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: shirish@hbcse.tifr.res.in
**
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
Physics Education •
September − October 2007
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
213
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
6) A 3 m measuring tape
2) An aluminum frame with two G-clamps
7) A pair of vernier calipers
3) A steel wire
8) A 5 kg mass
4) Three identical Allen keys
9) A S-hook
5) A stopwatch
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
214
Physics Education •
September − October 2007
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
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 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.
Physics Education •
September − October 2007
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.
215
4 Theory
A pair of Vernier Calliper
Measuring Tape
Allen Key
Stopwatch
Figure 4.
Part A: 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
Itotal = Irod+2×Iring
M
Irod =
rod
12
(L2+b2)
(1)
(2)
Iring =
The moment of inertia of a ring about an axis
parallel to its plane and passing through its
centre is given by
2
Iring =
216
mh
m
+ ( R12 + R22 )
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,
mh 2 m 2
+ ( R1 + R22 ) +mr2
12
4
(3)
Total moment of inertia of the pendulum is
given by
Itotal = Irod + 2×Iring =
Physics Education •
M rod
(L2+b2) +
12
September − October 2007
⎛ ⎛ mh 2 m 2
⎞
2
2⎞⎟
⎜2 × ⎜
⎜ ⎝ 12 + 4 ( R1 + R2 ) + 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
⎛1 1⎞
on the body becomes equal to α ⎜⎜ + ⎟⎟θ .
⎝ l1 l 2 ⎠
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 2θ
dt 2
+ kθ = 0.
(5)
⎛ 1 1⎞
where, k = α ⎜ + ⎟ , is the torque per unit
⎝ l1 l2 ⎠
twist.
Physics Education •
September − October 2007
Figure 7.
The period of oscillation of the torsional
pendulum is, then, given by
T = 2π
I
k
(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
217
IA
2
d θ1
dt
2
=− α
θ1
l1
−α
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.
(θ1 −θ 2 )
l3
and so its equation of motion would be
IA
2
d θ1
dt
2
⎛1 1⎞
θ2
⎜ l + l ⎟⎟θ1−α l =0
⎝ 1 3⎠
3
+α ⎜
(7)
Similarly, considering the net torque on
pendulum 2 we have
2
IB
d θ2
dt
2
⎛1 1⎞
θ
+ ⎟⎟θ 2 −α 1 =0
l3
⎝ l 2 l3 ⎠
+α ⎜⎜
(8)
The resultant motion of the system depends
upon the values of θ1 and θ2 at each instant.
Two simple cases are of importance.
The general equation with variable θ satisfying
both simultaneous equations (7) and (8) can be
obtained in the form
⎡ ⎛ 1 1⎞
⎛ 1 1⎞ ⎤
IAIBω4 − ⎢α ⎜ + ⎟ I A + α ⎜ + ⎟ I B ⎥ω 2
⎝ l1 l3 ⎠ ⎥⎦
⎢⎣ ⎝ l2 l3 ⎠
⎡ ⎛ 1 1 ⎞⎛ 1 1 ⎞ α2 ⎤
+ ⎢α 2 ⎜ + ⎟ ⎜ + ⎟ − 2 ⎥ = 0
⎢⎣ ⎝ l1 l3 ⎠ ⎝ l2 l3 ⎠ l3 ⎥⎦
where ω is used for
dθ
(9)
.
dt
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
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.
218
α ⎛1 1⎞
ω10 =
⎜ +
I A ⎜⎝ l1 l 3
⎟⎟
⎠
(10)
ω 20 =
1⎞
⎜⎜ + ⎟⎟
I B ⎝ l 2 l3 ⎠
(11)
α ⎛1
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.
Physics Education •
September − October 2007
Coupling Constant
The frequencies, ωout, ω10 and ωin, ω20 can be
expressed as
ω out =ω10 + 12 Δω
2
2
ω in =ω 20 − 12 Δω
2
2
2
2
(12)
Also calculate errors ΔIA and ΔIB.
