You are provided with a pendulum bob, a metre rule, a stopwatch, a retort stand with clamp, and other necessary apparatus. i. Suspend the pendulum bob from ...

Set-up and measurements

The pendulum bob is suspended from the clamp of the retort stand by a light thread. The length \(L = AC\) is measured from the point of suspension \(A\) to the centre of the bob \(C\). For each length the bob is displaced through a small angle and released so that it swings in a vertical plane, and the time \(t\) for 20 complete oscillations is taken with the stopwatch. The period is obtained from \(T = \dfrac{t}{20}\) and \(\sqrt{L}\) is evaluated with \(L\) in metres.

Table of readings

S/N	L / cm	L / m	t (20 osc.) / s	T / s	√L / m1/2
1	90	0.90	38.1	1.91	0.949
2	80	0.80	35.9	1.79	0.894
3	70	0.70	33.6	1.68	0.837
4	60	0.60	31.1	1.55	0.775
5	50	0.50	28.4	1.42	0.707

Graph of T against √L

Plotting \(T\) (vertical axis) against \(\sqrt{L}\) (horizontal axis) gives a straight line passing through the origin.

Slope of the graph

Taking two widely separated points on the line of best fit, \((0.707,\,1.42)\) and \((0.949,\,1.90)\):

\[ s = \frac{\Delta T}{\Delta \sqrt{L}} = \frac{1.90 - 1.42}{0.949 - 0.707} = \frac{0.48}{0.242} = 1.98\ \text{s\,m}^{-1/2}. \]

Evaluating g

For a simple pendulum \(T = 2\pi\sqrt{L/g} = \dfrac{2\pi}{\sqrt{g}}\sqrt{L}\), so the slope is \(s = \dfrac{2\pi}{\sqrt{g}}\) and therefore

\[ g = \frac{4\pi^{2}}{s^{2}} = \frac{4 \times (3.142)^{2}}{(1.98)^{2}} = \frac{39.48}{3.92} = 10.1\ \text{m s}^{-2}. \]

Two precautions

1. The bob was displaced through a small angle (less than about \(10^\circ\)) and made to oscillate in a vertical plane, so that conical (circular) oscillation was avoided and the motion remained simple harmonic.
2. The timing was started and stopped as the bob passed the central rest position, and the experiment was shielded from draught, so that counting and timing errors were reduced.

(b)(i) Period for L = 75 cm

For \(L = 0.75\ \text{m}\), \(\sqrt{L} = 0.866\ \text{m}^{1/2}\). Reading up from the graph (equivalently \(T = s\sqrt{L}\)):

\[ T = 1.98 \times 0.866 = 1.72\ \text{s}. \]

(b)(ii) Positions of maximum velocity and maximum acceleration

The bob swings between the two extreme positions \(P\) and \(Q\) about the central (lowest) equilibrium position \(O\).

• Maximum velocity occurs at \(O\), the equilibrium position, where the displacement is zero and the bob moves fastest.
• Maximum acceleration occurs at the extreme positions \(P\) and \(Q\), where the displacement from \(O\) is greatest and the restoring force is largest.