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Sloshing Mode of Water^ Experiment 1
1.1 Introduction
The lowest mode of oscillation in a closed body of liquid is called the sloshing mode. The sloshing phenomenon is very important in the design of the tank that holds the liquid. The violent sloshing force The sloshing mode can be easily created and studied. Sloshing also occurs naturally in lakes. Its loads to impact on the tank roof and walls which may damage the tank. is a phenomenon known as seiches. To experience the sloshing oscillation, partly fil little. Water in the pan will slosh. A better way is to place a pan on a horizontal surface, fill itl a rectangular pan with water. Push the pan a to the brim, and then overfill it so that water bulges above the level of the rim. Gently nudge the pan. After the higher modes have died out, you are left with the sloshing mode which oscillates with very little damping. If you observe carefully, you will notice that the water surface remains practically flat.
In this laboratory exercise you will study t you will measure the frequency of the mode and the damping time. The goal is to determinehe sloshing mode of water in a tank. In particular, the functional dependence of frequency of oscillation on the water depth and tank size.
1.2 Simple Theory of Sloshing Oscillation
Before performing the experiment it is a good idea to develop a simply theory of oscillation. Consider a rectangular tank of base size 2L x b and height H. We fill the tank with water to height h we place the tank on a horizontal surface. The water level can be seen to oscillate about ho (ho < H) and gently rock the tank around an axis normal to the surface 2L x H. Then,o. We would like to determine the frequency of this oscillation. To simplify the analysis we assume that the surface of the water is flat. A snapshot of the oscillation is depicted in Figure 1.
The strategy is to find the equation of motion. Since we are not dealing with a point mass, we need to find the equation of motion or Newton’s second law of motion for the center of mass of the water inside the tank. The center of mass is defined as
r^! c = (^) M^1 (^) "^! r dm [1]
where M is the mass of the water inside the tank. The mass M is equal to the product of the density of water, denoted here as ρ and the volume of water in the tank: M = ρ(2Lbho) = (^2) calculate the center of mass in a different way. To define the position of the center of mass weρLbho. The integration is quite tedious and is shown partly in the Appendix. We will need a coordinate system. We most corner, as shown in Figure 2. set the coordinate origin at the bottom of the tank in the left-
At equilibrium the water surface is flat and horizontal. The center of mass is located at the center of the body with length 2L, height ho and width b. Therefore,
r " c = L! i + ho! j + bk! When the water is sloshing, some amount of water is shifted from one side, as shown in Figure^2121
- The center of mass is given by r " (^) c t = xc! i + yc! j + bk! ( ) 21
This result shows that the out each other in a simple way, -of-equilibrium displacements of the center of mass are related to
! x =! r^ # " i^ $ = L 3 h^ ho! y =! r^ # "^ ˆ j = h 6 o ( (^) hho )^2 and! y = 23 L^ ho 2^ (! x )^2
The result suggests that as the water moves out of equilibrium as shown in Figure 1 the center of mass increases in height by Δy. The change in the gravitational potential energy of the body of water is
! U = mg (! y )= mg 23 L^ h 2 o^ (! x )^2 The force in the x-direction is equal to
F (^) x = "##((^!!^ U x )^ )=" 3 mg L^ ho 2! x
The negative sign indicates the force is a restoring force. The fact that the restoring force is linear in horizontal displacement suggests that the force is elastic in its nature. The equation of motion in the x-direction is
m d^2 dt (^! 2^ x^ )= F x =" 3 mg L^ h^ o 2 (! x )=" m # o^2 (! x )
The center of mass of the water oscillates about its equilibrium position with an angular frequency,
! o = L^1 3 gho
The frequency of oscillation is equal to f =^ " 2!^ o^^ = 2!^1 L 3 gho [2]
The following table shows the frequency of oscillations for typical ho and L. L (cm) 20 ho 10 (cm) f (Hz)1. 2025 2010 1.931. 2530 2010 1.540. 30 20 1.
The frequency is in the range of one or two cycles per second. It can be measured wi watch. th a stop
gravitational potential energy of the body of water by movement out of equilibrium induces a^ We now have a simple picture of the oscillation phenomenon. The change in the horizontal force that pulls or pushes the water back to equilibriu the real surface of the water is not flat – in profile it is closer to a sine function!m. It is useful to mention that A better theory yields " (^) o = 2! L gho which is 2 " 3! 0. 907 of the angular frequency calculated with the assumption that the water surface is flat. be about 10% higher than the actual frequency. Therefore, if we assume the water surface is flat, the calculate frequency would
1.3 Damping
The amplitude of the sloshing mode decreases with time due to damping wh unless the experiment is specially crafted. To understand the effect of damping let us brieflyich is significant review the properties of damped oscillator. For simplicity we assume that the damping force is proportional to velocity. The equation of motion becomes
m d dt^22^ x + m "^ dx dt + m! o^2 x = 0
where τ is the damping time. If the effect of damping is small, the damping time would be long. Mathematically, for a large τ the damping term, m!^ dxdt , is small. When damping is relatively small It takes the form,, x(t) varies as a sine or cosine function and its amplitude decreases with time.
