physics through teaching lab ixphysics.unipune.ac.in/~phyed/24.3/24.3_pathare.pdf8) a 5 kg mass 9) a...

Physics Education • September − October 2007 213

PHYSICS THROUGH TEACHING LAB IX

Coupled Torsion Pendulum

S.R. PATHARE,* A.M. SHAKER,** A.K. MISHRA*, C.S. DIGHE*** *Homi Bhabha Centre for Science Education (TIFR)

V.N. Purav Marg, Mankhurd. Mumbai 400088 e.mail: [email protected]

**Department of Physics, K.J. Somaiya College, Vidyavihar, Mumbai 400477

***Department of Physics

University of Mumbai, Mumbai

ABSTRACT 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 2007. 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. A 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

214 Physics Education • September − October 2007

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

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.

Apparatus

1) Two torsion pendulums

2) An aluminum frame with two G-clamps

3) A steel wire

4) Three identical Allen keys

5) A stopwatch

6) A 3 m measuring tape

7) A pair of vernier calipers

8) A 5 kg mass

9) A S-hook

10) A retort stand with clamps

Description of Apparatus 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 Allen screws. At 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


The pendulums can be clamped to the steel wire by tightening the two Allen screws. After 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. 2) The Frame

Figure 2. 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.

3. Assembly

For clamping the steel wire–pendulum assembly you may follow these steps: a) The frame is clamped to the table using

two G-clamps.

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 A on the frame.

f) Tighten the screws on brass strip A. 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 3. 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.


4 Theory

Figure 4.

Part A: Moment of inertia of a dumbbell-shaped 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

Itotal = Irod+2×Iring (1)

Irod = 12

rodM

(L2+b2) (2)

The moment of inertia of a ring about an axis parallel to its plane and passing through its centre is given by

Iring = mh m R R2

12

22

12 4+ +( )

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,

Iring = mh m R R2

12

22

12 4+ +( ) +mr2 (3)

Total moment of inertia of the pendulum is given by

Itotal = Irod + 2×Iring = M rod

12 (L2+b2) +

A pair of Vernier Calliper Measuring Tape Allen Key Stopwatch


212 4

2

12

22 2× + + +

⎛

⎝⎜

⎞

⎠⎟

⎛

⎝⎜⎜

⎞

⎠⎟⎟

mh m R R mr( ) (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 l1 and l2, then the net torque acting

on the body becomes equal to θα ⎟⎟⎠

⎞⎜⎜⎝

⎛+

21

11

ll.

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 ddt

2

2θ + kθ = 0. (5)

where, k = α1 1

1 2l l+

⎛

⎝⎜

⎞

⎠⎟ , is the torque per unit

twist.

Figure 7. The period of oscillation of the torsional

pendulum is, then, given by

T Ik

= 2π (6)

Part C: Normal Modes of Vibration and Coupling Constant

When two torsion pendulums A and B of moments of inertia IA and IB and of lengths l1 and l2 respectively, are coupled by a similar wire segment of length l3, the torque due to twist in the coupling segment will be equal to the difference in the twists at its two ends. If θ1 and θ2 be the twists in l1 and l2 (fixed at the ends) then the twist in l3 is (θ1 − θ2) and the restoring torque on pendulum A from l3 would be −(θ1 − θ2)(α/l3).[We follow the convention that θ1 is positive when the twist is anticlockwise and θ2 is positive if it is clockwise when seen facing the fixed ends.] Similarly the torque on pendulum B would be − (θ2 − θ1)(α/l3). The net torque on pendulum A (which we can refer as pendulum 1) would be


3

21

1

12

1 )(2

lldt

dI A

θθα

θα

θ −−−=

and so its equation of motion would be

0112

3

21

312

1 =−++ ⎟⎟⎠

⎞⎜⎜⎝

⎛llldt

dIAθαθαθ (7)

Similarly, considering the net torque on pendulum 2 we have

011

3

12

322

22

=−++ ⎟⎟⎠

⎞⎜⎜⎝

⎛llldt

dI B

θαθα

θ (8)

The resultant motion of the system depends upon the values of θ1 and θ2 at each instant. Two simple cases are of importance.

Figure 8.

Case 1) Both θ1 and θ2 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 2) During oscillation θ1 and θ2 are out of phase and as a consequence the restoring torque and the frequency is increased.

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

IAIBω4 − α α ω1 1 1 1

2 3 1 3

2

l lI

l lIA B+

⎛

⎝⎜

⎞

⎠⎟ + +

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+ αα2

1 3 2 3

2

32

1 1 1 1 0l l l l l

+⎛

⎝⎜

⎞

⎠⎟ +⎛

⎝⎜

⎞

⎠⎟ −

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= (9)

where ω is used for dt

dθ .

