To find the spring constant (k) of a spring experimentally, a student commits 2% positive error in the measurement of time and 1% negative error in measurement of mass. The percentage error in determining value of k is :
- 0.03
- 0.01
- 0.04
- 0.05
The Correct Option is D
Approach Solution - 1
The time period \( T \) of a spring is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}}. \]
Squaring both sides:
\[ T^2 \propto \frac{m}{k}. \]
Taking percentage errors:
\[ \frac{\Delta T^2}{T^2} = \frac{\Delta m}{m} - \frac{\Delta k}{k}. \]
Substituting the relationship \( \frac{\Delta T^2}{T^2} = 2 \frac{\Delta T}{T} \), we get:
\[ 2 \frac{\Delta T}{T} = \frac{\Delta m}{m} - \frac{\Delta k}{k}. \]
Rewriting for \( \frac{\Delta k}{k} \):
\[ \frac{\Delta k}{k} = \frac{\Delta m}{m} - 2 \frac{\Delta T}{T}. \]
Given:
\( \frac{\Delta T}{T} = 2\% \) (positive error),
\( \frac{\Delta m}{m} = -1\% \) (negative error).
Substitute the values:
\[ \frac{\Delta k}{k} = (-1\%) - 2(2\%) = -1\% - 4\% = -5\%. \]
Hence, the magnitude of the percentage error in \( k \) is:
\[ \left| \frac{\Delta k}{k} \right| = 5\%. \]
Final Answer: \( 5\% \) (Option 4)
Approach Solution -2
The problem requires us to calculate the percentage error in the spring constant, \( k \), given the percentage errors in the measurement of the time period and the mass.
Concept Used:
1. Time Period of a Spring-Mass System: The time period \( T \) of a mass \( m \) oscillating on a spring with a spring constant \( k \) is given by the formula:
\[ T = 2\pi\sqrt{\frac{m}{k}} \]2. Propagation of Errors: For a physical quantity \( Q \) that depends on other measured quantities \( A, B, C, \dots \) according to the relation \( Q = c A^a B^b C^c \dots \), where \( c \) is a constant, the maximum fractional error in \( Q \) is the sum of the fractional errors in \( A, B, C, \dots \), each multiplied by the magnitude of its corresponding power. The percentage error is 100 times the fractional error.
\[ \frac{\Delta Q}{Q} \times 100\% = \left(|a|\frac{\Delta A}{A} + |b|\frac{\Delta B}{B} + |c|\frac{\Delta C}{C} + \dots\right) \times 100\% \]When calculating the maximum possible error, we add the magnitudes of the individual percentage errors, regardless of whether they are positive or negative.
Step-by-Step Solution:
Step 1: Express the spring constant \( k \) in terms of mass \( m \) and time period \( T \).
We start with the formula for the time period:
\[ T = 2\pi\sqrt{\frac{m}{k}} \]To solve for \( k \), we first square both sides of the equation:
\[ T^2 = (2\pi)^2 \left(\frac{m}{k}\right) = 4\pi^2 \frac{m}{k} \]Now, we rearrange the equation to isolate \( k \):
\[ k = 4\pi^2 \frac{m}{T^2} = 4\pi^2 m^1 T^{-2} \]Step 2: Formulate the equation for the percentage error in \( k \).
Using the rule for propagation of errors, the fractional error in \( k \) is:
\[ \frac{\Delta k}{k} = |1|\frac{\Delta m}{m} + |-2|\frac{\Delta T}{T} \]The percentage error in \( k \) is therefore:
\[ \% \text{ error in } k = \left(\frac{\Delta k}{k} \times 100\%\right) = \left(\frac{\Delta m}{m} \times 100\%\right) + 2 \left(\frac{\Delta T}{T} \times 100\%\right) \]Step 3: Substitute the given error values into the equation.
We are given:
- Percentage error in time (\( T \)) = 2% (positive)
- Percentage error in mass (\( m \)) = 1% (negative)
For calculating the maximum possible error, we use the magnitudes of the percentage errors.
\[ \% \text{ error in } m = 1\% \] \[ \% \text{ error in } T = 2\% \]Final Computation & Result:
Step 4: Calculate the total percentage error in \( k \).
Substituting the values from Step 3 into the error formula from Step 2:
\[ \% \text{ error in } k = (1\%) + 2 \times (2\%) \] \[ \% \text{ error in } k = 1\% + 4\% \] \[ \% \text{ error in } k = 5\% \]Thus, the percentage error in determining the value of \( k \) is 5%.
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