(13)
Part B: Determination of the Torsion
Constant, α
where,
1/ 2
4
⎡⎛
⎤
⎞
4κ
2
2
2
⎟
⎥
Δω =(ω10 −ω 20 ) ⎢⎜⎜ 1+ 2
−
1
2 ⎟
⎢⎣⎝ (ω10 −ω 20 ) ⎠
⎥⎦
(14)
with the abbreviation
κ2 =
α
(15)
l3 I A I B
where κ is called the coupling constant.
(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.
Physics Education •
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
If ω10 = ω20, Eq. (14) reduces to
Δω2 = 2κ2
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).
September − October 2007
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.
219
Figure 9
Figure 10
Figure 11
Case 2: ω10 ≈ ω20
Make l1 = l2 = 10.0 cm. Measure ωout, ωin, ω10
and ω20. Hence calculate κ.
Figure 12
beat frequency. Also calculate the beat
frequency from ωout and ωin obtained in part C.
Typical Measurements and Calculations
Part D: Beats
Part A: Moment of Inertia of Pendulums
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
DIMENSIONS OF THE PENDULUMS:
220
FOR PENDULUM A
FOR PENDULUM B
For rod:
Mass of the rod = 136.1g
Length of the rod = 17.0 cm
Breadth of the rod, b = 0.97cm
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 = 0.68cm
Outer radius, R2 =1.88cm
Thickness of the ring, h =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
Physics Education •
September − October 2007
IA = Irod + 2×Iring =
M rod
12
2
Error in I
2
(L + b ) +
2
⎛ ΔI ⎞ ⎛ ΔI ring ⎞
ΔI
⎟
= ⎜ rod ⎟ + ⎜⎜
I
⎝ I rod ⎠ ⎝ I ring ⎟⎠
⎛ mh
⎞
m
2×⎜
+ ( R12 + R22 ) + mr 2 ⎟
4
⎝ 12
⎠
2
=
ΔI rod
=
I rod
1361
.
(17 2 + 0.97 2 ) + 2 ×
12
⎞
⎛ 79.4 × 0.97 2 79.4
⎜
+
(188
. 2 + 0.68 2 ) + 79.4 × 8.012 ⎟
12
4
⎠
⎝
ΔI ring
I ring
= 1.37 × 104 g-cm2
IB = Irod + 2×Iring =
2
2
2
ΔL ⎞ ⎛
Δb ⎞
⎛ ΔM ⎞ ⎛
⎟
⎟ + ⎜2 ×
⎜
⎟ + ⎜2 ×
⎝ M ⎠ ⎝
L⎠ ⎝
b ⎠
2
=
2
2
2
⎛
ΔR1 ⎞
Δh ⎞
⎛ Δm ⎞
⎛
⎟ +
⎜
⎟ + ⎜2 ×
⎟ + ⎜2 ×
⎝ m⎠
⎝
h ⎠
R1 ⎠
⎝
M rod 2
( L + b2 ) +
12
2
2
⎛
ΔR2 ⎞
Δr ⎞
⎛
⎟ + ⎜2 × ⎟
⎜2 ×
⎝
R2 ⎠
r ⎠
⎝
⎛ mh
⎞
m
2×⎜
+ ( R12 + R22 ) + mr 2 ⎟
4
⎝ 12
⎠
2
Substituting the values from the above table:
136.3
=
(17 2 + 0.97 2 ) + 2×
13
ΔI A ΔI B
=
= 0.045
IA
IB
⎛ 79.4 × 0.97 2 79.4
⎞
⎜
+
(187
. 2 + 0.67 2 ) + 79.4 × 8.012 ⎟
12
4
⎝
⎠
∴ ΔI A = ΔI B = 0.06 × 10 4 g⋅cm2
∴ I A = I B = (137
. ± 0.06) × 104 g⋅cm2
= 1.37 × 104 g⋅cm2
Graph of T against l1
2.0
1.9
1.8
T in s
1.7
1.6
1.5
1.4
1.3
1.2
10
20
30
40
50
60
70
l1 in cm
Physics Education •
September − October 2007
221
Part B: Determination of the Torsion constant, α
Sr.
