Question

Time period of a spring-block system is given by \(T = 2\pi\sqrt{\dfrac{m}{k}}\). If mass of the block is given by \(m = 10\,\text{g} \pm 10\,\text{mg}\) and time period is measured using a stopwatch having least count of \(2\,\text{s}\) and was found to be \(60\,\text{s}\) for \(50\) oscillations, then find the % error in measurement of \(k\).

• A. \(5.6%\)
• B. \(6.8%\)
• C. \(7.7%\)
• D. \(5.9%\)

Hint:

If a quantity depends on time squared, its percentage error gets multiplied by {2}. Always check the power carefully in error analysis.

Answer 1

The Correct Option is B

Concept:
For a spring–mass system: \[ T = 2\pi\sqrt{\frac{m}{k}} \Rightarrow k = \frac{4\pi^2 m}{T^2} \] Using error propagation: \[ \frac{\Delta k}{k} = \frac{\Delta m}{m} + 2\frac{\Delta T}{T} \]
Step 1: Error in Mass
Given: \[ m = 10\,\text{g} = 10{,}000\,\text{mg}, \quad \Delta m = 10\,\text{mg} \] \[ \frac{\Delta m}{m} = \frac{10}{10{,}000} = 0.001 = 0.1% \]
Step 2: Error in Time Period
Total time for 50 oscillations: \[ t = 60\,\text{s} \] Least count of stopwatch: \[ \Delta t = 2\,\text{s} \] Time period: \[ T = \frac{60}{50} = 1.2\,\text{s} \] Error in one time period: \[ \Delta T = \frac{2}{50} = 0.04\,\text{s} \] \[ \frac{\Delta T}{T} = \frac{0.04}{1.2} = \frac{1}{30} \approx 0.0333 = 3.33% \]
Step 3: Percentage Error in \(k\)
\[ \frac{\Delta k}{k} = 0.1% + 2(3.33%) = 0.1% + 6.66% = 6.76% \] \[ \boxed{% \text{ error in } k \approx 6.8%} \]