Indian Institute of Science Education andResearch

Thiruvananthapuram

SUMMER PROJECT 2015

Study of Parametric Standing Waves inFluid Filled Tibetan Singing Bowl

Author:Sandra B.IMS13118

IISER-TVM

Supervisor:Dr. S. Shankaranarayanan

School of Physics,IISER-TVM

27 July 2015

Dr. S. ShankaranarayananSchool of Physics,IISER-TVM

27 July 2015

To whom it may concern:

This is to certify that the summer project entitled ”STUDY OF PARAMETRICSTANDING WAVES IN FLUID FILLED TIBETAN SINGING BOWL” has beensuccessfully 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 mysupervision and guidance. She reported as a summer project fellow on 11 May 2015 and workedunder my supervision till 24 July 2015. Her project work spanned over a period of 10 weeks.

Yours sincerely

Dr. S. Shankaranarayanan

1. INTRODUCTION

According to the Tibetan oral tradition, the existence of Singing Bowls dates back to 560 -480B.C. The tradition was brought from India to Tibet in the 8th century A.D. Its compositionis 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 bystriking or rubbing its rim with a wooden or leather wrapped mallet. When the bowl is filledwith water,excitation of the bowl results in surface wave patterns on the water surface andmore vigorous forcing ultimately leads to creation of droplets via wave breaking. Faraday in1831, demonstrated that vertical vibration of horizontal fluid layer leads to parametric standingwaves oscillating with half the forcing frequency above critical acceleration.Increased forcingresults in more complex wave patterns and finally leads to surface fracture and the ejection ofdroplets.

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 dependentforce 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, itabsorbs energy at a rate proportional to the energy it already has. Without the compensatingenergy 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 parametricresonance,of driven simple harmonic oscillators, the amplitude grows linearly in time regardlessof the initial state.

2.2. IN TIBETAN SINGING BOWL

Striking or rubbing the fluid filled bowl with a leather wrapped mallet excites wall vibrationsand concomitant waves are formed on the fluid surface. Excitation cause vibration of the rimof the bowl and produce a rich sound. The vibrational frequency depends on the materialproperties, geometry and characteristics of the contained fluid. Tapping of the bowl excitesa number of vibrational modes and while rubbing with leather mallet, the rest of the modesexcept (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

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momentum to the bound liquid. The net driving acceleration that produces the standing gravitywaves is

g −∆ω02cos(ω0t)

where, g is the acceleration due to gravity, ∆ is the maximum amplitude of the oscillating rimand w0 is the angular frequency of the bowl. The water surface are produced due to the energytransfer 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 ofthe fluid is called Faraday waves.

3.1. Ideal FluidThe Euler equation for the ideal fluid is

δ~v

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−∆ω20cos(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 asgradient of a scalar function φ(x, y, z, t). Hence

~v = ∇φ = δφ

δxi+ δφ

δyj + δφ

δzk (4)

Hence the equations of motion has the integral

δφ

δt+ 1

2(u2 + v2 + w2) = −Pρ

+ (g −∆ω20cos(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. Thepressure at the free surface of water is given by P = σ(k1 + k2) where σ is the surface tension

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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 δφδn

= 0 and that at the base is δφδz

= 0.Since the deflection and slope of free surface are very small, we can neglect the square andproduct terms from equations (5) and (7) which gives

δφ

δt= −σ

ρ(k1 + k2) + (g −∆ω2

0cos(ω0t))a (8)

andδa

δt= w = δφ

δz(9)

From membrane theory, k1 = δ2a

δx2 and k2 = δ2a

δy2 hence equation (8) becomes

σ

ρ(δ

2a

δx2 + δ2a

δy2 ) + δφ

δt/z=0 − (g −∆ω2

0cos(ω0t))a = 0 (10)

Since δφ

δn= 0, δa

δn= 0 and δ

δn(δ

2a

δx2 + δ2a

δy2 ) = 0,hence a,φ and δ2a

δx2 + δ2a

δy2 can be expanded interms of complete orthogonal set of eigen functions Sm(x, y).