Eq. (9) can be solved by substituting the values of IA, IB, α, l1, l2 and l3. The roots of the equation are angular frequencies ωin and ωout 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

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

3110

11

llI A

αω (10)

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

3220

11

llI B

αω (11)

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.


Coupling Constant

The frequencies, ωout, ω10 and ωin, ω20 can be expressed as

2210

2out 2

1 ωωω Δ+= (12)

2220

2in 2

1 ωωω Δ−= (13)

where,

)( 220

210

2 ωωω −=Δ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

−+ 1

)(

41

2/1

220

210

4

ωω

κ

(14)

with the abbreviation

κα2

3=

l I IA B (15)

where κ is called the coupling constant.

If ω10 = ω20, Eq. (14) reduces to

Δω2 = 2κ2 (16)

Part D: Beats

If the coupling between the two pendulums is small, ωin and ωout 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 A: Moment of Inertia Make necessary geometrical measurements of the pendulums. Calculate the moment of inertia of both the pendulums.

MA = Mass of rod A (including Allen screws), MB = Mass of rod B (including Allen screws), mA = Mass of ring A (each ring), mB = Mass of ring B (each ring).

Also calculate errors ΔIA and ΔIB.

Part B: Determination of the Torsion Constant, α

Set up the pendulum A assembly. Clamp rod A at l1 = 10.0 cm. (Do not clamp rod B). Measure and note down the period of oscillation of pendulum A when the oscillating body is at l1 from the top. Change l1 in steps and study the period of oscillation of the pendulum A. Plot period of oscillation against l1. Also plot a suitable graph to determine the torsion constant α. Calculate error Δα. Part C: Normal Modes of Vibration and Coupling Constant

Case 1: ω10 ≠ ω20

Clamp rod A at l1 = 20.0 cm and rod B at l2 = 30.0 cm. Oscillate this coupled system in two normal modes i.e. in phase and out of phase. Calculate ωout and ωin from your observations.

Solve Eq.(9) by substituting the values of IA, IB, α, l1, l2 and l3 to get ωout and ωin.

Using the retort stand – clamp, fix pendulum B. Oscillate pendulum A. Calculate ω10 from your observations. Also calculate ω10 by substituting α, IA, l1 and l3 in Eq.(10).

Release rod B and clamp rod A similarly. Oscillate pendulum B. Calculate ω20 from your observations. Also calculate ω20 by substituting α, IB, l2 and l3 in Eq.(11).

From these values calculate Δω2 and hence find the coupling constant κ using Eqs.(12), (13) and (14). Also, calculate κ from Eq.(15) by substituting α, l3, IA, IB.


Figure 9 Figure 10 Figure 11 Figure 12 Case 2: ω10 ≈ ω20

Make l1 = l2 = 10.0 cm. Measure ωout, ωin, ω10 and ω20. Hence calculate κ.

Part D: Beats

At l1 = l2 = 10.0 cm, twist the pendulum A by holding pendulum B steady. Release pendulum A. Allow pendulum A to oscillate. While pendulum A is oscillating, release pendulum B. Observe the beat phenomenon. Measure the

beat frequency. Also calculate the beat frequency from ωout and ωin obtained in part C.

Typical Measurements and Calculations

Part A: Moment of Inertia of Pendulums DIMENSIONS OF THE PENDULUMS:

FOR PENDULUM A 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 ring=79.4g Inner radius, R1 = 0.68cm Outer radius, R2 =1.88cm Thickness of the ring, h =0.97cm

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 ring=79.4g Inner radius, R1 = 0.67cm Outer radius, R2 =1.87cm Thickness of the ring, h =0.97cm


IA = Irod + 2×Iring = )(12

22rod bLM

+ +

212 4

2

12

22 2× + + +

⎛

⎝⎜

⎞

⎠⎟

mh m R R mr( )

= 136112

17 0 972 2. ( . )+ + 2 ×

79 4 0 9712

79 44

188 0 68 79 4 8 012

2 2 2. . . ( . . ) . .×+ + + ×

⎛

⎝⎜

⎞

⎠⎟

= 1.37 × 104 g-cm2

IB = Irod + 2×Iring = M

L brod

122 2( )+ +

212 4

2

12

22 2× + + +

⎛

⎝⎜

⎞

⎠⎟

mh m R R mr( )

