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Indian Institute of Science Education and
Research
Thiruvananthapuram
SUMMER PROJECT 2015
Study of Parametric Standing Waves in
Fluid Filled Tibetan Singing Bowl
Author:
Sandra B.
IMS13118
IISER-TVM
Supervisor:
Dr. S. Shankaranarayanan
School of Physics,
IISER-TVM
27 July 2015

School of Physics,
IISER-TVM
27 July 2015
To whom it may concern:
This is to certify that the summer project entitled ”STUDY OF PARAMETRIC
STANDING WAVES IN FLUID FILLED TIBETAN SINGING BOWL” has been
successfully completed by Ms. Sandra B. as a second year student of BS-MS dual degree pro-
gramme at Indian Institute of Science Education and Research, Thiruvananthapuram under my
supervision and guidance. She reported as a summer project fellow on 11 May 2015 and worked
under my supervision till 24 July 2015. Her project work spanned over a period of 10 weeks.
Yours sincerely

1. INTRODUCTION
According to the Tibetan oral tradition, the existence of Singing Bowls dates back to 560 -480
B.C. The tradition was brought from India to Tibet in the 8th century A.D. Its composition
is doubted to comprise an 8 metal alloy of copper and tin with traces of iron, lead, zinc, gold,
silver and mercury. The Tibetan Singing Bowl is a type of standing bell which is played by
striking or rubbing its rim with a wooden or leather wrapped mallet. When the bowl is filled
with water,excitation of the bowl results in surface wave patterns on the water surface and
more vigorous forcing ultimately leads to creation of droplets via wave breaking. Faraday in
1831, demonstrated that vertical vibration of horizontal fluid layer leads to parametric standing
waves oscillating with half the forcing frequency above critical acceleration.Increased forcing
results in more complex wave patterns and finally leads to surface fracture and the ejection of
droplets.
Figure 1: Tibetan Singing Bowl (Credit:wikipedia)
2. PARAMETRIC RESONANCE
2.1. INTRODUCTION
Parametric oscillation is seen in driven harmonic oscillators. For parametric resonance, the driv-
ing frequency should be twice the natural frequency, for which a continuous time dependent
force is to applied. The parameters that maybe varied are its resonant frequency and damp-
ing dissipation rate.If the driving frequency is twice the natural frequency of the oscillator, it
absorbs energy at a rate proportional to the energy it already has. Without the compensating
energy loss mechanism by dissipation rate, the oscillation amplitude grows exponentially. How-
ever, if the initial amplitude is zero, no parametric resonance can happen. So in non parametric
resonance,of driven simple harmonic oscillators, the amplitude grows linearly in time regardless
of the initial state.
2.2. IN TIBETAN SINGING BOWL
Striking or rubbing the fluid filled bowl with a leather wrapped mallet excites wall vibrations
and concomitant waves are formed on the fluid surface. Excitation cause vibration of the rim
of the bowl and produce a rich sound. The vibrational frequency depends on the material
properties, geometry and characteristics of the contained fluid. Tapping of the bowl excites
a number of vibrational modes and while rubbing with leather mallet, the rest of the modes
except (2, 0) fundamental mode are suppressed. This is called mode lock in. During this stick-
slip process, one of the nodes follow the point of contact with the mallet, imparting angular
1

momentum to the bound liquid. The net driving acceleration that produces the standing gravity
waves is
g − ∆ω0
2
cos(ω0t)
where, g is the acceleration due to gravity, ∆ is the maximum amplitude of the oscillating rim
and w0 is the angular frequency of the bowl. The water surface are produced due to the energy
transfer from the bowl the water in contact with the walls of the bowl.
3. FARADAY WAVES IN TIBETAN SINGING BOWL
When the bowl is vibrated vertically, the non linear standing waves formed on the surface of
the fluid is called Faraday waves.
3.1. Ideal Fluid
The Euler equation for the ideal fluid is
δv
δt
+ (v. ).v +
P
ρ
− g = 0 (1)
and
.v = 0 (2)
where v = ui+vj +wk, P is the pressure, ρ is the density of the fluid and g = gz −∆ω2
0cos(ω0t).
The equations of motion of the fluid can be written in the form
δu
δt
+ u
δu
δx
+ v
δu
δy
+ w
δu
δz
=
−1
ρ
δP
δx
δv
δt
+ u
δv
δx
+ v
δv
δy
+ w
δv
δz
=
−1
ρ
δP
δy
δw
δt
+ u
δw
δx
+ v
δw
δy
+ w
δw
δz
=
−1
ρ
δP
δz
+ (g − ∆ω2
0cos(ω0t))
δu
δx
+
δv
δy
+
δw
δz
= 0 (3)
Since the flow is irrotational, the vorticity w = × v = 0 and hence v can be represented as
gradient of a scalar function φ(x, y, z, t). Hence
v = φ =
δφ
δx
i +
δφ
δy
j +
δφ
δz
k (4)
Hence the equations of motion has the integral
δφ
δt
+
1
2
(u2
+ v2
+ w2
) =
−P
ρ
+ (g − ∆ω2
0cos(ω0t))z (5)
The equation for free surface of water is
z = a(x, y, t) (6)
The equation of undisturbed free surface is z = 0 and that at base of the bowl is z = h. The
pressure at the free surface of water is given by P = σ(k1 + k2) where σ is the surface tension
2

