User:Fria Hossein/sandbox

MECHANICAL VIBRATIONS The measurements of the periodic oscillation respected to the equilibrium point called mechanical vibrations. All motions can be derived from newton’s second law of motion. F=ma	1 ∑_i▒〖F=m (d^2 x)/〖dt〗^2 〗	2 In figure 1, the mass suspended at the end of the spring and the weight stretched by the length L to reach the equilibrium position of the system.

Figure 1 Mechanical vibrations [www.math.psu.edu]. X (t) –displacement as a function of time. From Hooks law At equilibrium: mg=KL	3

At motion: mx ̈+γx ̇+Kx=F(t) 4

This is the second order of differential equation. γ Damping constant spring constant (t) applied force function,u ̇ initial velocity of the mass.

We have four force two are acting on the object other two may affect or may not. Gravitational force acting on the objectF_g=mg. Spring forceF_s=K(L+u). Damping force works to contract any movants F_d=γu ̇ If the object is moving downward the velocity u ̇ will be positive then the F_d will be negative acting to pull the object back up and vice versa. An external force F(t)=mx ̈+mg-k(L+x)+γx ̇ this is called d the force function. When the objective at equilibrium there is only two forces acting on the objective one is the gravitational force and the second is the spring force. Where mg=kLthe above equation becomes F(t)=m(x ̈+kL-KL-kx+γx ̇ ) ̈

Free vibration [undamped]. γ=0,F(t)=0 The equation 4 becomes mx ̈+Kx=0 x(t)=Acos⁡(ωt)+Bsin(ωt) Plug in the above equation to equation 5 (-mω^2+k)(Acos⁡ωt+Bsinωt)=0 As⁡(ωt)+Bsin(ωt)≠0 ,,then,,,(-mω^2+k)=0	5 (mω^2=k) ω=√(k/m)   Undamped Figure 2 spring mass system in horizontal position mx ̈+bx ̇+kx=F(t) b=F(t)=0 x(t)=ACosωt+BSinωt

x_0=x(t=0) a=x_0 x ̇=-AωCosωt+BwSinωt V_0=x ̇(t=0) x ̇(t=0)=bω=V_0 x(t)=x_(0 ) Cosωt+V_0/m Sinωt

T=2π/ω 〖tan〗^(-1) (b/a)=〖tan〗^(-1) (V_0/(X_0 ω)) Figure 3 graphical representative motion of harmonic oscillator [www.wordpress.com] φ Delay line angle φ=ω∆τ ∆τ Delay time A homogeneous equation mx ̈+bx ̇+kx=0 Let x=Ae^st and substitute this in the above equation ms^2+bs+k=0 (ms^2)/m+bs/m+k/m=0 s^2+b/m s+k/m=0 s=-b/2m±√(b^2/(4m^2 )-k/m) s=-(b/2mω)ω±ω√(b^2/(4m^2 ω^2 )-ω) The damping ratio ƺ=(b/2mω)……….ƺ-damping ratio

s=-ƺω±ω√(ƺ^2-ω) In free vibration, the total energy stays same, and the forced vibration occurs when the object forced to vibrate at a particular frequency.in free vibration, the objects are free to vibrate and will have one or more natural frequency. Objects are forced to vibrate at its natural frequency resonance will occur, and will observe the large amplitude. Figure 4 graphical representative on object forced to vibrate f_0-resonance frequency Acceleration of an object it directly proportional to its displacement from it is equilibrium position. Acceleration at any point                      a=(2πf)^2 x X-displacement from central position. x=ACos2πft Velocity v=±2πf√(A^2-x^2 )

If we have two equal bodies with mass m_1 and m_2 attached with 3 springs and each spring having constant k, see figure below. Figure 5 two mass-spring system at horizontal position with The equation of motion becomes: m_1 ((x_1 ) ̈=-kx_1+k(x_2-x_1 ) ) ̈=-2kx_1+kx_2 m_2 ((x_2 ) ̈=-kx_2+k(x_1-x_2 )=) ̈-2kx_2+kx_1 x_1 (t)=A_1 e^iωt x_2 (t)=A_2 e^iωt -ω^2 m_1 A_1 e^iωt=-2kA_1 e^iωt+kA_2 e^iωt 〖-ω^2 m〗_2 A_2 e^1ωt=-2kA_2 e^iωt+〖kA〗_1 e^iωt (ω^2 m-2k) A_1+kA_2=0 (ω^2 m-2k) A_2+kA_1=0 (■((ω^2 m-2k)&k@k&(ω^2 m-2k)))(A_1¦A_2 )=0 〖〖(ω〗^2 m-2k)〗^2-k^2=0 Solving for ω ω^4 m^2-4kω^2 m+4k^2-k^2=0 ω^4 m^2-4kω^2 m+3k^2=0 ω^4-(4kω^2)/m+(3k^2)/m^2 =0 (ω^2-k/m)(ω^2-3k/m)=0 ω_1=√(k/m ),and ω_2=√(3k/m)