x ( t )= Ae! t^ /^2 "^ cos( # t + $ ) where # 2 = # o^2! (^41) " 2
Constants A and value of damping time. When the α are determined by initial conditions. It is relatively easy to estimate the oscillation reaches a maximum amplitude, we set the time to be t = 0. We simply set the amplitude of that oscillation as x(0) = x oscillations for several full periods. After n number of oscillations, t = nT where T is the period.o. We then observe the Note that T = 2π/ωo. We have, xo = A cosα and x ( nT )= x n = Ae $ nT^ /^2!^ cos( 2 # n +")= e $^ nT /^2! xo It follows,
Average^3 Calculate the average time for the system to go through 20 oscillations. Use this average time to calculate the period (T) and then the frequency where f = 1/T. Enter the value for the frequency in column 4 of the above table. Calculate the average amplitude of the system after 20 oscillations. Use this average amplitude to calculate τ according to Equation 3. Enter of the value of τ in column 5 of the above table.
Initial conditions: h Trial Time for 20o = 15 cm, Amplitude = 1.0 cm
1 Oscillations (s)^ Amplitude after 20^ Oscillations (cm)^ f (Hz)^ τ^ (s) (^23) Average Calculate the average time for to calculate the period (T) and then the frequency where f = 1/T. Enter the value for the the system to go through 20 oscillations. Use this average time frequency in column 4 of the above table. Calculate the average amplitude of the system after 20 oscillations. Use this average amplitude to calculate τ according to Equation 3. Enter of the value of τ in column 5 of the above table. Initial conditions: h Trial Time for 20o = 20 cm, Amplitude = 1.0 cm
1 Oscillations (s)^ Amplitude after 20^ Oscillations^ (cm)^ f (Hz)^ τ^ (s) (^23) Average Calculate the average time for the system to go through 20 oscillations. Use this average time to calculate the period (T) and then the frequency where f = 1/T. Enter the value for the frequency in co Calculate the average amplitude of the system after 20 oscillations. Use this average amplitudelumn 4 of the above table. to calculate τ according to Equation 3. Enter of the value of τ in column 5 of the above table.
1.6 Data Analysis
Summarize your data and calculations in the following table ho (cm) 5 f (Hz) (^) τ (s) 10
Plot fvs ho
1.7 Report
1.8 Appendix: Determination of center of mass
The center of mass is defined as
!
" r = " rdm c (^) M^1 where M is the mass of t always flat. There is no motion in the zhe system. We assume that the surface of the water inside the tank is-direction. We first calculate the center of mass when the fluid is in equilibrium according to the coordinate system shown in Figure 2. We ca defined, the limits of integration can be set. Then, we haven reduce the vector, r! c , into its x-, y-, and z-components. Once the coordinate system is
" r = x! i + y! j + z k! c c c c = (^)! L
xc M xdm
2 (^10) = (^)!
yc M hoydm
(^10) = (^)! b
zc M^10 zdm
We begin by calculating x elemental or differential volume. We take dV as a thin plate parallel to the yc. Note that dm = ρdV where ρ is the density of water and dV is the-z plane with height ho, width b, and thickness dx. o L o^ L
x c M^1^ Lx dm M^1 x (! bh o dx )^! M^ bh^ x 2 L sin ceM 2! Lbh
2 0
2 2 = (^) " 02 = " 0 = = =
1 00 10 y 200 h 0
M
y Lbdy^ Lb
y M y dm M
c =^! h =! h^ o^ " =^ " h^ =
1010 z 20 b
M
z Lh dz^ Lh
z M z dm M
c =^!^ b =!^ b^ " o =^ "^ o b = r " = L! i +^ ho! j +^ bk! c (^) 2 2 for water in equilibrium
Now, we will determine the center of mass when the water is out of equilibrium, as shown in Figure 1. We will first prove that zc is still ½ b. Let A be the area occupied by the water on the x-y plane. We have
ho h o
y c M^10 ydm M^10 y " dV
We take dV to be a thin plate parallel to the x-z plane, as shown in Figure 4.
If the plate is below y = h Therefore, the integration has two parts,o – h, dV = 2Lbdy. If the plate is above y = ho – h, dV = (2L – x)bdy.
h h h h
h h c
o o
o
y M^1 y ( 2 Lbdy ) M^1 y ( 2 L x ) b dy
From Equation A.1 we obtain x = (L/h)(y + h substituting x into the second integral and using M = 2 - ho). Integrating the first integral andρLbho, we obtain
h h h h
c o o
o o
y # Lb ( h h )^2 # h^ Lb ( h h ) y y^2 ) dy
o c o h y h^ h^2 6
1 = 2 +
In summary, we have
x c = L + 31 ( h^ h o ) L
c o h o ho y h (^ h )^2 6
1 = 2 +
2 z^ b c =
The center of mass derived from direct integration is the same as the one derived using geometry.
