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 : 

(1) 3% 

(2) 1% 

(3) 4% 

(4) 5%

Answer 1

The correct option is (4) 5%

To determine the spring constant k of a spring experimentally, we can use the formula derived from Hooke's Law and the period of oscillation for a mass-spring system:

\(T = 2 \pi \sqrt\frac{m}{k}\)

Here, T is the period of oscillation, m is the mass, and k is the spring constant. Rearranging the formula to solve for k, we get:

\(k = \frac{4\pi^2 m}{T^2}\)

To find the error in k, we have to consider the errors in both the measurements of T and m. Let's denote the percentage errors as follows:

ΔT/T⋅100% = 2% (positive error)

Δm/m⋅100% = −1% (negative error)

According to the rules of error propagation, the relative error in k can be found by adding the relative errors in the measurements, each multiplied by the respective powers to which they affect k. Since T is squared in the denominator and m is linear in the numerator, the calculation is as follows:

\(\frac{\Delta k}{k} = |-2 . \frac{\Delta T}{T}| + |1. \frac{\Delta m}{m}|\)

Substituting the percentage errors:

\(\frac{\Delta k}{k} = |-2.0.02| + |1.(-0.01)|\)

\(\frac{\Delta k}{k} = 0.04 + 0.01\)

\(\frac{\Delta k}{k} = 0.05\)

Thus, the percentage error in determining the value of k is:

\(\frac{\Delta k}{k}. 100\% = 5\%\)

Answer 2

Correct option is :  (4) 5%

\( T=2 \pi \sqrt{\frac{m}{k}}\)

\(\mathrm{T}^{2} \propto \frac{\mathrm{m}}{\mathrm{k}}\)

\(\frac{2 \Delta \mathrm{T}}{\mathrm{T}} \%=\frac{\Delta \mathrm{m}}{\mathrm{m}} \%-\frac{\Delta \mathrm{k}}{\mathrm{k}} \%\)

\(\frac{\Delta \mathrm{k}}{\mathrm{k}} \%=\frac{\Delta \mathrm{m}}{\mathrm{m}} \%-\frac{2 \Delta \mathrm{T}}{\mathrm{T}} \%\)

\(\frac{\Delta \mathrm{k}}{\mathrm{k}} \%=(-1) \%-2(2) \%=|-5 \%|=5 \%\)