TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with a pendulum bob, a metre rule, a stopwatch, a retort stand with clamp, and other necessary apparat...
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 the clamp as illustrated in the diagram.
ii. Adjust the pendulum such that AC=L= 90 cm
iii. Displace the pendulum bob slightly such that it oscillates in a vertical plane.
iv. Measure and record the time t for 20 complete oscillations.
v. Evaluate T and \(\sqrt L\)
vi.Repeat the procedure for four others values of L= 80 cm,70 cm, 60 cm, and 50cm.
vii. Tabulate your readings
viii. Plot a graph with T on the vertical axis and \(\sqrt L\) on the horizontal axis.
ix. Determine the slope, s, of the graph. x. Evaluate g= \(\frac{4\pi^{2}}{5^{2}}\)
xi. State two precautions taken to ensure accurate results.
(b) i. Determine from your graph, the period of the pendulum for L= 75 cm.
ii. A simple pendulum bob is set into simple harmonic motion. Sketch a diagram of the setup and indicate on it; the positions of: (a) maximum velocity. (b) maximum acceleration of the bob
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.
Straight line through the origin; slope s = 1.98 s m^(-1/2).
Slope of the graph
Taking two widely separated points on the line of best fit, \((0.707,\,1.42)\) and \((0.949,\,1.90)\):
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.
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 \(O\).
Simple pendulum in SHM: maximum velocity at O (equilibrium), maximum acceleration at extremes P and Q.
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.
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.
Straight line through the origin; slope s = 1.98 s m^(-1/2).
Slope of the graph
Taking two widely separated points on the line of best fit, \((0.707,\,1.42)\) and \((0.949,\,1.90)\):
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.
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 \(O\).
Simple pendulum in SHM: maximum velocity at O (equilibrium), maximum acceleration at extremes P and Q.
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.
