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 :

• A. 0.03
• B. 0.01
• C. 0.04
• D. 0.05

Answer 1

Answer: (D)

0.05

Answer 2

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)