of water and k1 andk2 represent the principal curvatures of the surface.[2]
The kinematic surface condition at free surface is
D
Dt
(a(x, y, t) − z) =
δa
δt
+ u
δa
δx
+ v
δa
δy
− w = 0 (7)
The normal velocity at the wall is
δφ
δn
= 0 and that at the base is
δφ
δz
= 0.
Since the deﬂection and slope of free surface are very small, we can neglect the square and
product terms from equations (5) and (7) which gives
δφ
δt
=
−σ
ρ
(k1 + k2) + (g − ∆ω2
0cos(ω0t))a (8)
and
δa
δt
= w =
δφ
δz
(9)
From membrane theory, k1 =
δ2
a
δx2
and k2 =
δ2
a
δy2
hence equation (8) becomes
σ
ρ
(
δ2
a
δx2
+
δ2
a
δy2
) +
δφ
δt
/z=0 − (g − ∆ω2
0cos(ω0t))a = 0 (10)
Since
δφ
δn
= 0,
δa
δn
= 0 and
δ
δn
(
δ2
a
δx2
+
δ2
a
δy2
) = 0,hence a,φ and
δ2
a
δx2
+
δ2
a
δy2
can be expanded in
terms of complete orthogonal set of eigen functions Sm(x, y).
(
δ2
δx2
+
δ2
δy2
+ k2
m)Sm(x, y) = 0 (11)
where
δSm
δn
= 0 and k2
m is the eigen value. The required expansion for a(x, y, t),
δ2
a
δx2
+
δ2
a
δy2
and
φ(x, y, z, t) are
a(x, y, t) =
∞
0
am(t)Sm(x, y) (12)
δ2
a
δx2
+
δ2
a
δy2
= −
∞
0
k2
mam(t)Sm(x, y) (13)
φ = −
∞
1
cosh(km(h − z))
kmsinh(kmh)
Sm(x, y) (14)
Hence equation (10) becomes
Sm(x, y)
kmtanh(kmh)
+ [
d2
am(t)
dt2
+ kmtanh(kmh)(k2
m
σ
ρ
+ g − ∆ω2
0cos(ω0t) am(t))] = 0 (15)
Since Sm(x, y) are linearly independent,
d2
am(t)
dt2
+ kmtanh(kmh)(k2
m
σ
ρ
+ g − ∆ω2
0cos(ω0t) am(t)) = 0 (16)
Equation (16) can be re-written as,
d2
am(t)
dt2
+ ω2
m(1 − 2γcos(ω0t)) am(t) = 0 (17)
3

where,
ω2
m = (gkm +
σ
ρ
k3
m) tanh(kmh) (18)
and
2γ =
∆ω2
0
g + k2
m
σ
ρ
=
Γ
1 + B−1
0
(19)
where Γ =
∆ω2
0
g
and B0 =
ρg
σk2
m
.
3.1.1. S(r, θ)
The surface deformation be a(x,y,t)which can be written as
a(x, y, t) =
m
am(t)Sm(x, y)
where, Sm(x, y) is the container’s eigenmode and am(t) being the oscillating amplitude of the
eigenmode m. From equation (11) we have
( 2
+ k2
m)Sm(x, y) = 0 (20)
Let
x = rcos(θ)y = rsin(θ)
So
a(r, θ, t) = S(r, θ)am(t)
Let R be the radius of the singing bowl and hence we can apply two boundary conditions:
F(R) = 0 (21)
G(θ) = G(θ + 2π) (22)
This equations is Helmholtz equation in S(r, θ).
2
S + k2
mS = 0 (23)
where,
S(r, θ) = F(r)G(θ)
In cylindrical coordinate system, 2
=
1
r
∂
∂r
(r
∂
∂r
) +
1
r2
∂2
∂θ2
. So by the method of variable
separation we get,
r
F(r)
∂
∂r
(r
∂F(r)
∂r
) + k2
mr2
= −
1
G(θ)
∂2
G(θ)
∂θ2
= µ (24)
which are separated into two independent equations (25) and (31) which can be solved for F(r)
and G(θ). The solution of the equation
∂2
G(θ)
∂θ2
+ µG(θ) = 0 (25)
is
G(θ) = c1sin(
√
µθ) + c2cos(
√
µθ)
4

