杨氏模量实验数据处理

	1	2	3	4	5	6	平均值
L/mm	159.5	159.3	159.5	159.4	159.6	159.4	159.5
D/mm	5.987	5.977	5.975	5.987	5.979	5.980	5.980
M/g	37.433

不确定度的计算：

\[u_{m} = u_{B} = \frac{\mathrm{\Delta}_{仪} \ast m}{\sqrt{3}} = 0.6mg\]

\[u_{A} = \sqrt{\frac{\sum_{i = 1}^{6}{(l_{i} - l)}^{2}}{6 \ast 5}} = 0.048mm\ \ \ \ u_{B} = \frac{\mathrm{\Delta}_{仪} \ast l}{\sqrt{3}} = 0.12mm\ \ \ u_{l} = \sqrt{{u_{A}}^{2} + {u_{B}}^{2}} = 0.13mm\]

\[u_{A} = \sqrt{\frac{\sum_{i = 1}^{6}{(d_{i} - d)}^{2}}{6 \ast 5}} = 0.0021mm\ \ \ \ u_{B} = \frac{\mathrm{\Delta}_{仪} \ast d}{\sqrt{3}} = 0.0023mm\ \ \ u_{d} = \sqrt{{u_{A}}^{2} + {u_{B}}^{2}} = 0.0031mm\]

X/mm	5	10	15	20	25	30	35	40	45	50
X/L	2.56%	5.12%	7.67%	10.23%	12.79%	15.35%	×	20.46%	23.01%	25.58%
f/Hz	730.1	729.1	728.1	727.2	726.9	725.7	×	726.5	726.8	726.9

外延法作拟合图曲线如下：

利用代码找到最小值：

也就是说共振频率约为726.2Hz

\[u_{f} = u_{B} = \frac{\mathrm{\Delta}_{仪} \ast f}{\sqrt{3}} = 0.06Hz\]

代入公式：

\[E = 1.6067 \ast \frac{L^{3} \ast m \ast f^{2}}{d^{4}} = 1.0064 \ast 10^{11}Pa\]

\[u_{E} = E\sqrt{{(3\frac{u_{L}}{L})}^{2} + {(4\frac{u_{d}}{d})}^{2} + {(\frac{u_{m}}{m})}^{2} + {(2\frac{u_{f}}{f})}^{2}} = 1.6452 \ast 10^{9}Pa\]

因此该棒的杨氏模量为：

\[E = (1.0064 \pm 0.0164) \ast 10^{11}\ Pa\]