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EXP-1 Simple Pendulum
mechanical engineering (mec11)
Adithya Institute of Technology
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EXP-1 SIMPLE PENDULUM
AIM:
To study the oscillation of simple pendulum
PROCEDURE: Fix the balls with nylon ropes in to the hooks provided at the top of beam of the frame, and measure the length of pendulum as shown. Oscillate the pendulum and measure the time required for 30 oscillations. Repeat the procedure by changing the ball and changing the length
OBSERVATION: Sr. No. Ball Size Length L m ,Time for 30 Oscillations
CALCULATIONS: For Simple pendulum Let, t = time period sec/cycle Therefore 0 sec
Experimentally, texp= (time for 30 oscillations) / 30 Compare the values obtained practically and theoretically. Also plot a graph of t v/s L
CONCLUSION: Time period of the simple pendulum is proportional to the square root of the length L of simple pendulum
Sample Calculations Experiment – Simple Pendulum
Observation Sr. No. Ball Size Length Time for 30 oscillations 1 Big 359 mm 35 sec Experimental Time Period: - 35/30 = 1. Theoretically t = 2 x 3 x (0.359/9)0 = 1.
Hence Verified
EXP-2 COMPOUND PENDULUM
AIM:
To determine the radius of gyration of a compound pendulum
INTRODUCTION: A rigid body is allowed to oscillate in vertical plane about the axis of suspension under the action of gravitational force. This body is called a compound pendulum. The unit is provided with a compound pendulum with a simple design as shown in figure.
PROCEDURE: Fix the brass bush in any of the holes of pendulum and mount the pendulum over the suspension shaft which is fitted at the top disc. Oscillate the pendulum and measure the time required for 5 oscillations. Repeat the procedure by putting the bush in different holes.
OBSERVATION: h = distance of c. of axis of suspension Sr. h m Time for 5 oscillation
CALCULATIONS: Experimentally, t= (time for 5 oscillations) / 5 Let, m = Mass of compound Pendulum = 1 kg k = radius of gyration about an axis through c. perpendicular to plane of oscillations Now, Tme 2 + h / g ) 0. s Therefore, k = 2 g – h 2 And equivalent length of simple pendulum, L= ( k
- h ) h the equivalent length can be verified by setting the simple pendulum to L
CONCLUSION: The radius of gyration of a compound pendulum = ________
SAMPLE CALCULATIONS Experiment – CompoundPendulum
Observation: Sr. No. h (m) Time for 5 Oscillations 1 0 5. Time Period = 5/5 = 1 sec
For beam, Radius of gyration, kb = Lb 2
- hb
2
3 12
Where, Lb = 0
hb = 0
when weight are added away from center for each weight radius of gyration
kw = a
2
- (r /2) where, Distance of weight from beam centre, a = 0 or 0 m Radius of weight, r = 0 for the weight at the centre kw = r / 2 where , r = Radius of the weight = 0 m Now, Total moment of inertia of system IT = I beam + Iweights = (mbeam x kb 2 ) + (mw1 x k w1 x n1) + (mw2 x k w2 x n2) + --- Where, n1 = no. of weights at kw n2 = no. of weights at kw Now, IT = m (keff.) Where, keff = Effective radius of gyration m = Total mass of this systems = Mass of beam + Mass of weights Mass of beam = 0 kg
CONCLUSION: keff is approximately equal to calculated ‘k’
Sample Calculations Experiment – Bifilar Suspension Sr. No. Length of String Distance of weights from centre Time for 10 ‘a’ oscillations 200 gm 400 gm 1 0 0 ( x 2 ) 0 ( x 2 ) 7 sec Texp = 7/10 = 0 sec
fn = 1/0 = 1 s- Therefore K = 0 x (9.81/0)0 = 0. 2 x 3 x 1. Steps to determine keff Radius of gyration Beam = kb = {(0. /3) + (0. /12)}0 = 0. Radius of gyration 200 gm weight Kw1 = {0.
- (0. /2)}0 = 0. Radius of gyration 400 gm weight Kw2 = {0.
- (0. /2)}0 = 0. Since weight of 100 gm are attached at center, Kw3 = (0.02/20) = 0. Total moment of Inertia of System IT = Ibeam + Iweights = (0 x 0. ) + (0 X 0. x 2) + (0 X 0. x 2) + (0 X 0. x 2) = 0. Therefore keff = (0.01567/1)0 = 0.
Conclusion: Keff is Approximately equal to K
EXP-1 Simple Pendulum
Course: mechanical engineering (mec11)
University: Adithya Institute of Technology