On applying the boundary condition (22), we get the condition that
sin
√
µ(θ + 2π) = sin
√
µθ ⇒
√
µ = m ∈ Z,
thus
µ = m2
(26)
and the solution becomes
G(θ) = c1sin(mθ) + c2cos(mθ) (27)
On simplifying the second equation
r
F(r)
∂
∂r
(r
∂F(r)
∂r
) + k2
mr2
= µ
we get it to be in the form of Bessel equation as shown below
r2 ∂2
F(r)
∂r2
+ r
∂F(r)
∂r
+ (k2
mr2
− µ)F(r) = 0
applying the condition (26), the equation becomes
r2 ∂2
F(r)
∂r2
+ r
∂F(r)
∂r
+ (k2
mr2
− m2
)F(r) = 0 (28)
and the solution is
F(r) = AJm(kmr) + BYm(kmr)
But since the Neumann function Ym(kmr) blows up at r = 0, the solution becomes
F(r) = Jm(kmr) (29)
applying the boundary condition (1), we get
Jm(kmR) = 0
Thus km = km
n the value of km for which Jm(kmR) = 0. Hence the ﬁnal solution for S(r, θ) is
S(r, θ) = Jm(km
n r[c1sin(mθ) + c2cos(mθ)] (30)
Consider a water ﬁlled tibetan singing bowl of:
radius R = 7.5cm
height h = 10cm
density of water ρ = 1000kg/m3
surface tension of water σ = 72 ∗ 10−5
N/m2
external frquency with which the bowl is excited f0 = 188Hz
angular frequency of the bowl ω0 = 1180.64s−1
maximal acceleration of the rim normalized by the gravitational acceleration Γ =
∆ω2
0
g
= 6.2
maximum amplitude of vibrating rim ∆ = 4.359 ∗ 10−3
m
5

Figure 2: S(r, θ) v/s r − θ plot with km
n = 1.776cm−1
with m = 1 and n = 4
3.1.2. am(t)
Taking equation (17),
d2
am
dt2
+ ω2
m(1 − 2γcos(ω0t)) am = 0 (31)
with
ω2
m = (gkm +
σ
ρ
k3
m) tanh(kmh) (32)
γ =
Γ
2(1 + B−1
0 )
(33)
where B0 =
ρg
σk2
m
, Γ =
∆ω2
0
g
and km =
2π
λm
.
Let T =
ω0t
2
So the equation (31) becomes
d2
am
dT2
+
4kmtanh(kmh)
ω2
0
[k2
m
σ
ρ
+ g − ∆ω2
0cos(2T)] am = 0 (34)
which can be written in the form
d2
am
dT2
+ (a − 2qcos(2T)) am = 0 (35)
where
a =
4kmtanh(kmh)
ω2
0
(k2
m
σ
ρ
+ g) (36)
and
q = 2
km∆ω2
0tanh(kmh)
ω2
0
(37)
The value of km is obtained by solving the transcendental equation
ω2
m = (gkm +
σ
ρ
k3
m) tanh(kmh)
6

