JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Units & Measurement

The time period of a simple harmonic oscillator is

T = 2 π \frac{k}{m}

. The 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
______ %.

(A)3.43
(B)3.35
(C)7.60
(D)6.76✓ Correct

Answer: (D)

Solution

The period of a spring oscillator is

T = 2 π \frac{m}{k}

, so

k = \frac{4 π ^{2} m}{T ^{2}}

and

k \propto \frac{m}{T ^{2}}

. The fractional error is therefore

\frac{Δ k}{k} = \frac{Δ m}{m} + 2 \frac{Δ T}{T}

. Here

m = 10

g with

Δ m = 10

mg, giving

\frac{Δ m}{m} = \frac{0.01}{10} = 0.001

. Fifty oscillations take

60

s, so

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

s, and the watch resolution of

2

s over

50

oscillations gives

Δ T = \frac{2}{50} = 0.04

s. Hence

\frac{Δ k}{k} = 0.001 + 2 \times \frac{0.04}{1.2} = 0.001 + 0.0667 = 0.0677

, a percentage error of

6.76%

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Units & Measurement

Solution

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