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

To find the percentage error in the measurement of the spring constant \(k\), we utilize the given formula for the time period of a spring-block system:

T = 2\pi\sqrt{\frac{m}{k}}

Given data:

• Mass of the block, m = 10 \, \text{g} \pm 10 \, \text{mg} which means \Delta m = 0.01 \, \text{g}.
• Number of oscillations = 50, and the measured time for these oscillations is \(60 \, \text{s}\).
• Least count of the stopwatch = \(2 \, \text{s}\).

First, calculate the time period for a single oscillation:

T = \frac{60 \, \text{s}}{50} = 1.2 \, \text{s}

The absolute error in the time period measurement is given by the least count:

\Delta T = \frac{2 \, \text{s}}{50} = 0.04 \, \text{s}

Percentage error in T is:

\frac{\Delta T}{T} \times 100 = \frac{0.04}{1.2} \times 100 = 3.33\%

The formula for \(T\) implies:

T^2 = \frac{4\pi^2 m}{k}

Differentiating this, we get:

\frac{\Delta k}{k} = 2 \frac{\Delta T}{T} + \frac{\Delta m}{m}

Substituting the known values:

• 2 \times \frac{3.33}{100} = 6.66\%
• \frac{0.01}{10} \times 100 = 0.1\%

The percentage error in k is:

\frac{\Delta k}{k} = 6.66\% + 0.1\% = 6.76\% \approx 6.8\%

Therefore, the percentage error in the measurement of \(k\) is approximately 6.8%.