where ωm =
ω0
2
.
Equation (35) has the form of Mathieu equation and solution of the equation after applying
the boundary condition
dam
dT
/T=0 = 0 and with the value of
km = 1665.3m−1
a = 1.00001
q = 0.14
is
am(T) = MathieuC[1.00001, 0.14, T] (38)
10 20 30 40 50 60 70
T
- 20
- 10
10
20
a m
am ( T) V / s T =ω0 t/ 2
Figure 3: am(T) v/s T plot for Ideal Fluid
3.2. Viscous Fluid
The Navier Stokes equation for compressible and viscous fluid with irrotational flow is
δv
δt
+ (v. ).v +
P
ρ
− g +
η
ρ
2
v = 0 (39)
under the condition that
∆ρ
ρ
<< 1 where ρ is the density of the fluid, η is the dynamic viscosity
of the fluid, v = ui + vj + wk and g = gz − ∆ω2
0cos(ω0t). The equations of motion of the fluid
can be written in the form
δu
δt
+ u
δu
δx
+ v
δu
δy
+ w
δu
δz
=
−1
ρ
δP
δx
+
η
ρ
(
δ2
u
δx2
+
δ2
u
δy2
+
δ2
u
δz2
)
δv
δt
+ u
δv
δx
+ v
δv
δy
+ w
δv
δz
=
−1
ρ
δP
δy
+
η
ρ
(
δ2
v
δx2
+
δ2
v
δy2
+
δ2
v
δz2
)
δw
δt
+ u
δw
δx
+ v
δw
δy
+ w
δw
δz
=
−1
ρ
δP
δz
+
η
ρ
(
δ2
w
δx2
+
δ2
w
δy2
+
δ2
w
δz2
) + (g − ∆ω2
0cos(ω0t)) (40)
Since the flow is irrotational, the vorticity w = × v = 0 and hence v can be represented as
gradient of a scalar function φ(x, y, z, t). Hence
v = φ =
δφ
δx
i +
δφ
δy
j +
δφ
δz
k
7

Hence the equations of motion has the integral
δφ
δt
+
1
2
(u2
+ v2
+ w2
) =
−P
ρ
+
η
ρ
(
δ2
φ
δx2
+
δ2
φ
δy2
+
δ2
φ
δz2
) + (g − ∆ω2
0cos(ω0t))z (41)
The equation for free surface of water is
z = a(x, y, t) (42)
The equation of undisturbed free surface is z = 0 and that at base of the bowl is z = h. The
pressure at the free surface of water is given by P = σ(k1 + k2) where σ is the surface tension
of water and k1 andk2 represent the principal curvatures of the surface. Also
δ2
φ
δz2
= 0 since
δ
δz
(
δa
δt
) = 0
From equations (7), (8), (9) and (10) we arrive at the equation
σ
ρ
(
δ2
a
δx2
+
δ2
a
δy2
) +
δφ
δt
/z=0 −
η
ρ
(
δ2
φ
δx2
+
δ2
φ
δy2
) − (g − ∆ω2
0cos(ω0t))a = 0 (43)
Since
δφ
δn
= 0,
δa
δn
= 0 and
δ
δn
(
δ2
a
δx2
+
δ2
a
δy2
) = 0,hence a,φ and
δ2
a
δx2
+
δ2
a
δy2
can be expanded in
terms of complete orthogonal set of eigen functions Sm(x, y).
(
δ2
a
δx2
+
δ2
a
δy2
+ k2
m)Sm(x, y) = 0
where
δSm
δn
= 0 and k2
m is the eigen value. The required expansion for a(x, y, t),
δ2
a
δx2
+
δ2
a
δy2
and
φ(x, y, z, t) are
a(x, y, t) =
∞
0
am(t)Sm(x, y)
δ2
a
δx2
+
δ2
a
δy2
= −
∞
0
k2
mam(t)Sm(x, y)
φ = −
∞
1
cosh(km(h − z))
kmsinh(kmh)
Sm(x, y)
and
δ2
φ
δx2
+
δ2
φ
δy2
= −
∞
1
cosh(km(h − z))
kmsinh(kmh)
kmSm(x, y) (44)
Hence equation (43) becomes
Sm(x, y)
kmtanh(kmh)
+ [
d2
am(t)
dt2
+
k2
mη
ρ
dam(t)
dt
+ kmtanh(kmh)(k2
m
σ
ρ
+ g − ∆ω2
0cos(ω0t) am(t))] = 0
(45)
Since Sm(x, y) are linearly independent,
d2
am(t)
dt2
+
k2
mη
ρ
dam(t)
dt
+ kmtanh(kmh)(k2
m
σ
ρ
+ g − ∆ω2
0cos(ω0t) am(t)) = 0 (46)
Equation (46) can be re-written as,
d2
am(t)
dt2
+ 2β
dam(t)
dt
+ ω2
m(1 − 2γcos(ω0t)) am(t) = 0 (47)
8

