JEE Main 2026 — Oscillations Question with Solution

The given formula for time period is
$T=2\pi \sqrt{\frac{k}{m}}$.
Squaring both sides, we get $T^2=4{\pi }^2\frac{k}{m}$, which gives $k=\frac{4{\pi }^{2}m}{T^2}$.
The relative error in $k$ is given by $\frac{\Delta k}{k}=\frac{\Delta m}{m}+2\frac{\Delta T}{T}$.
Given: $m=10$ g, $\Delta m=10$ mg $=0.01$ g.
Total time for $n=50$ oscillations is $t=60$ s with resolution $\Delta t=2$ s.
Since $T=t/n$, the relative error in $T$ is the same as in $t$: $\frac{\Delta T}{T}=\frac{\Delta t}{t}$.
Substituting the values: $\frac{\Delta k}{k}=\frac{0.01}{10}+2\left(\frac{2}{60}\right)$.
$\frac{\Delta k}{k}=0.001+\frac{2}{30}=0.001+\mathrm{0.0666...}$
Percentage error $=\left(0.001+\frac{1}{15}\right)\times 100=0.1+\frac{100}{15}=0.1+\mathrm{6.666...}=6.766...\%$.
Rounding to two decimal places, we get $6.76\%$.