Question
mnol=50g
xnol=19cm
M=(m+mnol) plus or minus 0.05cm 250, 450, 650,850,1050
loading =25,30,35,40,45,45
unloading =25,30,35,40,45
Mean,pointer (x plus or minus 0.05cm)= 25,30,35,40,45
Extension e=(x-xnol) plus or minus 0.05cm 6cm , 11cm, 16cm, 2cm, 26cm
From the information given above, use the formula Error in S=S×the root of (error in M/change in M)^2 + (Error in e/change in e)^2 to estimate the error
xnol=19cm
M=(m+mnol) plus or minus 0.05cm 250, 450, 650,850,1050
loading =25,30,35,40,45,45
unloading =25,30,35,40,45
Mean,pointer (x plus or minus 0.05cm)= 25,30,35,40,45
Extension e=(x-xnol) plus or minus 0.05cm 6cm , 11cm, 16cm, 2cm, 26cm
From the information given above, use the formula Error in S=S×the root of (error in M/change in M)^2 + (Error in e/change in e)^2 to estimate the error
Answers
GPT-4o mini
To estimate the error in the calculated quantity using the given formula, we first need to identify and summarize the relevant data:
1. **Weight \( M \):** This is the total weight applied, which is \( M = m + m_{nol} \). You will typically derive \( m \) from the specified values \( 250g, 450g, 650g, 850g, 1050g \) combined with \( m_{nol} = 50g \). For this example, let's define \( m \) (actual loading):
- Loading Mass \( m \): \( 250g, 450g, 650g, 850g, 1050g \)
- Total Mass \( M = m + 50g \)
- Corresponding Masses: \( 300g, 500g, 700g, 800g, 1100g \)
2. **Extension \( e \)**: Calculated as \( e = x - x_{nol} \)
- Given \( x_{nol} = 19 \, cm \) and the mean pointer readings: \( 25, 30, 35, 40, 45 \)
- Extensions: \( e = \{ 25 - 19, 30 - 19, 35 - 19, 40 - 19, 45 - 19 \} = \{ 6, 11, 16, 21, 26 \} \)
However, you stated some data showing extensions with "plus or minus 0.05 cm," which represents measurement uncertainty:
- Calculated Extensions: \( e \approx \{ 6 \ldots \, 11 \ldots \, 16 \ldots \, 21 \ldots \, 26 \} \)
3. **Errors**:
- Error in \( M \) (Weight) \( \text{Error in } M = 0.05 \, \text{cm} \) (assuming it is the same for all).
- Error in \( e \) \( \text{Error in } e = 0.05 \, \text{cm} \).
4. **Change in \( M \)** and **Change in \( e \)**:
- Calculate the average change in \( M \) and \( e \). The change in \( M \) will simply be the difference between successive weights, whereas for \( e \), we can calculate the change in the extension for each corresponding weight.
**Example calculation of change and following error:**
Let's assume that we want to calculate the error for a specific mass point, say \( 300g \) (where \( M = 300g \) and \( e = 6 \)).
1. **Change in \( M \)**:
- \( \text{Change in } M \) can be approximated as the difference between a point (\( 300g \)) and the next point (\( 500g \)).
- So, \( \Delta M = 500g - 300g = 200g \).
2. **Change in \( e \)**:
- Corresponding differences in \( e \): \( e = 11 - 6 = 5 \).
Therefore, we can compute:
\[
\text{Error in } S = S \cdot \sqrt{ \left( \frac{\text{Error in } M}{\Delta M} \right)^2 + \left( \frac{\text{Error in } e}{\Delta e} \right)^2}
\]
If we were estimating \( S \):
- Calculated \( S \) depends on your system.
- Let's say \( S = 100 \, \text{(example value)} \).
