Coupled Torsion Pendulum

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1 PHYSICS THROUGH TECHING LB IX Coupled Torsion Pendulum S.R. PTHRE, *.M. SHKER, **.K. MISHR *, C.S. DIGHE *** * Homi Bhabha Centre for Science Education (TIFR) V.N. Purav Marg, Mankhurd. Mumbai e.mail: shirish@hbcse.tifr.res.in ** Department of Physics, K.J. Somaiya College, Vidyavihar, Mumbai *** Department of Physics University of Mumbai, Mumbai BSTRCT This is an edited version of the experiment set for experimental examination conducted at the Physics Olympiad Orientation cum Selection Camp held at Homi Bhabha Centre for Science Education (TIFR), Mumbai in May 007. In this experiment, a coupled torsion pendulum system is studied by observing its normal modes of vibration. With appropriate adjustments, energy exchange between the coupled pendulums demonstrating the phenomenon of beats can also be observed. Introduction If one end of an elastic wire is held fixed and a torque is applied at its other end to twist it about its axis, a restoring torque due to shear is generated internally in the wire. body attached to the free end of the wire, on removing the twisting torque, executes torsional oscillations. Such a system is called a torsion pendulum. If both the ends of the wire are fixed and a massive body is attached to the wire at some point between the ends, the system can be made to execute torsional oscillations by first rotating the body about the Physics Education September October

axis along the wire and then releasing it. Here the twists in the upper and lower segments of wire are in opposite directions as seen from the body but their torques act in the same direction.

The middle segment of the wire between the two bodies acts as a coupling pparatus 1) Two torsion pendulums ) n aluminum frame with two G-clamps 3) steel wire 4) Three identical llen keys 5) stopwatch

2 axis along the wire and then releasing it. Here the twists in the upper and lower segments of wire are in opposite directions as seen from the body but their torques act in the same direction. If there are two massive bodies at two points of the wire between the fixed ends, the system becomes a coupled system of two torsion pendulums. The middle segment of the wire between the two bodies acts as a coupling pparatus 1) Two torsion pendulums ) n aluminum frame with two G-clamps 3) steel wire 4) Three identical llen keys 5) stopwatch agent. Each of the two pendulums executes simple harmonic motion when the torsion displacements are small but the whole system has a complex motion. The oscillatory motion of the system appears relatively simple when it oscillates in one of the normal modes. By proper choice of initial conditions the system can be made to oscillate in the normal modes. 6) 3 m measuring tape 7) pair of vernier calipers 8) 5 kg mass 9) S-hook ) retort stand with clamps Description of pparatus 1) The Pendulums The oscillating body of each torsion pendulum consists of a dumbbell shaped body, made by passing two identical heavy rings on a split rod of rectangular cross-section at its ends symmetrically. The split rod is held together using two llen screws. t the centre of the rod is a groove through which the wire of the pendulum can be passed. By loosening these screws the body can be made to slide along the wire (Refer Figure1). Figure1 14 Physics Education September October 007

The pendulums can be clamped to the steel wire by tightening the two llen screws. fter clamping the rod on the steel wire, two ring masses are slid at its ends.

c) Remove the rings from the ends of the rods and loosen the allen screws of the strips. d) Insert the steel wire through the rod. (Do not tighten the screws.

3 The pendulums can be clamped to the steel wire by tightening the two llen screws. fter clamping the rod on the steel wire, two ring masses are slid at its ends. Care should be taken to mount these masses symmetrically on the rod. ) The Frame b) Insert the steel wire through brass strip B on the frame. c) Remove the rings from the ends of the rods and loosen the allen screws of the strips. d) Insert the steel wire through the rod. (Do not tighten the screws. The number of rods to be inserted depends on the experimental part that you are performing) e) Insert the steel wire through brass strip on the frame. f) Tighten the screws on brass strip. g) Now attach 5 kg mass at the loop end of the steel wire using a S-hook to straighten the wire between the ends. Figure. The aluminum frame is provided to support the steel wire-pendulum system. The frame has legs on both the sides. The frame is clamped to the table using two G-clamps. The frame has two long brass strips on which the clamping arrangement for steel wire is made. Figure ssembly For clamping the steel wire pendulum assembly you may follow these steps: a) The frame is clamped to the table using two G-clamps. h) Tighten the screws on brass strip B. i) Remove 5 kg mass and S-hook. j) Now tighten the allen screws of the pendulum by keeping the body at the desired position. Put on the rings on the sides of the rod. Physics Education September October