( δ2

δx2 + δ2

δy2 + k2m)Sm(x, y) = 0 (11)

where δSmδn

= 0 and k2m is the eigen value. The required expansion for a(x, y, t), δ

2a

δx2 + δ2a

δy2 andφ(x, y, z, t) are

a(x, y, t) =∞∑0am(t)Sm(x, y) (12)

δ2a

δx2 + δ2a

δy2 = −∞∑0k2mam(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) + [d

2am(t)dt2

+ kmtanh(kmh)(k2m

σ

ρ+ g −∆ω2

0cos(ω0t) am(t))] = 0 (15)

Since Sm(x, y) are linearly independent,

d2am(t)dt2

+ kmtanh(kmh)(k2m

σ

ρ+ g −∆ω2

0cos(ω0t) am(t)) = 0 (16)

Equation (16) can be re-written as,

d2am(t)dt2

+ ω2m(1− 2γcos(ω0t)) am(t) = 0 (17)

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where,ω2m = (gkm + σ

ρk3m) tanh(kmh) (18)

and2γ = ∆ω2

0

g + k2m

σ

ρ

= Γ1 +B−1

0(19)

where Γ = ∆ω20

gand B0 = ρg

σk2m

.

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 theeigenmode m. From equation (11) we have

(∇2 + k2m)Sm(x, y) = 0 (20)

Letx = rcos(θ)y = rsin(θ)

Soa(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).

∇2S + k2mS = 0 (23)

where,S(r, θ) = F (r)G(θ)

In cylindrical coordinate system, ∇2 = 1r

∂

∂r(r ∂∂r

) + 1r2

∂2

∂θ2 . So by the method of variableseparation we get,

r

F (r)∂

∂r(r∂F (r)

∂r) + k2

mr2 = − 1

G(θ)∂2G(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

∂2G(θ)∂θ2 + µG(θ) = 0 (25)

isG(θ) = c1sin(√µθ) + c2cos(

√µθ)

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On applying the boundary condition (22), we get the condition that

sin√µ(θ + 2π) = sin

√µθ ⇒ √µ = m ∈ Z,

thusµ = m2 (26)

and the solution becomesG(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∂2F (r)∂r2 + r

∂F (r)∂r

+ (k2mr

2 − µ)F (r) = 0

applying the condition (26), the equation becomes

r2∂2F (r)∂r2 + r

∂F (r)∂r

+ (k2mr

2 −m2)F (r) = 0 (28)

and the solution isF (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 = kmn the value of km for which Jm(kmR) = 0. Hence the final solution for S(r, θ) is

S(r, θ) = Jm(kmn r[c1sin(mθ) + c2cos(mθ)] (30)

Consider a water filled tibetan singing bowl of:radius R = 7.5cmheight h = 10cmdensity of water ρ = 1000kg/m3

surface tension of water σ = 72 ∗ 10−5N/m2

external frquency with which the bowl is excited f0 = 188Hzangular frequency of the bowl ω0 = 1180.64s−1

maximal acceleration of the rim normalized by the gravitational acceleration Γ = ∆ω20

g= 6.2

maximum amplitude of vibrating rim ∆ = 4.359 ∗ 10−3m

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Figure 2: S(r, θ) v/s r − θ plot with kmn = 1.776cm−1 with m = 1 and n = 4

3.1.2. am(t)

Taking equation (17),d2amdt2

+ ω2m(1− 2γcos(ω0t)) am = 0 (31)

withω2m = (gkm + σ

ρk3m) tanh(kmh) (32)

γ = Γ2(1 +B−1

0 )(33)

where B0 = ρg

σk2m

, Γ = ∆ω20

gand km = 2π

λm.

