Question

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

• A. \(7.60%\)
• B. \(6.76%\)
• C. \(3.43%\)
• D. \(3.35%\)

Hint:

For quantities involving squares, remember: percentage error doubles when a variable appears squared.

Answer 1

The Correct Option is B

Step 1: Express \(k\) in terms of \(m\) and \(T\)

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

Step 2: Write relative error relation

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

Step 3: Error in mass

Given:
\[ \frac{\Delta m}{m}=10\%=0.10 \]

Step 4: Error in time measurement

Total time for 50 oscillations:
\[ t=60\,\text{s} \]
Resolution of watch \(=2\,\text{s}\Rightarrow \Delta t=2\,\text{s}\)

\[ \frac{\Delta t}{t}=\frac{2}{60}=\frac{1}{30} \]

Time period:
\[ T=\frac{t}{50}=\frac{60}{50}=1.2\,\text{s} \] \[ \Delta T=\frac{\Delta t}{50}=\frac{2}{50}=0.04\,\text{s} \] \[ \frac{\Delta T}{T}=\frac{0.04}{1.2}=\frac{1}{30} \]

Step 5: Total percentage error

\[ \frac{\Delta k}{k} =0.10+2\left(\frac{1}{30}\right) =0.10+0.0667 =0.1667 \] \[ \text{Percentage error}=0.1667\times100 \approx 16.67\% \]

Final Answer:
\[ \boxed{16.67\%} \]