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

The Correct Option is D

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)