Let T = ω0t

2So the equation (31) becomes

d2amdT 2 + 4kmtanh(kmh)

ω20

[k2m

σ

ρ+ g −∆ω2

0cos(2T )] am = 0 (34)

which can be written in the formd2amdT 2 + (a− 2qcos(2T )) am = 0 (35)

wherea = 4kmtanh(kmh)

ω20

(k2m

σ

ρ+ g) (36)

andq = 2km∆ω2

0tanh(kmh)ω2

0(37)

The value of km is obtained by solving the transcendental equation

ω2m = (gkm + σ

ρk3m) tanh(kmh)

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where ωm = ω0

2 .Equation (35) has the form of Mathieu equation and solution of the equation after applyingthe boundary condition dam

dT/T=0 = 0 and with the value of

km = 1665.3m−1

a = 1.00001q = 0.14is

am(T ) = MathieuC[1.00001, 0.14, T ] (38)

10 20 30 40 50 60 70T

- 20

- 10

10

20

am

am ( T ) V / s T =ω 0 t / 2

Figure 3: am(T ) v/s T plot for Ideal Fluid

3.2. Viscous FluidThe Navier Stokes equation for compressible and viscous fluid with irrotational flow is

δ~v

δt+ (~v.∇).~v + ∇P

ρ− ~g + η

ρ∇2v = 0 (39)

under the condition that ∆ρρ

<< 1 where ρ is the density of the fluid, η is the dynamic viscosity

of the fluid, ~v = ui+ vj + wk and ~g = gz −∆ω20cos(ω0t). The equations of motion of the fluid

can be written in the formδu

δt+ u

δu

δx+ v

δu

δy+ w

δu

δz= −1

ρ

δP

δx+ η

ρ(δ

2u

δx2 + δ2u

δy2 + δ2u

δz2 )

δv

δt+ u

δv

δx+ v

δv

δy+ w

δv

δz= −1

ρ

δP

δy+ η

ρ(δ

2v

δx2 + δ2v

δy2 + δ2v

δz2 )

δw

δt+ u

δw

δx+ v

δw

δy+ w

δw

δz= −1

ρ

δP

δz+ η

ρ(δ

2w

δx2 + δ2w

δy2 + δ2w

δz2 ) + (g −∆ω20cos(ω0t)) (40)

Since the flow is irrotational, the vorticity w = ∇× ~v = 0 and hence ~v can be represented asgradient of a scalar function φ(x, y, z, t). Hence

~v = ∇φ = δφ

δxi+ δφ

δyj + δφ

δzk

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Hence the equations of motion has the integral

δφ

δt+ 1

2(u2 + v2 + w2) = −Pρ

+ η

ρ(δ

2φ

δx2 + δ2φ

δy2 + δ2φ

δz2 ) + (g −∆ω20cos(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. Thepressure 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 δ2φ

δz2 = 0 sinceδ

δz(δaδt

) = 0From equations (7), (8), (9) and (10) we arrive at the equation

σ

ρ(δ

2a

δx2 + δ2a

δy2 ) + δφ

δt/z=0 −

η

ρ(δ

2φ

δx2 + δ2φ

δy2 )− (g −∆ω20cos(ω0t))a = 0 (43)

Since δφ

δn= 0, δa

δn= 0 and δ

δn(δ

2a

δx2 + δ2a

δy2 ) = 0,hence a,φ and δ2a

δx2 + δ2a

δy2 can be expanded interms of complete orthogonal set of eigen functions Sm(x, y).

(δ2a

δx2 + δ2a

δy2 + k2m)Sm(x, y) = 0

where δSmδn

= 0 and k2m is the eigen value. The required expansion for a(x, y, t), δ

2a

δx2 + δ2a

δy2 andφ(x, y, z, t) are

a(x, y, t) =∞∑0am(t)Sm(x, y)

δ2a

δx2 + δ2a

δy2 = −∞∑0k2mam(t)Sm(x, y)

φ = −∞∑1

cosh(km(h− z))kmsinh(kmh) Sm(x, y)

andδ2φ

δx2 + δ2φ

δy2 = −∞∑1

cosh(km(h− z))kmsinh(kmh) kmSm(x, y) (44)

