Question 1: Units , Dimensions and Error Analysis

Question

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 10g with an accuracy of $10mg$, and the time for $50 oscillations $ of the spring is found to be 60s using a watch of 2s resolution. The percentage error in the determination of spring constant (k) is ______ %.

Accepted Answer

The formula for the time period of a simple harmonic oscillator is: $\$
$T= 2\pi \sqrt{\frac{k}{m}}$ $\$
$\Rightarrow T^{2} = (2\pi)^{2}(\frac{k}{m})$ $\$

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

When quantities are multiplied or divided, their relative (percentage) errors are added . Constants do not contribute to error.$\$

$\frac{\Delta k}{k} 	imes 100 = \left(\frac{\Delta m}{m} + 2 \cdot \frac{\Delta T}{T}ight) 	imes 100 $ $\$
Given Data $\$
Mass: $m = 10	ext{g}$ $\$
Error in mass: $\Delta m = 10	ext{mg} = 0.01	ext{g} $ $\$

$	ext{Percentage error in mass} = \frac{0.01}{10} 	imes 100 = 0.1\%$ $\$
Total time for 50 oscillations: $t = 60,	ext{s}$ $\$
Error in time (resolution): $\Delta t = 2,	ext{s}$ $\$
$	ext{Percentage error in time} ={\frac{\Delta t}{t}} 	imes100\%= \frac{2}{60} 	imes 100\% = 3.33\% $ $\$

$	ext{Percentage error in } k = (\% 	ext{ error in } m) + 2 	imes (\% 	ext{ error in } T)$ $\$
$= 0.1 + 2 	imes 3.33 = 6.76\% $$\$

Final Answer:
$\boxed{6.76\%}$$\$

Correct option: (D)

Answer

3.43

Answer

7.60

Answer

3.35

Answer

6.76

Units , Dimensions and Error Analysis

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