where,
2β =
k2
mη
ρ
(48)
ω2
m = (gkm +
σ
ρ
k3
m) tanh(kmh)
and
2γ =
∆ω2
0
g + k2
m
σ
ρ
=
Γ
1 + B−1
0
where Γ =
∆ω2
0
g
and B0 =
ρg
σk2
m
.
3.2.1. am(t)
Taking equation (47),
d2
am
dt2
+ 2β
dam(t)
dt
+ ω2
m(1 − 2γcos(ω0t)) am = 0
Let
am(t) = amd(t)e−βt
(49)
Thus the equation(47) becomes
d2
amd
dt2
+ (ω2
m − β2
)(1 −
2ω2
m
ω2
m − β2
γcos(ω0t)) amd = 0
Hence
d2
amd
dt2
+ ω2
md(1 − γdcos(ω0t)) amd = 0 (50)
where
ω2
md = ω2
m − β2
(51)
and
γd =
ω2
m
ω2
md
γ (52)
since
ω2
m = (gkm +
σ
ρ
k3
m)tanh(kmh)
equation (50) can be written in the form
d2
amd
dt2
+ kmtanh(kmh)(
ω2
md
ω2
m
k2
m
σ
ρ
+
ω2
md
ω2
m
g − ∆ω2
0cos(ω0t) amd(t)) = 0 (53)
Let T =
ω0t
2
So the equation (53) becomes
d2
amd
dT2
+
4kmtanh(kmh)
ω2
0
[
ω2
md
ω2
m
k2
m
σ
ρ
+
ω2
md
ω2
m
g − ∆ω2
0cos(2T)] amd = 0 (54)
which can be written in the form
d2
amd
dT2
+ (ad − 2qdcos(2T)) amd = 0 (55)
9

where
ad =
4kmtanh(kmh)
ω2
0
[
ω2
md
ω2
m
k2
m
σ
ρ
+
ω2
md
ω2
m
g] (56)
and
qd = 2
km∆ω2
0tanh(kmh)
ω2
0
(57)
The value of km is obtained by solving the transcendental equation
ω2
md = (gkm +
σ
ρ
k3
m)tanh(kmh) −
k4
mη2
4ρ2
(58)
where ωmd =
ω0
2
.
Equation (55) has the form of Mathieu equation and solution of the equation after applying
the boundary condition
dam
dT
/T=0 = 0 and with the value of
km = 1665.58m−1
ω2
md
ω2
m
= 0.99
a = 1.00001
q = 0.14
β = 1.2345s−1
is
am(T) = MathieuC[0.990, 0.14, T]
Hence the solution for the equation
d2
am
dt2
+ 2β
dam(t)
dt
+ ω2
m(1 − 2γcos(ω0t)) am = 0
is
am(T) = amd(T)e
−2βT
ω0 (59)
am(T) = MathieuC[0.990, 0.14, T]e−2.09×10−3T
(60)
10 20 30 40 50 60 70
T
- 20
- 10
10
20
A T
A ( T) vs T =wt / 2
Figure 4: am(T) v/s T plot for Viscous Fluid
10

4. CONCLUSION
When a fluid filled tibetan singing bowl is subjected to vertical vibrations, it undergoes para-
metric resonance. This parametric resonance leads to non linear standing gravity waves called
faraday waves with steadily growing amplitude on the free surface of the fluid. The surface
deformation is a combination of Bessel function and Mathieu function. The amplitude of the
waves decay as we go radially outwards from the centre of the bowl. At the centre of the bowl,
the amplitude is maximum resembling a fountain at the centre. It is also evident from the cal-
culation that there is a steady growing amplitude with time leading to resonance. The growth
of amplitude with time was studied in the case of ideal fluid and viscous fluid with
δρ
ρ
<< 1,
for which density is approximated to be a constant. It was seen that in the case of viscous fluid,
there is a decrese in amplitude of the waves along with resonance compared to ideal fluid for
the same forcing acceleration. The same study can be done for viscous and compressible fluids
for which desity can not be taken to be a constant, where basically a more general form of the
Navier Stokes need to be used where density of the fluid is not a constant.
11

References
[1] Denis Terwagne, John W M Bush Tibetan singing bowls .IOP (2011).
[2] Horace Lamb, Hydrodynamics 6th
ed.. Cambridge University Press, (1932).
[3] T. B. Benjamin, F. Ursell The stability of the plane free surface of the liquid in vertical
periodic motion.(1954).
[4] Krishna Kumar,Laurette S. Tuckerman Parametric instability of the interface between two
ﬂuids.Cambridge University Press, (1994).
[5] Arfken G., Weber H. Mathematical methods for Physicists 7th
ed.
[6] D. Blandford, Kip S. Thorne Applications of Classical Physics(2012)
http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/
12

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Study of Parametric Standing Waves in Fluid filled Tibetan Singing bowl

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Study of Parametric Standing Waves in Fluid filled Tibetan Singing bowl