The moment of inertia of the dumbbell shaped assembly is given as I total I rod + I ring (1) M rod I rod 1 (L +b ) () The moment of inertia of a ring about an axis parallel to its plane and passing

4 4 Theory pair of Vernier Calliper Measuring Tape llen Key Stopwatch Figure 4. Part : Moment of inertia of a dumbbellshaped body about an axis perpendicular to its length and passing through its centre Figure 5. Figure 6. The moment of inertia of the dumbbell shaped assembly is given as I total I rod + I ring (1) M rod I rod 1 (L +b ) () The moment of inertia of a ring about an axis parallel to its plane and passing through its centre is given by I ring mh + m ( R + R ) where m is the mass of the ring. For a ring rotating about an axis passing through the center of the dumbbell, using parallel axis theorem, I ring mh + m ( R + R ) +mr (3) 1 4 Total moment of inertia of the pendulum is given by I total I rod + I ring M rod 1 1 (L +b ) + 16 Physics Education September October 007

5 mh m + ( R 1 + R ) + mr 1 4 (4) Part B: Determination of Torsion Constant α When a wire, with one end fixed, is twisted by applying a torque at the other end, the angle of twist is found to be directly proportional to the magnitude of the torque as well as to the length of the wire. The restoring torque is equal to the applied torque when the twist is constant. So, we conclude that the restoring torque is directly proportional to the angle of twist θ and inversely proportional to the length l of the wire and the expression giving the restoring torque can be written as α(θ/l), α being the constant of proportionality. Its value depends upon modulus of rigidity of the material of the wire and the cross-section of the wire. When a body clamped to a stretched wire is rotated by a small angle θ about an axis along the wire, it twists the two segments of the wire in opposite directions. If the segments of the wire between the body and the fixed ends are of lengths l 1 and l, then the net torque acting 1 1 on the body becomes equal to α + θ. l1 l Let I be the moment of inertia of the system about the axis of twist. Then the equation of motion of the system of torsional pendulum can be written as I d θ + kθ 0. (5) dt where, k α 1 1 +, is the torque per unit l1 l twist. Figure 7. The period of oscillation of the torsional pendulum is, then, given by I T π (6) k Part C: Normal Modes of Vibration and Coupling Constant When two torsion pendulums and B of moments of inertia I and I B and of lengths l 1 and l respectively, are coupled by a similar wire segment of length l 3, the torque due to twist in the coupling segment will be equal to the difference in the twists at its two ends. If θ 1 and θ be the twists in l 1 and l (fixed at the ends) then the twist in l 3 is (θ 1 θ ) and the restoring torque on pendulum from l 3 would be (θ 1 θ )(α/l 3 ).[We follow the convention that θ 1 is positive when the twist is anticlockwise and θ is positive if it is clockwise when seen facing the fixed ends.] Similarly the torque on pendulum B would be (θ θ 1 )(α/l 3 ). The net torque on pendulum (which we can refer as pendulum 1) would be Physics Education September October

6 I d θ dt θ θ θ 1 1 ( 1 ) α α l1 and so its equation of motion would be I d θ α + dt l l θ θ α 0 (7) Similarly, considering the net torque on pendulum we have I B d θ θ α θ α dt l (8) The resultant motion of the system depends upon the values of θ 1 and θ at each instant. Two simple cases are of importance. When one of the above conditions is satisfied the system is said to oscillate in one of the normal modes. By choosing the initial conditions the system can be made to oscillate in the desired normal mode. The general equation with variable θ satisfying both simultaneous equations (7) and (8) can be obtained in the form I I B ω 4 α 1 1 α I + + I B ω l l1 + α α + l l l + l l (9) dθ where ω is used for. dt Eq. (9) can be solved by substituting the values of I, I B, α, l 1, l and l 3. The roots of the equation are angular frequencies ω in and ω out of normal modes. In Eqs. (7) and (8) if the third terms were not present, the equations would be independent of one another and we would have independent harmonic oscillations at frequencies α 1 1 ω + () I l1 Figure 8. Case 1) Both θ 1 and θ are in phase during oscillation. The net restoring torque is reduced by the coupling and as a consequence the frequency of oscillation also is reduced. Case ) During oscillation θ 1 and θ are out of phase and as a consequence the restoring torque and the frequency is increased. α 1 1 ω 0 + (11) I B l These are the frequencies with which each mass would vibrate if the other were held fixed. The third term in Eqs. (7) and (8) represents the coupling between the motions of the two masses. 18 Physics Education September October 007