Hence equation (43) becomes

∑ Sm(x, y)kmtanh(kmh) + [d

2am(t)dt2

+ k2mη

ρ

dam(t)dt

+ kmtanh(kmh)(k2m

σ

ρ+ g−∆ω2

0cos(ω0t) am(t))] = 0

(45)Since Sm(x, y) are linearly independent,

d2am(t)dt2

+ k2mη

ρ

dam(t)dt

+ kmtanh(kmh)(k2m

σ

ρ+ g −∆ω2

0cos(ω0t) am(t)) = 0 (46)

Equation (46) can be re-written as,

d2am(t)dt2

+ 2βdam(t)dt

+ ω2m(1− 2γcos(ω0t)) am(t) = 0 (47)

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where,2β = k2

mη

ρ(48)

ω2m = (gkm + σ

ρk3m) tanh(kmh)

and2γ = ∆ω2

0

g + k2m

σ

ρ

= Γ1 +B−1

0

where Γ = ∆ω20

gand B0 = ρg

σk2m

.

3.2.1. am(t)

Taking equation (47),

d2amdt2

+ 2βdam(t)dt

+ ω2m(1− 2γcos(ω0t)) am = 0

Letam(t) = amd(t)e−βt (49)

Thus the equation(47) becomes

d2amddt2

+ (ω2m − β2)(1− 2ω2

m

ω2m − β2γcos(ω0t)) amd = 0

Henced2amddt2

+ ω2md(1− γdcos(ω0t)) amd = 0 (50)

whereω2md = ω2

m − β2 (51)and

γd = ω2m

ω2md

γ (52)

sinceω2m = (gkm + σ

ρk3m)tanh(kmh)

equation (50) can be written in the form

d2amddt2

+ kmtanh(kmh)(ω2md

ω2m

k2m

σ

ρ+ ω2

md

ω2m

g −∆ω20cos(ω0t) amd(t)) = 0 (53)

Let T = ω0t

2So the equation (53) becomes

d2amddT 2 + 4kmtanh(kmh)

ω20

[ω2md

ω2m

k2m

σ

ρ+ ω2

md

ω2m

g −∆ω20cos(2T )] amd = 0 (54)

which can be written in the form

d2amddT 2 + (ad − 2qdcos(2T )) amd = 0 (55)

9

wheread = 4kmtanh(kmh)

ω20

[ω2md

ω2m

k2m

σ

ρ+ ω2

md

ω2m

g] (56)

andqd = 2km∆ω2

0tanh(kmh)ω2

0(57)

The value of km is obtained by solving the transcendental equation

ω2md = (gkm + σ

ρk3m)tanh(kmh)− k4

mη2

4ρ2 (58)

where ωmd = ω0

2 .Equation (55) has the form of Mathieu equation and solution of the equation after applyingthe boundary condition dam

dT/T=0 = 0 and with the value of

km = 1665.58m−1

ω2md

ω2m

= 0.99a = 1.00001q = 0.14β = 1.2345s−1

isam(T ) = MathieuC[0.990, 0.14, T ]

Hence the solution for the equation

d2amdt2

+ 2βdam(t)dt

+ ω2m(1− 2γcos(ω0t)) am = 0

is

am(T ) = amd(T )e−2βTω0 (59)

am(T ) = MathieuC[0.990, 0.14, T ]e−2.09×10−3T (60)

10 20 30 40 50 60 70T

- 20

- 10

10

20

A T

A ( T ) vs T =wt / 2

Figure 4: am(T ) v/s T plot for Viscous Fluid

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4. CONCLUSIONWhen 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 calledfaraday waves with steadily growing amplitude on the free surface of the fluid. The surfacedeformation is a combination of Bessel function and Mathieu function. The amplitude of thewaves 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 growthof 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 forthe same forcing acceleration. The same study can be done for viscous and compressible fluidsfor which desity can not be taken to be a constant, where basically a more general form of theNavier Stokes need to be used where density of the fluid is not a constant.

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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 verticalperiodic motion.(1954).

[4] Krishna Kumar,Laurette S. Tuckerman Parametric instability of the interface between twofluids.Cambridge University Press, (1994).

[5] Arfken G., Weber H. Mathematical methods for Physicists 7thed.

[6] D. Blandford, Kip S. Thorne Applications of Classical Physics(2012)http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/

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