Question 30: 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 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 ________%.

Accepted Answer

The formula for the time period of simple harmonic oscillator is
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$.
$$\Rightarrow $$ $$ \mathrm{T}^2=(2 \pi)^2\left(\frac{\mathrm{k}}{\mathrm{~m}}ight) $$
$\Rightarrow \mathrm{k}=\frac{\mathrm{T}^2 \cdot \mathrm{~m}}{4 \pi^2}$
Using propagation of error, when quantities are multiplied or divided, their relative errors (percentage errors) are added. The constants have no error.
$$ \frac{\Delta \mathrm{k}}{\mathrm{k}} 	imes 100=\left(\frac{\Delta \mathrm{m}}{\mathrm{~m}}+2 \cdot \frac{\Delta \mathrm{~T}}{\mathrm{~T}}ight) 	imes 100 $$
The measured value of mass is, $\mathrm{m}=10 \mathrm{~g}$
Accuracy/Error in measurement of mass is, $\Delta \mathrm{m}=10 \mathrm{mg}=0.01 \mathrm{~g}$
So, percentage error in measurement of mass $\frac{\Delta \mathrm{m}}{\mathrm{m}} 	imes 100 \%=\frac{0.01}{10} 	imes 100=0.1 \%$
The percentage error in the time period ( $T$ ) is the same as the percentage error in the total time ( $t$ ) measured for n oscillations.
Total time measured is, $\mathrm{t}=60 \mathrm{~s}$
Resolution/Error in measurement of time is, $\Delta t=2 \mathrm{~s}$
So, percentage error in measurement of time is, $\frac{\Delta t}{t} 	imes 100 \%=\frac{2}{60} 	imes 100 \%=3.33 \%$
So, using propagation formula:
Percentage error in $\mathrm{k}=(\%$ error in m$)+2 	imes(\%$ error in t$)=0.1 \%+2 	imes 3.33 \%=6.76 \%$
Therefore, the correct option is (D).

Answer

3.43

Answer

7.60

Answer

3.35

Answer

6.76

Units , Dimensions and Error Analysis

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