TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with a stopwatch, a meter rule, a split cork, retort stand and clamp, a pendulum bob, a piece of threa...

TEST OF PRACTICAL KNOWLEDGE QUESTION

You are provided with a stopwatch, a meter rule, a split cork, retort stand and clamp, a pendulum bob, a piece of thread, and other necessary apparatus.

i. Place the retort stand on a laboratory stool. Clamp the split cork.

ii. Suspend the pendulum bob from the split cork such that the point of support P of the bob is at height \(H = 100\text{cm}\) above the floor Q. The bob should not touch the floor and H should be kept constant throughout the experiment.

iii. Adjust the length of the thread such that the center A of the bob is at a height \(y = \text{AQ} = 20\text{cm}\) from the floor.

iv. Displace the bob such that it oscillates in a horizontal plane.

v. Take the time t for 20 complete oscillations.

vi. Determine the period T of oscillation and evaluate T

vii. Repeat the procedure for four other values of \(y = 30\text{cm}, 40\text{cm}, 50\text{cm},\) and \(60\text{cm}\). In each case, determine T and T.

viii. Tabulate the results.

ix. Plot a graph of T on the vertical axis and y on the horizontal axis, starting both axes from the origin (0,0).

x. Determine the slope, s, of the graph and the intercept c on the vertical axis.

xi. If in this experiment SR= c, calculate R.

x. State two precautions taken to ensure accurate results.

(b) i. The bob of a simple pendulum is displaced a small distance from the equilibrium position and then released to perform simple harmonic motion Identify where its:
\((\propto)\) kinetic energy is maximum
\((\beta)\) acceleration is maximum

ii. An object of weight 120N vibrates with a period of 4.0s when hung from a spring. Calculate the force per unit length of the spring. [\(g = 10\text{ms}^{-2}\), \(\pi = 3.142\)]

(a) Pendulum experiment

The length of the pendulum is \(l=H-y\), where \(H=100\,\text{cm}\). For each setting, the time for 20 oscillations is measured and

\[T=\frac{t}{20},\qquad T^2=\left(\frac{t}{20}\right)^2.\]

S/N	\(y\) (cm)	Time, \(t\), for 20 oscillations (s)	Period, \(T=t/20\) (s)	\(T^2\) (s2)
1	20	50.50	2.525	6.376
2	30	47.50	2.375	5.641
3	40	43.50	2.175	4.731
4	50	40.00	2.000	4.000
5	60	35.80	1.790	3.204

Graph of \(T^2\) against \(y\):

The graph has a negative gradient. From the line of best fit, using two well-separated points on the line, \((20\,\text{cm},6.39\,\text{s}^2)\) and \((60\,\text{cm},3.20\,\text{s}^2)\):

\[\text{gradient}=\frac{3.20-6.39}{60-20}=-0.0798\,\text{s}^2\text{cm}^{-1}.\]

Hence, the magnitude of the slope is \(S=0.080\,\text{s}^2\text{cm}^{-1}\). The intercept on the \(T^2\)-axis is \(c\approx 8.00\,\text{s}^2\).

Since \(SR=c\),

\[R=\frac{c}{S}=\frac{8.00}{0.080}=100\,\text{cm}.\]

Precautions:

1. The bob was given a small angular displacement and released gently.
2. The support was kept rigid and the bob was allowed to oscillate freely without draughts or touching the floor.

(b)

(i) The kinetic energy is maximum at the equilibrium position. The acceleration is maximum at either extreme position, where the displacement is maximum.

(ii) \(W=120\,\text{N}\), so

\[m=\frac{W}{g}=\frac{120}{10}=12\,\text{kg}.\]

For a mass-spring system,

\[T=2\pi\sqrt{\frac{m}{k}},\qquad k=\frac{4\pi^2m}{T^2}.\]

\[k=\frac{4(3.142)^2(12)}{(4.0)^2}=29.6\,\text{N m}^{-1}.\]

The force per unit length (spring constant) is \(\boxed{29.6\,\text{N m}^{-1}}\).

Read lesson note on Simple Harmonic Motion (WAEC)