= 136 313

17 0 972 2. ( . )+ + 2×

79 4 0 9712

79 44

187 0 67 79 4 8 012

2 2 2. . . ( . . ) . .×+ + + ×

⎛

⎝⎜

⎞

⎠⎟

= 1.37 × 104 g⋅cm2

Error in I

Δ Δ ΔII

II

II

=⎛

⎝⎜

⎞

⎠⎟ +

⎛

⎝⎜⎜

⎞

⎠⎟⎟

rod

rod

ring

ring

2 2

ΔII

rod

rod= Δ Δ ΔM

ML

Lb

b⎛⎝⎜

⎞⎠⎟

+ ×⎛⎝⎜

⎞⎠⎟

+ ×⎛⎝⎜

⎞⎠⎟

2 2 2

2 2

ΔII

ring

ring=

Δ Δ Δ

Δ Δ

mm

hh

RR

RR

rr

⎛⎝⎜

⎞⎠⎟

+ ×⎛⎝⎜

⎞⎠⎟

+ ×⎛

⎝⎜

⎞

⎠⎟ +

×⎛

⎝⎜

⎞

⎠⎟ + ×⎛

⎝⎜⎞⎠⎟

2 21

1

2

2

2

2 2

2 2

2 2

Substituting the values from the above table:

Δ ΔII

II

A

A

B

B= = 0 045.

∴ = = ×Δ ΔI IA B 0 06 104. g⋅cm2

∴ = = ± ×I IA B ( . . )137 0 06 104 g⋅cm2

Graph of T against l1

10 20 30 40 50 60 701.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.0

T in

s

l1 in cm


Part B: Determination of the Torsion constant, α

t for 10 oscillations Sr. No.

l1 in cm

l2 in cm

t1 s t2 s t3 s

Mean t in s T

ts=

10

T2 in s2

l1l2 cm2

1 10 70.4 12.62 12.66 12.72 12.67 1.267 1.60 704

2 15 65.4 14.75 14.88 14.94 14.86 1.486 2.21 981

3 20 60.4 16.62 16.5 16.53 16.55 1.655 2.74 1208

4 25 55.4 17.75 17.82 17.81 17.79 1.779 3.17 1385

5 30 50.4 18.72 18.62 18.62 18.65 1.865 3.48 1512

6 32 48.4 18.81 18.88 18.71 18.80 1.880 3.53 1548.8

7 34 46.4 18.97 19.04 18.93 18.98 1.898 3.60 1577.6

8 36 44.4 19.06 19.16 19.12 19.11 1.911 3.65 1598.4

9 38 42.4 19.12 19.16 19.22 19.167 1.917 3.67 1611.2

10 40 40.4 19.22 19.37 19.22 19.27 1.927 3.71 1616

11 42 38.4 19.09 19.22 19.15 19.15 1.915 3.67 1612.8

12 44 36.4 18.97 19.1 19.06 19.04 1.904 3.63 1601.6

13 46 34.4 18.97 19.09 19.03 19.03 1.903 3.62 1582.4

14 48 32.4 18.94 18.87 18.79 18.87 1.887 3.56 1555.2

15 50 30.4 18.66 18.72 18.63 18.67 1.867 3.49 1520

16 55 25.4 17.97 17.84 17.97 17.93 1.793 3.21 1397

17 60 20.4 16.78 16.75 16.72 16.75 1.675 2.81 1224

18 65 15.4 15.06 15.12 15.15 15.11 1.511 2.28 1001

19 70 10.4 12.85 12.94 12.78 12.86 1.286 1.65 728


Rearranging Eq.(5) and (6),

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

21

11

llακ and

κπ

IT 2=

T I

l l

I l ll l

22

1 2

21 2

1 2

41 1

4=

+⎛⎝⎜

⎞⎠⎟

=+

π

α

πα

T I l l I l l22

1 2

2

1 2480 4 201

= =π

απ

α. .

Graph of T2 against l1l2:

Slope = πα

2

201I

. = .00230 ± 0.00001 s2/cm2

απ

=× ×

×

2 4137 10201 0 0023

.. .

= 2.922×106 dyne⋅cm2

Error in α

22

slope

slope⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛ Δ+

Δ=

Δ

A

A

I

I

α

α

Since the error in the slope is very less,

045.0=Δ

=Δ

A

A

I

I

α

α

Δα = 0.13×106 dynes⋅cm2

∴ α = (2.92±0.13)×106 dynes⋅cm2

Part C

Here, l1=20.0 cm, l2=30.0 cm and l3=30.4 cm

Solving Eq.(9) for ωin and ωout:

