The time period of a simple harmonic oscillator is T=2 π √ k m. Measured value of mass (m) of the object is 10 g with an accuracy of 10 mg and time…

The given formula for time period is

$T = 2\pi \sqrt{\frac{k}{m}}$T=2πmk​​.

Squaring both sides, we get $T^2 = 4\pi^2 \frac{k}{m}$T2=4π2mk​, which gives $k = \frac{4\pi^2 m}{T^2}$k=T24π2m​.

The relative error in $k$k is given by $\frac{\Delta k}{k} = \frac{\Delta m}{m} + 2\frac{\Delta T}{T}$kΔk​=mΔm​+2TΔT​.

Given: $m = 10$m=10 g, $\Delta m = 10$Δm=10 mg $= 0.01$=0.01 g.

Total time for $n=50$n=50 oscillations is $t = 60$t=60 s with resolution $\Delta t = 2$Δt=2 s.

Since $T = t/n$T=t/n, the relative error in $T$T is the same as in $t$t: $\frac{\Delta T}{T} = \frac{\Delta t}{t}$TΔT​=tΔt​.

Substituting the values: $\frac{\Delta k}{k} = \frac{0.01}{10} + 2\left(\frac{2}{60}\right)$kΔk​=100.01​+2(602​).

$$\frac{\Delta k}{k} = 0.001 + \frac{2}{30} = 0.001 + 0.0666...$$kΔk​=0.001+302​=0.001+0.0666...

Percentage error $= \left(0.001 + \frac{1}{15}\right) \times 100 = 0.1 + \frac{100}{15} = 0.1 + 6.666... = 6.766...\%$=(0.001+151​)×100=0.1+15100​=0.1+6.666...=6.766...%.

Rounding to two decimal places, we get $6.76\%$6.76%.