Oscillations

JEE Main level, single correct, medium

105 of 194

Question 105 of 194

The time period of a simple harmonic oscillator is T=2πkmT=2 \pi \sqrt{\frac{k}{m}}. Measured value of mass (m)(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)(k) is ____\_\_\_\_ \%.

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Answer: (D) 6.76

The given formula for time period is
T=2πkmT = 2\pi \sqrt{\frac{k}{m}}.
Squaring both sides, we get T2=4π2kmT^2 = 4\pi^2 \frac{k}{m}, which gives k=4π2mT2k = \frac{4\pi^2 m}{T^2}.
The relative error in kk is given by Δkk=Δmm+2ΔTT\frac{\Delta k}{k} = \frac{\Delta m}{m} + 2\frac{\Delta T}{T}.
Given: m=10m = 10 g, Δm=10\Delta m = 10 mg =0.01= 0.01 g.
Total time for n=50n=50 oscillations is t=60t = 60 s with resolution Δt=2\Delta t = 2 s.
Since T=t/nT = t/n, the relative error in TT is the same as in tt: ΔTT=Δtt\frac{\Delta T}{T} = \frac{\Delta t}{t}.
Substituting the values: Δkk=0.0110+2(260)\frac{\Delta k}{k} = \frac{0.01}{10} + 2\left(\frac{2}{60}\right).
Δkk=0.001+230=0.001+0.0666...\frac{\Delta k}{k} = 0.001 + \frac{2}{30} = 0.001 + 0.0666...
Percentage error =(0.001+115)×100=0.1+10015=0.1+6.666...=6.766...%= \left(0.001 + \frac{1}{15}\right) \times 100 = 0.1 + \frac{100}{15} = 0.1 + 6.666... = 6.766...\%.
Rounding to two decimal places, we get 6.76%6.76\%.

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