 0,0054\ s$$

Obliczenie niepewności standardowej dla u(l)

$$\overset{\overline{}}{l} = \frac{0,001}{\sqrt{3}} = 0,0006\ m$$

Obliczenie niepewności całkowitej u_c(g)

$$u_{c}\left( g \right) = \sqrt{\sum_{i = 1}^{1}{\left\lbrack \frac{\partial g}{\partial T}u\left( T \right) \right\rbrack^{2} + \sum_{i = 1}^{1}\left\lbrack \frac{\partial g}{\partial l}u\left( l \right) \right\rbrack^{2}}} =$$

$$= \sqrt{\left( \frac{- 8\pi^{2}L}{T^{3}} \bullet u(T) \right)^{2} + \left( \frac{4\pi^{2}}{T^{2}} \bullet u(l) \right)^{2}}$$

$$u_{c}\left( g_{1} \right) = \sqrt{\left( \frac{- 8 \bullet \left( 3,14 \right)^{2} \bullet 0,618}{\left( 1,51 \right)^{3}} \bullet 0,0054 \right)^{2} + \left( \frac{4 \bullet \left( 3,14 \right)^{2}}{\left( 1,51 \right)^{2}} \bullet 0,0006 \right)^{2}} = \sqrt{0,0059} = 0,076$$

$$u_{c}\left( g_{2} \right) = \sqrt{\left( \frac{- 8 \bullet \left( 3,14 \right)^{2} \bullet 0,637}{\left( 1,56 \right)^{3}} \bullet 0,0054 \right)^{2} + \left( \frac{4 \bullet \left( 3,14 \right)^{2}}{\left( 1,56 \right)^{2}} \bullet 0,0006 \right)^{2}} = \sqrt{0,0052} = 0,072$$

$$u_{c}\left( g_{3} \right) = \sqrt{\left( \frac{- 8 \bullet \left( 3,14 \right)^{2} \bullet 0,552}{\left( 1,47 \right)^{3}} \bullet 0,0054 \right)^{2} + \left( \frac{4 \bullet \left( 3,14 \right)^{2}}{\left( 1,47 \right)^{2}} \bullet 0,0006 \right)^{2}} = \sqrt{0,0055} = 0,074$$

$$u_{c}\left( g_{4} \right) = \sqrt{\left( \frac{- 8 \bullet \left( 3,14 \right)^{2} \bullet 0,79}{\left( 1,74 \right)^{3}} \bullet 0,0054 \right)^{2} + \left( \frac{4 \bullet \left( 3,14 \right)^{2}}{\left( 1,74 \right)^{2}} \bullet 0,0006 \right)^{2}} = \sqrt{0,041} = 0,064$$

3.2.Wyznaczenie logarytmicznego dekrementu tłumienia

D = βT

$D_{1} = In\frac{460}{425}$ =0,079 $\text{\ D}_{2} = In\frac{425}{395}$ =0,073 $D_{3} = In\frac{395}{345} =$ 0,144 $D_{4} = In\frac{345}{310} = 0,106$

$D_{5} = In\frac{310}{290} = 0,066$ $D_{6} = In\frac{290}{265} = 0,091$ $D_{7} = In\frac{265}{230} = 0,141$ $D_{8} = In\frac{230}{185} = 0,217$

$D_{9} = In\frac{185}{160} = 0,145$ $D_{10} = In\frac{160}{145} = 0,098$

Obliczenie średniego dekrementu

$$D_{sr} = \frac{\sum_{i = 1}^{10}D}{10} = \frac{0,079 + 0,073 + 0,144 + 0,106 + 0,066 + 0,091 + 0,141 + 0,217 + 0,145 + 0,098}{10} = 0,116$$

