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 = 100cm 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= AQ= 20cm 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 = 30cm, 40cm, 50cm, and 60cm. 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= 10ms\(^{-2}\), \(\pi\)=3.142]
Scale: Let 2cm represent 1 unit on the vertical axis and 1.5cm represent 10 units on the horizontal axis. Precautions - Avoided parallax error in reading stopwatch/clock/ meter rule. - Notcd/corrected/avoided zero error on stopwatch/ clock/meter rule. - Avoided draught/switch off fans. - Avoided conical oscillation Ensured that support was rigid/firm. - Ensured bob of the pendulum was free from table/did not touch table - Repeated reading shown on the table. - Small angular displacement (b) (\(\propto\)) The kinetic energy is maximum at the equilibrium position (\(\beta\)) The acceleration is maximum at the point of maximum displacement.
ii. The period T is given by T = 2\(\pi\)\(\sqrt\frac{M}{K}\) T\(^{2}\) = 4\(\pi^2\)\(\frac{M}{K}\) K = 4\(\pi^2\)\(\frac{M}{T^{2}}\) = \(\frac{4x(3.14)^2(\frac{120}{10})}{4}^2\) 29.62NM\(^{-1}\) OR T = 2\(\pi\)\(\sqrt\frac{e}{g}\) T\(^{2}\) = 4\(\pi^2\)\(\frac{e}{g}\) e = \(\frac{T^{2}}g{4\pi^2}\) = 4.05 k = \(\frac{f}{e}\) = \(\frac{120}{4.05}\) = 29.62NM\(^{-1}\)