No.
l1
in cm
l2
in cm
t for 10 oscillations
t1 s
t2 s
t3 s
Mean t in s
T=
t
10
s
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
222
Physics Education •
September − October 2007
Since the error in the slope is very less,
Rearranging Eq.(5) and (6),
Δα ΔI A
=
=0.045
α IA
⎛1 1⎞
I
⎟⎟ and T =2π
κ
⎝ l1 l 2 ⎠
κ =α ⎜⎜ +
Δα = 0.13×106 dynes⋅cm2
4π 2 I
4 π 2 I l1l2
=
T2 =
α l1 + l2
⎛ 1 1⎞
α⎜ + ⎟
⎝ l1 l2 ⎠
T2 =
∴ α = (2.92±0.13)×106 dynes⋅cm2
Part C
4π 2 I
π2 I
l1l2 =
l1l2
80.4α
201
.α
Here, l1=20.0 cm, l2=30.0 cm and l3=30.4 cm
Solving Eq.(9) for ωin and ωout:
Graph of T2 against l1l2:
Slope =
α=
⎡ ⎛1
π2 I
= .00230 ± 0.00001 s2/cm2
201
.α
IAIBω4 − ⎢α ⎜⎜
⎣ ⎝ l2
. × 104
π 2 × 137
= 2.922×106 dyne⋅cm2
201
. × 0.0023
Error in α
⎛ Δslope ⎞
= ⎜
⎟
α
⎝ slope ⎠
Δα
⎡
⎛1
⎣
⎝ l1
2
+ ⎢α ⎜⎜
+
⎛1 1⎞ ⎤
⎟⎟ I A + α ⎜⎜ + ⎟⎟ I B ⎥ω 2
l3 ⎠
⎝ l1 l3 ⎠ ⎦
+
2
1⎞ α ⎤
⎟⎟⎜⎜ + ⎟⎟ − 2 ⎥ = 0
l3 ⎠⎝ l 2 l3 ⎠ l3 ⎦
1⎞
1 ⎞⎛ 1
IAIB = 1.88 × 108 g2 ⋅cm4
2
⎛ ΔI ⎞
+⎜⎜ A ⎟⎟
⎝ IA ⎠
2
4.0
3.5
2.5
2
T in s
2
3.0
2.0
1.5
600
800
1000
1200
1400
1600
2
l1l2 in cm
Physics Education •
September − October 2007
223
⎛1
α ⎜⎜
⎝ l2
+
1⎞
⎛1 1⎞
⎟⎟ I A +α ⎜⎜ + ⎟⎟ I B =
l3 ⎠
⎝ l1 l3 ⎠
ω− =
2
59.6−26.9
3.76
= 8.7 rad/s
ω + =23.0 rad/s, ω − = 8.7 rad/s
2
1 ⎞
⎛ 1
+
2.92×10 ⎜
⎟ 1.37×104 +
⎝ 30.0 30.4 ⎠
6
2
ω+ = ωout = 4.80 rad/s, ω− = ωin = 2.95 rad/s
1 ⎞
⎛ 1
+
⎟ 1.37×104
⎝ 20.0 30.4 ⎠
Tout = 1.31s, Tin = 2.13s
2.92×106 ⎜
2
1 ⎞
⎛ 1
2.92×106× 1.37×104× ⎜
+
+
⎟
⎝ 30.0 30.4 20.0 ⎠
= 4.00×1010×0.149
Calculating ω10 and ω20:
ω10 =
= 5.96×109 g2⋅cm4 s–2
⎛ 1 1 ⎞⎛ 1 1 ⎞ α
α 2 ⎜⎜ + ⎟⎟⎜⎜ + ⎟⎟− 2 =
⎝ l1 l3 ⎠⎝ l2 l3 ⎠ l3
1
1 ⎞⎛ 1
1 ⎞
(2.92×10 ) ⎜
+
+
⎟⎜
⎟
⎝ 20.4 30.4 ⎠ ⎝ 30.0 30.4 ⎠
6 2
( 2.92 × 10 )
−
2
(30.4)
⎟
6
1 ⎞
⎛ 1
+
⎜
⎟ = 4.20 rad/s
4
1.37 × 10 ⎝ 20.0 30.4 ⎠
ω 20 =
6 2⎛
⎜
+
I A ⎜⎝ l1 l 3 ⎟⎠
2.92 × 10
=
2
α ⎛1 1⎞
