Question

ASSIGNMENT QUESTIONS AND PRORLEMS FOR TOPIC 3 ASSIGNMENT QUESTIONS AND PROBLEMS FOR TOPIC 3 3.1A student was supplied with a stop watch, two metre rules and a simple pendulum suspended from a ceiling and was asked to measure the height of the ceiling indirectly, He set the pendulum swinging through a small angle and measured the period of oscillation for different lengths of the pendulum. Since he was unabie to measure the length of the pendulum directly, he measured the height of the centre of the pendulum bob above the floor. He obtained the results tabulated below Time for 50 oscillations Height of bob above floo (s) (mm) 155.3 400 148.8 600 142.2 800 1000 134.0 1200 127.4 1400 119.2 110.5 1600 The period T of the pendulum of length is given by 7 T 2T (1) gravity. But f= H-hwhere H is the where g is the acceleration due to height of the ceiling and h is the height of the centre of the pendulum above the floor. Therefore - 20! r(H-h) T 2 (2) Plot a suitable graph intercepts using linear regression to fit the best straight line through the data points. Assume that g = 9.8 ms (g could be obtained from the slope of the graph). Having obtained your values for H, answer the following questions: to find the height of the ceiling H from the two (a) Which value of H do you consider to be the least accurate? Give reasons for your choice and explain how the accuracy could have been improved. (b) Why was the bob set swinging through a small angle? (c) Why was h measured to the centre of the bob? (d) Why was the completed number of oscillations chosen to be large? (e) Can you see any advantage in measuring the height of the ceiling in this way?

Answer 1

h(m)	T=t/50	T^2
0.4	3.106	9.647236
0.6	2.976	8.856576
0.8	2.844	8.088336
1	2.68	7.1824
1.2	2.548	6.492304
1.4	2.384	5.683456
1.6	2.21	4.8841

from H h1/2 Н- T = 2T g

.......................(1)

Н -Һ т? — 4т? T 2 д

or

T (4Tg)H-(4T2/g)h

so if we plot graph between T² (as Y) and h(as X) then

slope

- (4т?/9)

and intercept

c (42/g)H

thus from c we can find value of H

cg H 471

y-3.9699x + 11.232 12 10 8 4 2 0 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0 0.2 h(m) Zvl

from equation of graph:

c=11.232

H = 4T2 11.232 9.8 H 4 3.142

H2.79m l

Also from equation 1

gT2 - H-h 4T2

or

gT2 h H 4T2

h(m)	T=t/50	T^2	gT^2/4pi^2	H
0.4	3.106	9.647236	2.3925145	2.792514528
0.6	2.976	8.856576	2.1964308	2.796430848
0.8	2.844	8.088336	2.0059073	2.805907328
1	2.68	7.1824	1.7812352	2.7812352
1.2	2.548	6.492304	1.6100914	2.810091392
1.4	2.384	5.683456	1.4094971	2.809497088
1.6	2.21	4.8841	1.2112568	2.8112568