TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with a retort stand, clamp and boss, a pendulum bob, a piece of thread, and other necessary apparatus....

Answer Details

Practical: simple pendulum (\(T^{2}\) against \(L\))

The experiment establishes the relation \(T = 2\pi\sqrt{\dfrac{L}{g}}\), so that \(T^{2} = \dfrac{4\pi^{2}}{g}\,L\). A straight-line graph of \(T^{2}\) (vertical) against \(L\) (horizontal) is expected.

Method / expected readings. For each length the time \(t\) for 20 oscillations is measured and the period found from \(T = \dfrac{t}{20}\). Then \(T^{2}\) is evaluated for the five lengths (about 130, 110, 90, 70 and 50 cm). Tabulate \(L\), \(t\), \(T\), \(T^{2}\).

Graph and analysis. The points lie on a rising straight line through (or near) the origin.

• Slope \(s = \dfrac{\Delta T^{2}}{\Delta L} = \dfrac{4\pi^{2}}{g}\).
• Intercept \(C\) on the \(T^{2}\) axis (ideally near zero for an ideal pendulum).
• \(k_{1} = \dfrac{4\pi^{2}}{s}\) gives a value numerically close to the acceleration due to gravity, since \(s = \dfrac{4\pi^{2}}{g}\Rightarrow \dfrac{4\pi^{2}}{s}=g\approx 9.8\ \text{to}\ 10\ \text{m s}^{-2}\).
• \(k_{2} = \dfrac{C}{8}\) is evaluated from the measured intercept.

Two precautions:

• The pendulum was displaced through a small angle (less than about \(10^{\circ}\)) so that the motion is truly simple harmonic.
• Oscillations were counted from the central (rest) position and timed with the bob swinging in one vertical plane (no conical motion) to avoid parallax and miscount.

(b)(i) The period of oscillation is the time taken by the oscillating body to make one complete oscillation (one full to-and-fro swing back to its starting point moving in the same direction).

(b)(ii) The acceleration of free fall due to gravity, \(g\), is the constant acceleration with which a body falls freely under the action of gravity alone, air resistance being neglected. It is directed vertically downwards and has an average value of about \(9.8\ \text{m s}^{-2}\) near the Earth's surface.