Obliczenie niepewności wyznaczenia dekrementu

$$u\left( D \right) = \sqrt{\frac{\sum_{i = 1}^{10}{(D - \overset{\overline{}}{D})}}{10 \bullet 9}}$$

$$u\left( D \right) = \sqrt{\frac{\begin{matrix} \left( 0,079 - 0,116 \right)^{2} + \left( 0,073 - 0,116 \right)^{2}{+ \left( 0,144 - 0,116 \right)}^{2} + \left( 0,106 - 0,116 \right)^{2}{+ \left( 0,066 - 0,116 \right)}^{2} + \\ \left( 0,091 - 0,116 \right)^{2} + \left( 0,141 - 0,116 \right)^{2} + \left( 0,217 - 0,116 \right)^{2}{+ \left( 0,145 - 0,116 \right)}^{2}{+ \left( 0,098 - 0,116 \right)}^{2} \\ \\ \end{matrix}}{90}} = \sqrt{\frac{0,0192}{90} =}0,014$$

Obliczenie wielkości b

Okres drgań:

$$T_{1} = \frac{t_{1}}{10} = \frac{38,08}{11} = 3,46$$

$$T_{2} = \frac{t_{2}}{10} = \frac{38,1}{11} = 3,46$$

$$T_{3} = \frac{t_{3}}{10} = \frac{38,24}{11} = 3,47$$

$$T_{sr} = \frac{3,46 + 3,46 + 3,47}{3} = 3,46$$

$$b = \frac{2mD}{T}$$

b₁= 0,012 b₆= 0,014

b₂= 0,011 b₇= 0,021

b₃= 0,022 b₈= 0,033

b₄= 0,016 b₉= 0,022

b₅= 0,010 b₁₀= 0,015

Obliczenie niepewności u(b)

$$u_{B}\left( T \right) = \sqrt{\frac{\left( \frac{_{d}t}{10} \right)^{2} + \left( \frac{_{e}t}{10} \right)^{2}}{3}}$$

$$u_{B}\left( T \right) = \sqrt{\frac{\left( \frac{0,2}{10} \right)^{2} + \left( \frac{0,2}{10} \right)^{2}}{3}} = \sqrt{0,00026} = 0,016$$

$$u_{c}\left( b \right) = \sqrt{\sum_{i = 1}^{1}\left\lbrack \frac{\partial b}{\partial T}u\left( T \right) \right\rbrack^{2} + \sum_{i = 1}^{1}\left\lbrack \frac{\partial b}{\partial D}u\left( D \right) \right\rbrack^{2} + \sum_{i = 1}^{1}\left\lbrack \frac{\partial b}{\partial m}u\left( m \right) \right\rbrack^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sqrt{\left( \frac{- 2mD}{T^{2}}u(T) \right)^{2} + \left( \frac{2m}{T}u(D) \right)^{2} + \left( \frac{2D}{T}u(m) \right)^{2}}$$

$$u_{c}\left( b \right) = \sqrt{\left( \frac{- 2 \bullet 0,2685 \bullet 0,116}{11,9716}0,016 \right)^{2} + \left( \frac{2 \bullet 0,2685}{3,46}0,014 \right)^{2} + \left( \frac{2 \bullet 0,116}{3,46}0,289 \right)} = \sqrt{0,00038} = 0,019$$

Obliczenie wielkości β

$$\beta = \frac{D}{T}$$

β₁=0,0228 β₆=0,0263

β₂=0,0210 β₇=0,0407

β₃=0,0416 β₈=0,0627

β₄=0,0306 β₉=0,0419

β₅=0,0190 β₁₀=0,0283

Obliczenie niepewności u(β)

$$u_{c}\left( \beta \right) = \sqrt{\sum_{i = 1}^{1}\left\lbrack \frac{\partial\beta}{\partial T}u\left( T \right) \right\rbrack^{2} + \sum_{i = 1}^{1}\left\lbrack \frac{\partial\beta}{\partial D}u\left( D \right) \right\rbrack^{2} =}\sqrt{\left( \frac{- D}{T^{2}}u(T) \right)^{2} + \left( \frac{T}{T^{2}}u(D) \right)^{2}}$$

$$u_{c}\left( \beta \right) = \sqrt{\left( \frac{- 0,116}{\left( 3,46 \right)^{2}} \bullet 0,016 \right)^{2} + \left( \frac{3,46}{\left( 3,46 \right)^{2}} \bullet 0,014 \right)^{2}} = \sqrt{0,00001639} = 0,0041$$