Question 76: Units , Dimensions and Error Analysis

Question

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 :

Accepted Answer

To determine the spring constant $$k$$ of a spring experimentally, we can use the formula derived from Hooke's Law and the period of oscillation for a mass-spring system:
$$T = 2 \pi \sqrt{\frac{m}{k}}$$
Here, $$T$$ is the period of oscillation, $$m$$ is the mass, and $$k$$ is the spring constant. Rearranging the formula to solve for $$k$$, we get:
$$k = \frac{4 \pi^2 m}{T^2}$$
To find the error in $$k$$, we have to consider the errors in both the measurements of $$T$$ and $$m$$. Let's denote the percentage errors as follows:
$$\Delta T / T \cdot 100\% = 2\%$$ (positive error)
$$\Delta m / m \cdot 100\% = -1\%$$ (negative error)
According to the rules of error propagation, the relative error in $$k$$ can be found by adding the relative errors in the measurements, each multiplied by the respective powers to which they affect $$k$$. Since $$T$$ is squared in the denominator and $$m$$ is linear in the numerator, the calculation is as follows:
$$\frac{\Delta k}{k} = \left| -2 \cdot \frac{\Delta T}{T} \right| + \left| 1 \cdot \frac{\Delta m}{m} \right|$$
Substituting the percentage errors:
$$\frac{\Delta k}{k} = \left| -2 \cdot 0.02 \right| + \left| 1 \cdot (-0.01) \right|$$
$$\frac{\Delta k}{k} = 0.04 + 0.01$$
$$\frac{\Delta k}{k} = 0.05$$
Thus, the percentage error in determining the value of $$k$$ is:
$$\frac{\Delta k}{k} \cdot 100\% = 5\%$$
The correct answer is Option A: 5%

Answer

5%

Answer

3%

Answer

1%

Answer

4%

Units , Dimensions and Error Analysis

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 :

Choose the correct answer: