The time period of a simple harmonic oscillator is T = 2 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 i

Question

The time period of a simple harmonic oscillator is $T = 2 π \sqrt{\frac{k}{m}} .$ 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 ______ %. [2026]

Smart Achievers · Accepted Answer

(2)

$\frac{Δ K}{K} = \frac{2 Δ T}{T} + \frac{Δ m}{m}$

$T = \frac{60}{50} = 1.2 sec$

$Δ T = \frac{2}{50}$

$∴$ $\frac{Δ K}{K} = \frac{2 \times 2}{50 \times 1.2} + \frac{10 \times 10^{- 3}}{10} = 0.0676$

$∴$ $% Error = 6.76 %$

Solution

1 3.35

2 6.76

3 7.60

4 3.43

Ans.
(2)

$\frac{Δ K}{K} = \frac{2 Δ T}{T} + \frac{Δ m}{m}$

$T = \frac{60}{50} = 1.2 sec$

$Δ T = \frac{2}{50}$

$∴$ $\frac{Δ K}{K} = \frac{2 \times 2}{50 \times 1.2} + \frac{10 \times 10^{- 3}}{10} = 0.0676$

$∴$ $% Error = 6.76 %$

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