α ⎛1 1⎞
⎜
⎟
+
=
I B ⎜⎝ l 2 l3 ⎟⎠
2.92 × 10
6
1.37 × 10
4
1 ⎞
⎛ 1
+
⎜
⎟ = 3.76 rad/s
⎝ 30.0 30.4 ⎠
= 8.53×1012[5.49×10–3 – 1.08×10–3]
Coupling Constant
3.76×1010 g2⋅cm4 s–4
ω out =ω10 + Δω
2
1
2
2
Therefore, Eq.(9) becomes,
(1.88×108)ω4−(5.96×109)ω2+(3.76×1010) = 0
1.88ω4 − 59.6ω2 + 376 = 0
2
− b± b −4ac
2a
59.6 ± (59.6) − 4 × 188
. × 376
2 × 188
.
2 59.6+26.9
= 23.0 rad/s and
3.76
ω+ =
224
2
2
2
2
2
Δω 2 =[( 4.80) 2 −( 2.95) 2 ] − [(4.20) 2 − (3.76) 2 ]
= 10.84 (rad/s)2
2
=
1
2
Δω =(ω out −ω in )−(ω10 −ω 20 )
2
2
ω =
2
ω in =ω 20 + 2 Δω
2
=
59.6±26.9
3.76
From Eq.(14)
κ=
4
2
2 2
(ω10 −ω 20 ) ⎡⎛
4
Physics Education •
2
⎞ ⎤
⎢⎜⎜ 2 2 +1⎟⎟ −1⎥
⎢⎣⎝ (ω10 −ω 20 ) ⎠ ⎥⎦
2
( Δω )
September − October 2007
κ=
4
From Eq.(15)
2
⎞ ⎤
+1⎟⎟ −1⎥
⎢⎜⎜
2
2
⎢⎣⎝ ( 4.20 −3.76 ) ⎠ ⎥⎦
2
2 2
( 4.20 −3.76 ) ⎡⎛
4
10.84
2
κ =
α
2.92 × 10
6
= 7.01
=
l 3 I 1 I 2 30.4 (1.37 × 10 4 ) 2
κ = 2.65 rad/s
κ = 2.64 rad/s
Case 1
ω10 ≠ ω20
Observed
t1 (s)
t2 (s)
t3 (s)
t (s)
Calculated
T =
t
10
(s)
ω=
2π
ω (rad/s)
rad/s
T
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
t
s
2π
t1(s)
t2(s)
t3(s)
t(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
Physics Education •
September − October 2007
T =
10
ω=
(rad/s)
T
225
Part D: Beats
Observed
Time between 3 minima
Calculated
l1
cm
l2
cm
t1
s
t2
s
t3
s
t
s
10.0
10.0
26.59
26.75
26.53
26.62
ΔT=
t
3
s
8.87
Angular
Beat
Frequency
rad/s
0.71
Angular
Beat
Frequency
rad/s
0.71
Conclusion
Acknowledgement
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.
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)
2)
226
Symon K.R., Mechanics, 3rd edition, AddisonWesley, Readings, Mass., p.191 (1971).
Yee-Tak Yu, “The Double Torsion Pendulum
in a Liquid”, American Journal of Physics 10,
152 (1942).
Physics Education •
September − October 2007