7 Coupling Constant The frequencies, ω out, ω and ω in, ω 0 can be expressed as where, ω 1 out ω + Δω (1) 0 ω 1 in ω Δω (13) Δω ( ω ω 0 with the abbreviation κ α l I I 3 1/ 4 4κ ) 1+ 1 ( ω ω 0 ) B where κ is called the coupling constant. If ω ω 0, Eq. (14) reduces to (14) (15) Δω κ (16) Part D: Beats If the coupling between the two pendulums is small, ω in and ω out are nearly equal and then motion of each pendulum is a superposition of its two normal modes motions which leads to beats, the beat frequency being the difference between the two normal mode frequencies. EXPERIMENT Part : Moment of Inertia Make necessary geometrical measurements of the pendulums. Calculate the moment of inertia of both the pendulums. M Mass of rod (including llen screws), M B Mass of rod B (including llen screws), m Mass of ring (each ring), m B Mass of ring B (each ring). lso calculate errors and B. Part B: Determination of the Torsion Constant, α Set up the pendulum assembly. Clamp rod at l 1.0 cm. (Do not clamp rod B). Measure and note down the period of oscillation of pendulum when the oscillating body is at l 1 from the top. Change l 1 in steps and study the period of oscillation of the pendulum. Plot period of oscillation against l 1. lso plot a suitable graph to determine the torsion constant α. Calculate error Δα. Part C: Normal Modes of Vibration and Coupling Constant Case 1: ω ω 0 Clamp rod at l cm and rod B at l 30.0 cm. Oscillate this coupled system in two normal modes i.e. in phase and out of phase. Calculate ω out and ω in from your observations. Solve Eq.(9) by substituting the values of I, I B, α, l 1, l and l 3 to get ω out and ω in. Using the retort stand clamp, fix pendulum B. Oscillate pendulum. Calculate ω from your observations. lso calculate ω by substituting α, I, l 1 and l 3 in Eq.(). Release rod B and clamp rod similarly. Oscillate pendulum B. Calculate ω 0 from your observations. lso calculate ω 0 by substituting α, I B, l and l 3 in Eq.(11). From these values calculate Δω and hence find the coupling constant κ using Eqs.(1), (13) and (14). lso, calculate κ from Eq.(15) by substituting α, l 3, I, I B. Physics Education September October

8 Figure 9 Figure Figure 11 Figure 1 Case : ω ω 0 Make l 1 l.0 cm. Measure ω out, ω in, ω and ω 0. Hence calculate κ. Part D: Beats t l 1 l.0 cm, twist the pendulum by holding pendulum B steady. Release pendulum. llow pendulum to oscillate. While pendulum is oscillating, release pendulum B. Observe the beat phenomenon. Measure the FOR PENDULUM For rod: Mass of the rod 136.1g Length of the rod 17.0 cm Breadth of the rod, b 0.97cm For ring: Mass of the ring79.4g Inner radius, R cm Outer radius, R 1.88cm Thickness of the ring, h 0.97cm beat frequency. lso calculate the beat frequency from ω out and ω in obtained in part C. Typical Measurements and Calculations Part : Moment of Inertia of Pendulums DIMENSIONS OF THE PENDULUMS: FOR PENDULUM B For rod: Mass of the rod 136.3g Length of the rod 17.0 cm Breadth of the rod, b 0.97cm For ring: Mass of the ring79.4g Inner radius, R cm Outer radius, R 1.87cm Thickness of the ring, h 0.97cm 0 Physics Education September October 007

9 M rod I I rod + I ring ( L + b ) + 1 mh m + ( R 1 + R ) + mr ( ) ( ) g-cm rod I B I rod + I ring M ( L + b ) + 1 mh m + ( R 1 + R ) + mr ( ) ( ) g cm Error in I I I rod rod ring ring ring I + rod I rod I ring ΔM ΔL Δb M + L + b Δm Δh ΔR m h R ΔR Δr R + r Substituting the values from the above table: I B I B B g cm I IB (. 137 ± 0. 06) 4 g cm 1 Graph of T against l T in s l 1 in cm Physics Education September October 007 1

