Compound Pendulum Lab Report

ABSTRACT In this experiment, we used a compound pendulum to calculate the acceleration due to gravity & radius of gyration. The acceleration due to gravity is often determined by indirect methods – for example, using a simple pendulum or a compound pendulum. But when we determine g using a simple pendulum, the result is not very accurate because an ideal simple pendulum cannot be realized under laboratory conditions. So we used the compound pendulum to determine the acceleration due to gravity in the lab, called the Bar pendulum. Theory A bar pendulum is the simplest form of compound pendulum. It is in the form of a rectangular bar (with its length much larger than the breadth and the thickness) with holes (for fixing the knife edges) drilled along its length at equal separation. If a bar pendulum of mass M oscillates with a very small amplitude θ about a horizontal axis passing through it, the nits angular acceleration (d2 θ/dt2) is proportional to the angular displacement θ. The motion is simple harmonic and the time period T is given T =2 π

√

I Mgl

where I denotes the moment of inertia of the pendulum about the horizontal axis through its center of suspension and lis the distance between the center of suspension and C.G. of the pendulum.

According to the theorem of parallel axes, if I G is the moment of inertia of the pendulum about an axis through C.G., then the moment of inertia I about a parallel axis at a distance l from C.G. is given by I=IG+MI2 =Mk2+MI2 where k is the radius of gyration of the pendulum about the axis through C.G. Using Equation (2) in Equation (1), we get T =2 π

√

M k 2 + M l2 Mgl

T =2 π

√

k 2+l2 gl

T =2 π

√

k2 +l l g

T =2 π

√

L g

where L is the length of the equivalent simple pendulum, given by L=

k2 +l l

( )

Therefore, g=4 π

2

L T2

The point at a distance L from the centre of suspension along a line passing through the centre of suspension and C.G. is known as the centre of oscillation. Time period T will have minimum value when l = k Hence PQ = 2k Simplifying Equation l 2−¿+ k 2=0

This equation is a quadratic equation in l having two roots. If l 1 and l2 are the two values of l, then by the theory of quadratic equations l1+l2=L and l1l2=k2 So we can write the solutions as l=l1

k2 l=l2= l 1

Since both the sum and the product of the two roots are positive, for any particular value of l, there is a second point on the same side of C.G. and at a distance k2/l from it, about which the pendulum will have the same time period. If a graph is plotted with the time period as ordinate and the distance of the point of suspension from C.G. as abscissa, it is expected to have the shape shown in Figure, with two curves which are symmetrical about the C.G. of the bar. To find the length L of a simple pendulum with the same period, a horizontal line ABCDE can be drawn which cuts the graph at points A, B, D and E, all of which read the same time period. For A as the center of suspension, D is the center of oscillation (D is at distance of l1+l2=L from the centre of suspension A). Similarly, for B as the center of suspension, E is the center of oscillation.

Procedure 1. We balanced the bar on a sharp wedge and marked the position of its C.G. 2. We fixed the knife edges in the outermost holes at either end of the bar pendulum. The knife edges were horizontal and lied symmetrically with respect to centre of gravity of the bar. 3. We checked with spirit level that the glass plates were fixed on the suspension wall bracket were horizontal. The support was rigid. 4. We suspended the pendulum vertically by resting the knife edge at end A of the bar on the glass plate. 5. We adjusted the lower end of the bar and put a reference mark on the wall behind the bar to denote its equilibrium position. 6. We displaced the bar slightly to one side of the equilibrium position and let it oscillate with the amplitude not exceeding 5 degrees. We made sure that there was no air current in the vicinity of the pendulum. 7. We used the stop watch to measure the time for 30 oscillations. The time

was measured after the pendulum had had a few oscillations and the oscillations were become regular. 8. We measured the distance l from C.G. to the knife edge. 9. We recorded the results in Table. Repeating the measurement of the time for 30 oscillations, we took the mean. 10. We suspended the pendulum on the knife edge of side B and repeated the measurements in steps 6 to 9 as mentioned above. 11. We fixed the knife edges successively in various holes on each side of C.G. and in each case, measuring the time for 30 oscillations and the