IAIBω4 2

3132

1111ωαα ⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+++− BA I

llI

ll

+ ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛−++ 2

3

2

3231

2 1111

lllll

αα = 0

IAIB = 1.88 × 108 g2 ⋅cm4

600 800 1000 1200 1400 1600

1.5

2.0

2.5

3.0

3.5

4.0

T2 in s2

l1l2 in cm2


BA Ill

Ill ⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛+++

3132

1111αα =

2.92×106 ⎟⎠⎞

⎜⎝⎛

+4.30

1

0.30

11.37×104 +

2.92×106 ⎟⎠⎞

⎜⎝⎛

+4.30

1

0.20

11.37×104

2.92×106× 1.37×104×1

30 02

30 41

20 0. . .+ +⎛

⎝⎜⎞⎠⎟

= 4.00×1010×0.149

= 5.96×109 g2⋅cm4 s–2

23

2

3231

2 1111lllllαα −⎟⎟

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+ =

(2.92×106)2 120 4

130 4

130 0

130 4. . . .

+⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

2)4.30(

2)61092.2( ×−

= 8.53×1012[5.49×10–3 – 1.08×10–3]

3.76×1010 g2⋅cm4 s–4

Therefore, Eq.(9) becomes,

(1.88×108)ω4−(5.96×109)ω2+(3.76×1010) = 0

1.88ω4 − 59.6ω2 + 376 = 0

a

acbb

2

422 −±−=ω

= 59 6 59 6 4 188 376

2 188

2. ( . ) ..

± − × ×

×=

76.3

9.266.59 ±

76.3

9.266.592 +=+ω = 23.0 rad/s and

76.3

9.266.592 −=−ω = 8.7 rad/s

0.232 =+ω rad/s, 7.82 =−ω rad/s

ω+ = ωout = 4.80 rad/s, ω− = ωin = 2.95 rad/s

Tout = 1.31s, Tin = 2.13s

Calculating ω10 and ω20:

⎟⎟⎠

⎞⎜⎜⎝

⎛= +

3110

11

llI A

αω

= ⎟⎠⎞

⎜⎝⎛

+×

×

4.30

1

0.20

141037.1

61092.2 = 4.20 rad/s

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

3220

11

llI B

αω =

⎟⎠⎞

⎜⎝⎛

+×

×

4.30

1

0.30

1

1037.1

1092.24

6

= 3.76 rad/s

Coupling Constant

2210

2out 2

1ωωω Δ+=

2212

202in ωωω Δ+=

)220

210

2in

2out

2 ()( ωωωωω −−−=Δ

]2)95.2(2)80.4[(2 −=Δω − −[( . ) ( . ) ]4 20 3762 2

= 10.84 (rad/s)2

From Eq.(14)

=κ 4 11)(

)(

4

)(2

220

210

22220

210

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+

−

Δ−

ωω

ωωω


κ =

4

2

22

222

11)76.320.4(

84.10

4

)76.320.4(

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+

−

−

κ = 2.64 rad/s

From Eq.(15)

24

6

213

2

)1037.1(4.30

1092.2

×

×==

IIl

ακ = 7.01

κ = 2.65 rad/s

Case 1

ω10 ≠ ω20

Observed Calculated

t1 (s) t2 (s) t3 (s) t (s) 10

tT = (s)

T

πω

2= rad/s ω (rad/s)

ωin

21.00 21.04 21.06 21.03 2.10 2.99 2.95

ωout 12.97 12.90 12.94 12.93 1.29 4.86 4.80

ω10 14.90 14.81 14.75 14.82 1.48 4.24 4.20

ω20 16.53 16.40 16.47 16.46 1.64 3.82 3.76

Case 2

ω10 ≈ ω20

t1(s) t2(s) t3(s) t(s) Tt

=10

s T

πω

2= (rad/s)

ωin 13. 53 13.53 13.47 13.51 1.35 4.65

ωout 11.71 11.75 11.69 11.71 1.17 5.36

ω10 12.40 12.44 12.44 12.42 1.24 4.08

ω20 12.50 12.47 12.53 12.5 1.25 5.02


Part D: Beats

Observed

Time between 3 minima Calculated

l1 cm

l2 cm

t1 s

t2 s

t3 s

t s

ΔT= t3

s Angular

Beat Frequency

rad/s

Angular Beat

Frequency rad/s

10.0 10.0 26.59 26.75 26.53 26.62 8.87 0.71 0.71

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.

Acknowledgement

We are thankful to all the Olympiad students of 2007 batch who gave us reason to develop this experiment. We express our thanks to Prof. D.A. 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, 3rd edition, Addison-Wesley, Readings, Mass., p.191 (1971).

2) Yee-Tak Yu, “The Double Torsion Pendulum in a Liquid”, American Journal of Physics 10, 152 (1942).

physics through teaching lab ixphysics.unipune.ac.in/~phyed/24.3/24.3_pathare.pdf8) a 5 kg mass 9) a...

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