10 Part B: Determination of the Torsion constant, α Sr. No. l 1 in cm l in cm t for oscillations t 1 s t s t 3 s Mean t in s t T l 1 l T s in s cm Physics Education September October 007

11 Rearranging Eq.(5) and (6), 1 1 κ α + and T π I l1 l κ T T 4π I 4π I ll α l + l α + l l 1 4π I I ll 1 ll 80 4 π. α 01. α 1 Graph of T against l 1 l : 1 π I Slope.0030 ± s /cm 01. α 4 α π Error in α.9 6 dyne cm Since the error in the slope is very less, Δα α I Δα dynes cm α (.9±0.13) 6 dynes cm Part C Here, l cm, l 30.0 cm and l cm Solving Eq.(9) for ω in and ω out : I I B ω 4 α + I α ω + + I B l l α + α l1 l I I B g cm 4 Δα α Δslope slope + I T in s l 1 l in cm Physics Education September October 007 3

12 I I B l l l l α + + α g cm 4 s α α + l1 l + 3 l l 3 (.9 6 ) (.9 ) (30.4) [ ] 3.76 g cm 4 s 4 Therefore, Eq.(9) becomes, ( )ω 4 ( )ω +(3.76 ) ω ω b± b 4ac ω a 59. 6± ( 59. 6) ω rad/s and ± ω 8.7 rad/s 3.76 ω rad/s, ω 8. 7 rad/s ω + ω out 4.80 rad/s, ω ω in.95 rad/s T out 1.31s, T in.13s Calculating ω and ω 0 : α 1 ω + I l1 l 4 Physics Education September October rad/s α 1 1 ω 0 + I B l Coupling Constant 1 ω ω ω Δ out + 1 ωin ω0 + Δω out in Δ ω ( ω ω ) ( ω ω rad/s Δω [(4.80) (.95) ] [( 40. ) ( 376. ) ].84 (rad/s) From Eq.(14) ( ω ω ) 0 ( Δω ) κ ( 0 ) ω ω )

13 κ 4 4 ( ) κ.64 rad/s ( ) 1 From Eq.(15) α.9 κ l I I 30.4 (1.37 ) 3 1 κ.65 rad/s 6 Case 1 ω ω 0 Observed Calculated t 1 (s) t (s) t 3 (s) t (s) T t (s) π ω rad/s ω (rad/s) T ω in ω out ω ω Case ω ω 0 t 1 (s) t (s) t 3 (s) t(s) T t s π ω (rad/s) T ω in ω out ω ω Physics Education September October 007 5

14 Part D: Beats l 1 cm l cm t 1 s Observed Time between 3 minima t s t 3 s t s ΔT t 3 s ngular Beat Frequency rad/s Calculated ngular Beat Frequency rad/s Conclusion The experiment gives a simple arrangement of coupled pendulum system which is suitable for the undergraduate laboratories. The experiment covers different aspects of the coupled systems such as normal modes of vibration, coupling constant between two oscillating systems and beats. This experiment can also be extended to study some aspects like effect of change in coupling on the behavior of the system; more specifically, the effect of weak coupling and strong coupling between the oscillating bodies. cknowledgement We are thankful to all the Olympiad students of 007 batch who gave us reason to develop this experiment. We express our thanks to Prof. D.. Desai, Prof. H.C. Pradhan, Prof. R.M. Dharkar and Prof. Vijay Singh for their continuous guidance in the development process. We also thank our colleagues from HBCSE for helping us at various stages. References 1) Symon K.R., Mechanics, 3 rd edition, ddison- Wesley, Readings, Mass., p.191 (1971). ) Yee-Tak Yu, The Double Torsion Pendulum in a Liquid, merican Journal of Physics, 15 (194). 6 Physics Education September October 007

PHYSICS LAB Experiment 9 Fall 2004 THE TORSION PENDULUM

PHYSICS LAB Experiment 9 Fall 2004 THE TORSION PENDULUM PHYSICS 83 - LAB Experiment 9 Fall 004 THE TORSION PENDULUM In this experiment we will study the torsion constants of three different rods, a brass rod, a thin steel rod and a thick steel rod. We will