Q. The time period of a simple harmonic oscillator is T = 2π √(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 ______ %.

Correct Answer: 6.76

Solution

The time period of a simple harmonic oscillator is given by:

T = 2π √( m / k )

Rewriting for spring constant:

k ∝ m / T²

Taking percentage error:

Δk / k = Δm / m + 2 ΔT / T

Given mass:

m = 10 g,   Δm = 10 mg = 0.01 g

Percentage error in mass:

Δm / m = 0.01 / 10 = 0.001 = 0.1%

Time for 50 oscillations:

T_total = 60 s,   ΔT_total = 2 s

Percentage error in time:

ΔT / T = 2 / 60 = 0.0333 = 3.33%

Substituting values:

Δk / k = 0.1 + 2 × 3.33

Δk / k = 6.76%

Hence, percentage error in determination of spring constant is 6.76%.

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