Molecular vibrations can be treated using Newtonian mechanics to calculate the vibration frequencies. F=-kx E=hv E=(n+1/2)hv=(n+1/2)h/2π √(k/m) n=0,1,2,3……… The heavier particle vibrates slower than the lighter particle .interactions in colloid depend on the density distribution of the ionic fluid.in the suspension system, the total pressure cannot be written of the sum of simple pair interaction. ∆P(x)=P_o sin⁡(kx)Cos(ωt) In the solution, the Cation and anion having different masses so they have a difference acceleration and difference pressure .due to the change in acceleration the dynamic potential is created and the potential measured in electrolyte called IVP {Debye, 1933.Oshima 2005].the main force action in the colloid is pressure gradient [Oshima 2005].the electric field that generated in the colloid due to the vibration cause an increase in velocity of the counter ions and when the get closer to the particles this cause  change in ion directions. CVP from the very small particle approaches IVP and in the limit of very small Ka particles behave like an ions. Colloid particles much larger and carrying greater charge than electrolyte ions.in the suspension the electrical double layer created the potential difference around each particle rather than the motion of cations and anions. How the signal changes with changing atomic and ionic mass and with the atomic size When the colloid particle suspended in a solution are charged and surrounded by a counter charge in the fluid and gives the solution overall charge neutrality [Diebold.Gusev.etl.2004].the density of the particle and surrounding fluid are different from each other the passage of the ultrasound pulses through the solution cause the motion of the particles.[1].the ions and particles in colloid accelerate due to the passage of the ultrasound pulses through the fluid and this acceleration changes the electrical neutrality and produces the electric current in the electrolyte[2]. The ions have different masses in the solution and it may show different acceleration and potential while the ultrasound pulses pass through the fluid. [Debye.1933.], and this potential can be measured as IVP [3].in case of the effect of the atomic and ionic weight on the signal amplitude, by increasing the atomic weight the signal amplitude increased.The heavier the metal ion gives the larger IVP signal [4].The particle size does no effect on the signal amplitude while the concentration on the test at affixed frequency about 1Mhz [5]. The smaller particle having smaller electron cloud and the smaller electron cloud create a weaker electric dipole, therefore, the electric field made by this dipole weaker, and due to the difference in the mass of cation between the species the ionic vibration potential signal increases with increasing molecular weight. In the periodic table if we notice we see that in the crossing period moving from one element to the next element the electrons are being added to the same level of energy and proton add to the nucleus, and the electrons in the outer level attract more strongly and so pulled to the nucleus closer so the atomic size decreasing. In descending group from one element to another, the occupied energy level is increased and the atomic size increased.

Group 2 Period 2	Li	Be	B	C	N Mg Ca Sr Table 1: Group 2 and periodic 2 elements from the periodic table. In the periodic 2 elements when the atomic number increases the atomic radius decrease and the electronegativity increases so the ionization energy increases, and so the conductivity increase which results in a decrease in IVP. But in the group 2 when we move from one element to the next the occupied energy level increases and the ionization energy decreases so the electronegativity decreases and the conductivity decrease which result in increasing IVP. Table 1 periodic table [www.chemwiki.ucdavis.edu]

Ionization energy: The energy required to remove and electron from the neutral atom called ionization energy. Ionization increasing with increasing the electron negativity and its and inverse proportional to the ion vibration potential.and directly proportional to the conductivity. Electronegativity: An atom can attract and can bind with electrons. Figure 6 ionization energy from hydrogen to argon [www.chemwiki.ucdavis.edu]. Forces play an important role in interaction of colloid particles Electrostatic interactions:[charged colloid particle attract and repulse]. Excluded volume repulsion [impossibility of any overlaps between hard particles]. Vander Waals force [interaction between two dipoles that are permanent or induced]. Gravity force: [a significant driving force for particle transport to the surface]. Hydrodynamic interactions:[arises from disturbance induced in the fluid flow field by the presence of a particle]. Drag force: [particle moves in fluid drag force experienced and has two terms force and dimensional drag coefficient].a particle initially at rest and will accelerate until the drag force is balanced by the gravitational force. Gravitational force: This force depends on the particles volume and the density difference between the fluid and the particle.fro sphere the force is given by: F_g=(πd^3)/6 (ρ_l-ρ_s )g If the particle much less dense than the fluid the above equation gives negative and means that the particle moves upward. The smaller particle has greater attractive force than the gravitational force and the larger particle has smaller attractive force than the gravitational force.the smaller the particle has the larger the surface area, and small particle has greater interfacial energy per unit mass, and large particles achieve more stable low energies.the ions are more mobile than the particles and this cause charge separation. The interaction between particles can give the force of attraction or repulsion .the strength of interaction is roughly proportional the-the first power of the particle size .colloid interaction become less important as particle size increase. Figure 7 different particle size F_A Attractive force holding the particle to the wall. F_D Fluid drag caused by flow parallel to the surface. F_g Gravitationla attraction acting opposite the drag force. For the small particle the attractive force is greater than the gravitational force and for a larger particle is vice verse. Fria Hossein