Expt : 5 Simple pendulum
Aim :
To determine the acceleration due to gravity using a simple pendulum and to determine the length of seconds pendulum
Apparatus :
Retort stand, metallic bob, string, stop watch and a meter scale
Formula:
(a) Time period of oscillation of a simple pendulum
T = 2p √(L / g)
therefore acceleration is given by
g = 4p2 ( L / T2 )
(b) Seconds pendulum has a time period of 2 seconds
Diameter of the sphere
S.No. | MSR | VC | VC x LC | Actual reading (cm) |
1 | 1.9 | 4 | 0.04 | 1.94 |
2 | 1.9 | 3 | 0.03 | 1.93 |
3 | 1.9 | 5 | 0.05 | 1.95 |
4 | 1.9 | 4 | 0.04 | 1.94 |
5 | 1.9 | 3 | 0.03 | 1.93 |
Average : 1.94 cm
radius of the sphere = D/2
= 1.94 / 2
= 0.97 cm
L/T2 for simple pendulum
S.No. | length (L) | Time for 20 oscillations | T = t / 20 | T2 | L / T2 | ||
trial 1 | trial2 | mean ( t ) | |||||
1 | 50 | 28 | 29 | 28.5 | 1.41 | 2.01 | 24.8 |
2 | 60 | 30 | 31 | 30.5 | 1.54 | 2.40 | 24.9 |
3 | 70 | 33 | 33 | 33.0 | 1.66 | 2.77 | 25.2 |
4 | 80 | 35 | 36 | 35.5 | 1.78 | 3.20 | 25.0 |
5 | 90 | 38 | 38 | 38.0 | 1.90 | 3.62 | 24.8 |
6 | 110 | 41 | 42 | 41.5 | 2.08 | 4.36 | 25.1 |
Average : 25.05 cm s-2
Acceleration due to gravity is given by
g = 4 p2 ( L / T2 )
= 4 x 3.14 x 3.14 x 25.05
= 988.0 cm s-2
[Click here for the sample L - T2 graph]
[Click here for the sample L - T graph]
Acceleration due to gravity from graph
g = 4 x p2 x cot (q)
= 4 x 3.14 x 3.14 x 25
= 986.9 cm s-2
Result
(a) Acceleration due to gravity from table is : 988.0 cm s-2
(b) Length of seconds pendulum is : 100 cm
(c) Acceleration due to gravity from graph is : 986.9 cm s-2
(d) Percentage of error in measurement is : 0.816 %