In a simple pendulum experiment, a student measures the period T of the pendulum many times and obtains an average value T = (2.540 ± 0.005) s. The length L of the pendulum is measured to be L = (1.60 ± 0.01) m.

Calculate, using $g = \frac{{4{\pi ^2}L}}{{{T^2}}}$, the value of the acceleration of free fall, including its uncertainty. State the value of the uncertainty to one significant figure.

In a different experiment a student investigates the dependence of the period T of a simple pendulum on the amplitude of oscillations θ. The graph shows the variation of $\frac{T}{{{T_0}}}$ with θ, where T₀ is the period for small amplitude oscillations.

The period may be considered to be independent of the amplitude θ as long as $\frac{{T - {T_0}}}{{{T_0}}} < 0.01$. Determine the maximum value of θ for which the period is independent of the amplitude.

$g = \frac{{4{\pi ^2} \times 1.60}}{{{{2.540}^2}}} = 9.7907$

$\Delta g = g\left( {\frac{{\Delta L}}{L} + 2 \times \frac{{\Delta T}}{T}} \right) =$ «$9.7907\left( {\frac{{0.01}}{{1.60}} + 2 \times \frac{{0.005}}{{2.540}}} \right) =$» 0.0997

OR

1.0%

hence g = (9.8 ± 0.1) «m$\,$s⁻²» OR Δg = 0.1 «m$\,$s⁻²»

For the first marking point answer must be given to at least 2 dp.
Accept calculations based on

${g_{\max }} = 9.8908$

${g_{\min }} = 9.6913$

$\frac{{{g_{\max }} - {g_{\min }}}}{2} = 0.099 \approx 0.1$

[3 marks]

$\frac{T}{{{T_0}}} = 1.01$

θ_max = 22 «º»

Accept answer from interval 20 to 24.

[2 marks]