Using the details for \( \Delta M = 200g, \Delta e = 5 \, cm\):
Final formula calculation:
\[
\text{Error in } S = 100 \cdot \sqrt{ \left( \frac{0.05}{200} \right)^2 + \left( \frac{0.05}{5} \right)^2 }
\]
Calculating:
- First term:
\[
\left( \frac{0.05}{200} \right)^2 = (0.00025)^2 = 6.25 \times 10^{-8}
\]
- Second term:
\[
\left( \frac{0.05}{5} \right)^2 = (0.01)^2 = 0.0001
\]
\[
\text{Total} = \sqrt{ 6.25 \times 10^{-8} + 0.0001 } = \sqrt{0.0001000625} \approx 0.01
\]
Finally,
\[
\text{Error in } S \approx 100 \cdot 0.01 = 1.0
\]
You would repeat this process for various points to compute the combined errors you are interested in. Each calculation for other points can similarly be done using the \( M \) and \( e \) values accordingly.
1. **Weight \( M \):** This is the total weight applied, which is \( M = m + m_{nol} \). You will typically derive \( m \) from the specified values \( 250g, 450g, 650g, 850g, 1050g \) combined with \( m_{nol} = 50g \). For this example, let's define \( m \) (actual loading):
- Loading Mass \( m \): \( 250g, 450g, 650g, 850g, 1050g \)
- Total Mass \( M = m + 50g \)
- Corresponding Masses: \( 300g, 500g, 700g, 800g, 1100g \)
2. **Extension \( e \)**: Calculated as \( e = x - x_{nol} \)
- Given \( x_{nol} = 19 \, cm \) and the mean pointer readings: \( 25, 30, 35, 40, 45 \)
- Extensions: \( e = \{ 25 - 19, 30 - 19, 35 - 19, 40 - 19, 45 - 19 \} = \{ 6, 11, 16, 21, 26 \} \)
However, you stated some data showing extensions with "plus or minus 0.05 cm," which represents measurement uncertainty:
- Calculated Extensions: \( e \approx \{ 6 \ldots \, 11 \ldots \, 16 \ldots \, 21 \ldots \, 26 \} \)
3. **Errors**:
- Error in \( M \) (Weight) \( \text{Error in } M = 0.05 \, \text{cm} \) (assuming it is the same for all).
- Error in \( e \) \( \text{Error in } e = 0.05 \, \text{cm} \).
4. **Change in \( M \)** and **Change in \( e \)**:
- Calculate the average change in \( M \) and \( e \). The change in \( M \) will simply be the difference between successive weights, whereas for \( e \), we can calculate the change in the extension for each corresponding weight.
**Example calculation of change and following error:**
Let's assume that we want to calculate the error for a specific mass point, say \( 300g \) (where \( M = 300g \) and \( e = 6 \)).
1. **Change in \( M \)**:
- \( \text{Change in } M \) can be approximated as the difference between a point (\( 300g \)) and the next point (\( 500g \)).
- So, \( \Delta M = 500g - 300g = 200g \).
2. **Change in \( e \)**:
- Corresponding differences in \( e \): \( e = 11 - 6 = 5 \).
Therefore, we can compute:
\[
\text{Error in } S = S \cdot \sqrt{ \left( \frac{\text{Error in } M}{\Delta M} \right)^2 + \left( \frac{\text{Error in } e}{\Delta e} \right)^2}
\]
If we were estimating \( S \):
- Calculated \( S \) depends on your system.
- Let's say \( S = 100 \, \text{(example value)} \).
Using the details for \( \Delta M = 200g, \Delta e = 5 \, cm\):
Final formula calculation:
\[
\text{Error in } S = 100 \cdot \sqrt{ \left( \frac{0.05}{200} \right)^2 + \left( \frac{0.05}{5} \right)^2 }
\]
Calculating:
- First term:
\[
\left( \frac{0.05}{200} \right)^2 = (0.00025)^2 = 6.25 \times 10^{-8}
\]
- Second term:
\[
\left( \frac{0.05}{5} \right)^2 = (0.01)^2 = 0.0001
\]
\[
\text{Total} = \sqrt{ 6.25 \times 10^{-8} + 0.0001 } = \sqrt{0.0001000625} \approx 0.01
\]
Finally,
\[
\text{Error in } S \approx 100 \cdot 0.01 = 1.0
\]
You would repeat this process for various points to compute the combined errors you are interested in. Each calculation for other points can similarly be done using the \( M \) and \( e \) values accordingly.