Simple Pendulum Experiment (Class 11) Readings
Simple Pendulum Experiment (Class 11) with Calculations and Readings
Aim
i. To verify the relation between period (T) and length (l) of pendulum
ii. To find out the acceleration due to gravity
iii. To determine the length of seconds pendulum
iv. To find the period of the pendulum whose length is 105 cm.
Apparatus
Simple pendulum, Stopwatch, Meter scale, Wooden block etc..
Principle
For small amplitudes of oscillation of a simple pendulum,
l/T^{2} = a constant where, l —> Length of the pendulum
T—> Period of the pendulum
Also, T = 2π√(l/g) where, g —> Acceleration due to gravity at the place
Therefore, g = 4π^{2}(1/T^{2})
Procedure
i. Relation between I and T^{2}
The bob is placed between two wooden blocks and its diameter (d) is measured. Hence radius (r = d/2) can be calculated. The length of the string (I) is so adjusted that the distance between the point of suspension and the bottom of the bob is (50 + r) cm. Hence the length of the pendulum ‘l’ is 50 cm.
A mark is made on the edge of the table using a piece of chalk to indicate the equilibrium position of the pendulum, ie. , the position of the pendulum when it is at rest. The bob is pulled aside through a small distance and released. The pendulum executes oscillations, after a few oscillations, a stop watch is started just when the pendulum crosses the equilibrium position. When it next crosses the mark in the same direction, the oscillations are counted and when the pendulum passes the chalk mark in the same direction for the 20th time, the stop watch is stopped.
Thus the time taken for 20 oscillations is determined. This is repeated and the mean time taken for 20 oscillations is found. Hence the time for one oscillation ie., period 'T' is calculated. l/T^{2} is also calculated.
The experiment is repeated for lengths 60, 70, 80, 90, 100,110 cm. In each case l/T^{2} is calculated. It is found to be a constant. This is the relation between the period and length of a simple pendulum. Graph is drawn with 'l' along the Xaxis and 'T^{2}' along the Yaxis. A straight line graph is obtained.
ii. Acceleration due to gravity at the place
The mean value of l/T^{2} is determined. Hence 'g' can be calculated using the formula g = 4π^{2}(1/T^{2}). Note AB and BC from the graph. Hence g can be calculated using the equation, g = 4π^{2}.AB/BC
iii. Length of the seconds pendulum
For a seconds pendulum, T = 2 seconds. Then T^{2}= 4. From the graph, the value of 'l’ for which T^{2} = 4 is found (OD). This gives the length of the seconds pendulum.
iv. Period of the pendulum whose length is 105 cm
From the graph the square of the period corresponding to the length 105 cm is noted (OE). The square root of this ( √OE ) gives the period.
Observations and Readings
Radius of the bob, r = 0.95 cm
No: 
Distance between point of suspension and the bottom of the bob (l+r) 
Length of the pendulum (l) 
Time for 20 oscillations (t) 
Period of oscillation (T=t/20) 
T^{2} 
l/T^{2} 

1 
2 
mean 


cm 
cm 
s 
s 
s 
s 
s^{2} 
cm/s^{2} 
1 2 3 4 5 6 7 
50.95 60.95 70.95 80.95 90.95 100.95 110.95 
50 60 70 80 90 100 110 
28 30 31 34 35 38 40 
28 30 31 34 35 38 40 
28 30 31 34 35 38 40 
1.4 1.5 1.55 1.7 1.75 1.9 2 
1.96 2.25 2.40 2.89 3.06 3.61 4 
25.5 26.7 29.7 27.7 29.4 27.7 27.5 
Mean value of 1/T^{2} = 27.661 cm/s^{2} = 0.276 m/s^{2}
Therefore, Acceleration due to gravity, g = 4π^{2}(1/T^{2}) = 4 x (3.14)^{2} x 0.276 = 10.896 m/s^{2}
l – T^{2} graph
From graph:
AB = 43 cm = 0.43 m
BC = 1.5 s^{2}
G = 4π^{2}(AB/BC) = 4 x (3.14)^{2} x 0.43/1.5 = 11.30 m/s^{2}
Length of the seconds pendulum (OD) = 110 cm = 1.10 m
Period of the pendulum whose length is 105 cm = √OE = 1.92 s
Results
i. l/T^{2} is found to be a constant
ii. Acceleration due to gravity at the place,
(a) by calculation = 10.896 m/s^{2}
(b) from graph = 11.30 m/s^{2}
iii. Length of the seconds pendulum =1.10 m
iv. Period of the pendulum whose length is 105 cm = 1.92 s
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