Se. No

For One Side

For Second side

l (cm)

Time for 10 oscillation s (t)

T=t/10 (sec)

l (cm)

Time for 10 oscillation s (t)

T=t/10 (sec)

1

48.2

18.3

1.83

48.5

18.6

1.86

2

43.2

15.9

1.59

43.5

15.62

1.562

3

38.2

15.38

1.538

38.3

15.01

1.501

4

33.1

14.6

1.46

33.25

14.31

1.431

5

28

14.11

1.411

28.1

14.03

1.403

6

23

14.88

1.488

23.1

15.06

1.506

7

17.9

16.09

1.609

18.1

15.3

1.53

8

13

18.25

1.825

13.1

16.16

1.616

9

7.3

22

2.2

7.8

20.9

2.09

10

2.25

32.11

3.211

2.9

30.8

3.08

distance of the knife edges from C.G. Observations Table 1: Measurement of T and l Least count of stop-watch =0.01 sec.

Graph between time T and Distance from C.G:

3.5

3

2.5

2

1.5

1

0.5

0 -60

-40

-20

0

20

40

60

Acceleration due to gravity (g) We drew horizontal lines on the graph corresponding to two periods, T 1and T2, as shown in fig. For the line ABCDE L1=(AD+BE)/2=61.2 cm Hence, using the formula for gas given in Equation (5), g= 9.68 m/sec2 For the line A'B'C'D'E' L2=(A’D’+B’E’)/2= 62.3 cm Hence, g= 9.7 m/sec2 Mean value of g = 9.69 m/sec2 Radius of gyration (k): Let l1 = ½ (AC + CE) = ½ AE, and l2 = ½ (BC + CD) = ½ BD. Calculating the radius of gyration using the expression k =√l 1 l 2

k=0.278 m. Made the following table for calculated values of I 2 and lT2 corresponding to all the measurements recorded in Table. Table 2: Calculated values of l2 and lT2

Graph between l2 and lT2

1.4 1.2 1

0.92

0.8 0.6

Linear ()

0.59

Linear ()

0.4 0.2 0 0

0.05

0.1

0.15

0.2

0.25

l2(cm2) 2337.745 1879.245 1463.065 1100.58625 786.805 531.305 324.01 170.3 57.0625

S.No. 1 2 3 4 5 6 7 8 9

lt2(cm/s2) 128.32 109.65 92.32 73.345 58.53 51.66 44.35 38.75 34.7

Slope of the graph = 0.249 m/sec Intercept = 0.082 m Acceleration due to gravity g= 4π2x slope = 9.82 m/sec2 Radius of gyration, k= 0.286 cm

Estimation of error Maximum log error Using Equation g=4 π 2

L 2 T

---------------(5)

Taking logarithm on both sides and differentiating, we have Δg ΔL 2 ΔT = + g L T

Δg=g

( ΔLL + 2 TΔT )

where ΔL and ΔT are the least counts of distance and period axes of the graph between time period and distance from C.G.

Results The acceleration due to gravity, g= 9.82 m/s2 Actual value = 9.8 m/s2 Percentage error =2.04 % Maximum log error =6.7 cm/s2 The radius of gyration about the axis of rotation = 0.286 cm.

Q.1 Define a compound pendulum. Ans. It is a rigid body capable of vibrating abut a horizontal axis passing through it. Q.2 Why is a compound pendulum preferred for the determination of 'g'? Ans. Because it moves as one rigid body and we take into account the moment of inertia of the whole body. Q.3 Define a simple pendulum. Ans. It consists of a point mass suspended by a weightless, inextensible and flexible string from a fixed point about which the pendulum oscillates without friction. In practice, it is not possible to find a pendulum which .may meet such ideal conditions . Q.4 What is a second's pendulum?

Ans. A simple pendulum having a time period of two seconds. Q.5 Define an equivalent simple pendulum. Ans. A simple pendulum having the same time period as that of a compound pendulum is called equivalent simple pendulum. Q.6 Why do you place the knife edges symmetrically? Ans. The knife edges are kept symmetrical so that the centre of gravity does not shift.

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