The time period of a simple harmonic oscillator is $$T=2\pi \sqrt{\frac{k}{m}}$$. Measured value of mass (m) of the object is 10 g with an accuracy of 10 mg and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant (k) is________%.

The spring constant is given by the formula $$ k = \frac{4\pi^2 m}{T^2} $$, derived from $$ T = 2\pi \sqrt{\frac{k}{m}} $$.

The percentage error in $$ k $$ is calculated using the maximum possible relative error, which is the sum of the relative errors in mass and twice the relative error in time period, since $$ k \propto m T^{-2} $$. Thus,

$$ \frac{\Delta k}{k} = \left| \frac{\Delta m}{m} \right| + \left| 2 \frac{\Delta T}{T} \right| $$

The percentage error is then $$ \frac{\Delta k}{k} \times 100\% $$.

Given:

• Mass $$ m = 10 $$ g, with an accuracy of $$ \Delta m = 10 $$ mg. Convert to consistent units: $$ m = 10 $$ g = 10000 mg, so $$ \Delta m = 10 $$ mg. Thus,

$$ \frac{\Delta m}{m} = \frac{10}{10000} = 0.001 $$

• Time for 50 oscillations is $$ t = 60 $$ s, measured with a watch of resolution 2 s. The resolution indicates the least count, so the absolute error in time measurement is $$ \Delta t = 2 $$ s.

The time period $$ T $$ is calculated as:

$$ T = \frac{t}{n} = \frac{60}{50} = 1.2 \text{ s} $$

Since $$ T = \frac{t}{n} $$ and $$ n = 50 $$ is exact, the relative error in $$ T $$ is the same as in $$ t $$:

$$ \frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{2}{60} = \frac{1}{30} $$

Now, substitute into the error formula:

$$ \frac{\Delta k}{k} = \left| 0.001 \right| + \left| 2 \times \frac{1}{30} \right| = 0.001 + \frac{2}{30} = 0.001 + \frac{1}{15} $$

Compute $$ \frac{1}{15} = 0.0666667 $$, so:

$$ \frac{\Delta k}{k} = 0.001 + 0.0666667 = 0.0676667 $$

Percentage error:

$$ 0.0676667 \times 100 = 6.76667\% $$

Rounding to two decimal places gives 6.77%, but the closest option is 6.76%. Using fractions:

$$ \frac{\Delta k}{k} = \frac{1}{1000} + \frac{1}{15} = \frac{15 + 1000}{15000} = \frac{1015}{15000} = \frac{203}{3000} $$

Percentage error:

$$ \frac{203}{3000} \times 100 = \frac{203}{30} = 6.7666...\% $$

This value matches option D, 6.76%.

Thus, the percentage error in the determination of the spring constant is 6.76%.